Question 11 Mark
An electronic assembly consists of two subsystems, say, $A$ and $B.$ From previous testing procedures, the following probabilities are assumed to be known:
$P(A$ fails$) = 0.2$
$P(B$ fails alone$) = 0.15$
$P(A$ and $B$ fail$) = 0.15$
Evaluate the following probabilities $P(A$ fails alone$).$
AnswerLet us define events;
$A : A$ fails. and $B : B$ fails.
Given: $P (A) = 0.2$
Event failed by both, $P\left(A \cap B\right) = 0.15$
We have,
$P(A$ fails alone$) = P(A) - P\left(A \cap B\right)$
$= 0.2 - 0.15$
$= 0.05$ View full question & answer→Question 21 Mark
An electronic assembly consists of two subsystems, say, $A$ and $B$. From previous testing procedures, the following probabilities are assumed to be known:
$P(A$ fails$) = 0.2$
$P(B$ fails alone$) = 0.15$
$P(A$ and $B$ fail$) = 0.15$
Evaluate the following probabilities $P(A $ fails|$B$ has failed$).$
AnswerLet's define events;
$E_A: A$ fails
$E_B: B$ fails
Given that:
Event failed by $A, P(E_A) = 0.2$
Event failed by both, $P\left(E_{A} \cap E_{B}\right) = 0.15$
And, event failed by B alone = $P\left(E_{B}\right)-P\left(E_{A} \cap E_{B}\right)$
$0.15 = P (E_B) - 0.15$
$\therefore P (E_B) = 0.30$
Therefore, $P\left(E_{A} | E_{B}\right)=\frac{P\left(E_{A} \cap E_{B}\right)}{P\left(E_{B}\right)}$
$= \frac{0.15}{0.3}$
$= 0.5$
Which is the required solution.
View full question & answer→Question 31 Mark
If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\frac{1}{2}$).
AnswerThere are four entries in a determinant of $2 \times 2$ order. Each entry may be filled up in two ways with 0 or 1.
$\therefore $ Number of determinants that can be formed = $2^4 = 16$
The value of determinants is positive in the following cases:
$\left| {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right|,\left| {\begin{array}{*{20}{c}} 1&0 \\ 1&1 \end{array}} \right|,\left| {\begin{array}{*{20}{c}} 1&1 \\ 0&1 \end{array}} \right| = 3$
Therefore, the probability that the determinant is positive $ = \frac{3}{{16}}$
View full question & answer→Question 41 Mark
If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?
AnswerWe know that, in a leap year there are total 366 days, 52 weeks and 2 days.
Now, in 52 weeks there are total 52 Tuesdays.
$\therefore$ Probability that the leap year will contain 53 Tuesdays is equal to the probability of remaining 2 days will be Tuesdays.
Thus, the remaining two days can be:
(Monday and Tuesday), (Tuesday and Wednesday), (Wednesday and Thursday), (Thursday and Friday), (Friday and Saturday), (Saturday and Sunday) and (Sunday and Monday)
$\therefore$ Total Number of cases = 7
Cases in which Tuesday can come = 2
Hence, probability (leap year having 53 Tuesdays) = $\frac{2}{7}$
View full question & answer→Question 51 Mark
If A and B are any two events such that P(A) + P(B) - P(A and B) = P(A), then
AnswerIt is given in the question that,
A and B are any two events where,
P(A) + P(B) - P(A and B) = P(A)
P(A) + P(B) - P(A $\cap$ B) = P(A)
P(A $\cap$ B) = P(B)
$\therefore P(A | B)=\frac{P(A \cap B)}{P(B)}$ = 1
View full question & answer→Question 61 Mark
If P(A|B) > P(A), then which of the following is correct :
AnswerIt is given in the question that,
P (A|B) > P (A)
$\therefore \frac{P(A \cap B)}{P(B)}>P(A)$
$P(A \cap B)>P(A) \cdot P(B)$
$\frac{P(A \cap B)}{P(A)}>P(B)$
P(B|A) > P (B)
View full question & answer→Question 71 Mark
A and B are two events such that P (A) $\ne$ 0. Find P(B|A), if $A \cap B=\phi$.
AnswerGiven that $A \cap B = \phi \Rightarrow P(A \cap B)=0$
$\therefore P(B | A)=\frac{P(A \cap B)}{P(A)} = 0$
View full question & answer→Question 81 Mark
If A and B are two events such that P(A) $\ne$ 0 and P(B|A) = 1, then
AnswerIt is given in the question that,
A and B are two events where,
p(A) $\ne$ 0
And, P (B|A) = 1
$\therefore P(B | A)=\frac{P(B \cap A)}{P(A)}$
$1=\frac{P(B \cap A)}{P(A)}$
p(A) = p(B $\cap$ A)
$\therefore A \subset B$
View full question & answer→Question 91 Mark
A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?
AnswerGiven: P(A) = $\frac{{60}}{{100}}$, P(B) = $\frac{{40}}{{100}}$
Let D denotes a defective item :
$\therefore $ P(D|A) = $\frac{2}{{100}}$and P(D|B) $ = \frac{1}{{100}}$
P(B|D) =$\frac{{P\left( B \right)P\left( {D|B} \right)}}{{P\left( A \right)P\left( {D|A} \right) + P\left( B \right)P\left( {D|B} \right)}}$
=$\frac{{\frac{{40}}{{100}} \times \frac{1}{{100}}}}{{\frac{{60}}{{100}} \times \frac{2}{{100}} + \frac{{40}}{{100}} \times \frac{1}{{100}}}}$
=$\frac{{40}}{{120 + 40}} = \frac{{40}}{{160}} = \frac{1}{4}$
View full question & answer→Question 101 Mark
If A and B are two events such that A ⊂ B and P(B) ≠ 0, then which of the following is correct?
AnswerSince, $\space A\subset{B},\space\space \space A\cap{B} =A$ $P(A/B) =\frac{P(A\cap{B})}{P(B)} =\frac{P(A)}{P(B)}$
View full question & answer→Question 111 Mark
Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find P(B|A)
AnswerGiven: P(A) = 0.3 and P(B) = 0.4
Two events A and B are independent if $P(A \cap B)=P(A) . P(B) = 0.12$
As we know $P(B | A)=\frac{P(A \cap B)}{P(A)}$
$\Rightarrow P(B | A)=\frac{0.12}{0.3}$
$\Rightarrow$ P(B|A) = 0.4
View full question & answer→Question 121 Mark
Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find P(A|B)
AnswerGiven: P(A) = 0.3 and P(B) = 0.4
Two events A and B are independent if $ P(A \cap B)=P(A) . P(B)$
$\Rightarrow P(A \cap B) = 0.12$
As we know $P(A | B)=\frac{P(A \cap B)}{P(B)}$
$\Rightarrow P(A | B)=\frac{0.12}{0.4}$
$\Rightarrow$ P (A|B) = 0.3
View full question & answer→Question 131 Mark
Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find P(A $\cup$ B)
AnswerGiven: P(A) = 0.3 and P(B) = 0.4
As we know, P (A $\cup$ B) = P(A) + P(B) - P (A $\cap$ B)
Since events are independent so P (A $\cap$ B) = 0. Therefore,
P (A $\cup$ B) = 0.3 + 0.4
⇒ P (A $\cup$ B) = 0.7
View full question & answer→Question 141 Mark
Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find P(A $\cap$ B)
AnswerGiven: P(A) = 0.3 and P(B) = 0.4
When A and B are independent.
$\Rightarrow$ P (A $\cap$ B) = P(A) . P(B)
$\Rightarrow P(A \cap B)=0.3 \times 0.4$
$\Rightarrow$ P (A $\cap$ B) = 0.12
View full question & answer→Question 151 Mark
Given that the events A and B are such that P(A) = $\frac{1}{2}$, $P(A \cup B)=\frac{3}{5}$ and P(B) = p.
Find p if they are independent.
AnswerGiven: P(A) = $\frac{1}{2}$, P(A $\cup$ B) = $\frac{3}{5}$ and P(B) = p
When A and B are independent.
$\Rightarrow P(A \cap B)=P(A) \cdot P(B)$
$\Rightarrow P(A \cap B)=\frac{1}{2} \cdot p$
As we know, P (A $\cup$ B) = P(A) + P(B) - P (A $\cap$ B)
$\Rightarrow \frac{3}{5}=\frac{1}{2}+p-\frac{p}{2}$
$\Rightarrow \frac{p}{2}=\frac{3}{5}-\frac{1}{2}$
$\Rightarrow \mathrm{p}=2 \times \frac{1}{10}=\frac{1}{5}$
View full question & answer→Question 161 Mark
Given that the events A and B are such that P(A) = $\frac{1}{2}$, $P(A \cup B)=\frac{3}{5}$ and P(B) = p.
Find p if they are mutually exclusive.
AnswerGiven: P(A) = $\frac{1}{2}$, P(A $\cup$ B) = $\frac{1}{5}$ and P(B) = p
When A and B are mutually exclusive,
$\Rightarrow (A \cap B)=\phi$
$\Rightarrow P(A \cap B)=0$
As we know, P (A $\cup$ B) = P(A) + P(B) - P (A $\cap$ B)
$\Rightarrow \frac{3}{5}=\frac{1}{2}+p-0$
$\Rightarrow \mathrm{p}=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}$
View full question & answer→Question 171 Mark
Two events A and B will be independent, if
AnswerTwo events A and B will be independent, then $P(A\cap{B}) =P(A).P(B)$ $P(A'\cap{B' })=P(AUB)' = 1- P(AUB) \\=1-P(A)-P(B)-P(A).P(B) \\=[1-P(A)][1-P(B)]=P(A').P(B')$
View full question & answer→Question 181 Mark
The probability of obtaining an even prime number on each die, when a pair of dice is rolled is:
AnswerClearly, n(s) = 36. Favourable cases are {2, 2} Therefore required probability = $\frac{1}{36}$
View full question & answer→Question 191 Mark
In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random. If she reads English newspaper, find the probability that she reads Hindi newspaper.
AnswerLet H and E denote the number of students who read Hindi and English newspaper respectively.
Hence, P(H) = Probability of students who read Hindi newspaper = $\frac{60}{100}=\frac{3}{5}$
P(E) = Probability of students who read English newspaper = $\frac{40}{100}=\frac{2}{5}$
P (H $\cap$ E) = Probability of students who read Hindi and English both newspaper = $\frac{20}{100}=\frac{1}{5}$
P (H|E) = English newspaper reading has already occurred and the probability that she reads Hindi newspaper is to find.
As we know $P(H | E)=\frac{P(H \cap E)}{P(E)}$
$\Rightarrow \mathrm{P}(\mathrm{H} | \mathrm{E})=\frac{\frac{1}{5}}{\frac{2}{5}}=\frac{1}{5} \times \frac{5}{2}$
$\Rightarrow P(H | E)=\frac{1}{2}$
View full question & answer→Question 201 Mark
In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random. If she reads Hindi newspaper, find the probability that she reads English newspaper.
AnswerLet H and E denote the number of students who read Hindi and English newspaper respectively.
Hence, P(H) = Probability of students who read Hindi newspaper = $\frac{60}{100}=\frac{3}{5}$
P(E) = Probability of students who read English newspaper = $\frac{40}{100}=\frac{2}{5}$
P (H $\cap$ E) = Probability of students who read Hindi and English both newspaper = $\frac{20}{100}=\frac{1}{5}$
P (E|H) = Hindi newspaper reading has already occurred and the probability that she reads English newspaper is to find.
As we know $P(E | H)=\frac{P(H \cap E)}{P(H)}$
$\Rightarrow P(E | H)=\frac{\frac{1}{5}}{\frac{2}{5}}=\frac{1}{5} \times \frac{5}{3}$
$\Rightarrow P(E | H)=\frac{1}{3}$
View full question & answer→Question 211 Mark
In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random. Find the probability that she reads neither Hindi nor English newspapers.
AnswerGiven: Let H and E denote the number of students who read Hindi and English newspaper respectively.
Hence, P(H) = $\frac{60}{100}=\frac{3}{5}$ and P(E) = $\frac{40}{100}=\frac{2}{5}$
P (H $\cap$ E) = Probability of students who read Hindi and English both newspaper = $\frac{20}{100}=\frac{1}{5}$
We need to find the Probability that she reads neither Hindi nor English newspapers.
$\begin{equation} \text { i.e. } P\left(H^{\prime} \cap E^{\prime}\right) \end{equation}$
Where, $ ~P\left(H^{\prime} \cap E^{\prime}\right)=1- $Probability that she reads both the newspapers.
= 1 - P (H $\cup$ E)
= 1- [P(H) + P(E) - P (H $\cap$ E)]
= $1-\left[\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\right]$
= $1-\left[\frac{4}{5}\right]=\frac{1}{5}$
View full question & answer→Question 221 Mark
One card is drawn at random from a well shuffled deck of 52 cards.
In which of the following cases are the events E and F independent?
E: the card drawn is a king or queen
F: the card drawn is a queen or jack
AnswerGiven: A deck of 52 cards.
In a deck of 52 cards, 4 cards are queen, 4 cards are king and 4 cards are jack.
Hence, P(E) = The card drawn is either king or queen = $\frac{8}{52}=\frac{2}{13}$
P(F) = The card drawn is either queen or jack = $\frac{8}{52}=\frac{2}{13}$
There are 4 cards which are either king or queen and either queen or jack.
P(E ∩ F) = The card drawn is either king or queen and either queen or jack = $\frac{4}{52}=\frac{1}{13}$ ......(i)
And P(E).P(F) = $\frac{2}{13} \times \frac{2}{13}=\frac{4}{169}$ ......(ii)
From (i) and (ii)
$\Rightarrow \mathrm{P}(\mathrm{E} \cap \mathrm{F}) \neq \mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{F})$
Hence, E and F are not independent events.
View full question & answer→Question 231 Mark
One card is drawn at random from a well shuffled deck of 52 cards.
In which of the following cases are the events E and F independent?
E: the card drawn is black
F: the card drawn is a king
AnswerGiven: A deck of 52 cards.
In a deck of 52 cards, 26 cards are black and 4 cards are king and only 2 cards are black and King both.
Hence, P(E) = The card drawn is black = $\frac{26}{52}=\frac{1}{2}$
P(F) = The card drawn is a king = $\frac{4}{52}=\frac{1}{13}$
P(E $\cap$ F) = The card drawn is a black and king both = $\frac{2}{52}=\frac{1}{26}$ ......(i)
And P(E).P(F) = $\frac{1}{2} \times \frac{1}{13}=\frac{1}{26}$ ......(ii)
From (i) and (ii)
P (E $\cap$ F) = P(E).P(F)
Hence, E and F are independent events.
View full question & answer→Question 241 Mark
One card is drawn at random from a well shuffled deck of 52 cards.
In which of the following cases are the events E and F independent?
E: the card drawn is a spade
F: the card drawn is an ace
AnswerGiven: A deck of 52 cards.
In a deck of 52 cards, 13 cards are spade and 4 cards are ace and only one card is there which is spade and ace both.
Hence, P(E) = The card drawn is a spade = $\frac{13}{52}=\frac{1}{4}$
P(F) = The card drawn is an ace = $\frac{4}{52}=\frac{1}{13}$
P(E $\cap$ F) = The card drawn is a spade and ace both = $\frac{1}{52}$ .....(i)
And P(E).P(F) = $\frac{1}{4} \times \frac{1}{13}=\frac{1}{52}$ ...(ii)
From (i) and (ii)
P (E $\cap$ F) = P(E).P(F)
Hence, E and F are independent events.
View full question & answer→Question 251 Mark
Probability of solving specific problem independently by A and B are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that exactly one of them solves the problem.
AnswerGiven:
P(A) = Probability of solving the problem by A = $\frac{1}{2}$
P(B) = Probability of solving the problem by B = $\frac{1}{3}$
P(A') = $1 - \frac{1}{2} = \frac12$
and P(B') = $1 - \frac{1}{3} = \frac23$
Since, A and B are independent.
Now, P (exactly one of them solves) = Either problem is solved by A but not by B or vice versa
= P(A).P(B’) + P(A’).P(B)
= $\frac{1}{2} \cdot \frac{2}{3}+\frac{1}{2} \cdot \frac{1}{3}$
= $\frac{1}{3}+\frac{1}{6}=\frac{3}{6}$
⇒ P(A).P(B') + P(A').P(B) = $\frac{1}{2}$
View full question & answer→Question 261 Mark
Probability of solving specific problem independently by A and B are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that the problem is solved.
AnswerGiven:
P(A) = Probability of solving the problem by A = $\frac{1}{2}$
P(B) = Probability of solving the problem by B = $\frac{1}{3}$
Since, A and B both are independent.
$\Rightarrow$ P(A $\cap$ B) = P(A).P(B)
$\Rightarrow$ P (A $\cap$ B) = $\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}$
The problem is solved, i.e. it is either solved by A or it is solved by B.
= P(A $\cup$ B)
As we know, P (A $\cup$ B) = P(A) + P(B) - P (A $\cap$ B)
⇒ P (A $\cup$ B) = $\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{4}{6}$
$\Rightarrow \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{2}{3}$
View full question & answer→Question 271 Mark
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that one of them is black and other is red.
AnswerGiven: A box containing 10 black and 8 red balls.
Total number of balls in box = 18
Probability of getting a black ball in first draw = $\frac{10}{18}=\frac{5}{9}$
As the ball is replaced after first throw,
Hence, Probability of getting a red ball in second draw = $\frac{8}{18}=\frac{4}{9}$
Now, Probability of getting first ball is black and second is red = $\frac{5}{9} \times \frac{4}{9}=\frac{20}{81}$
Probability of getting a red ball in first draw = $\frac{8}{18}=\frac{4}{9}$
As the ball is replaced after first throw,
Hence, Probability of getting a black ball in second draw = $\frac{10}{18}=\frac{5}{9}$
Now, Probability of getting first ball is red and second is black = $\frac{4}{9} \times \frac{5}{9}=\frac{20}{81}$
Therefore, Probability of getting one of them is black and other is red :
= Probability of getting first ball is black and second is red + Probability of getting first ball is red and second is black
= $\frac{20}{81}+\frac{20}{81}$ = $\frac{40}{81}$
View full question & answer→Question 281 Mark
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that first ball is black and second is red.
AnswerGiven: A box containing 10 black and 8 red balls.
Total number of balls in box = 18
Probability of getting a black ball in first draw = $\frac{10}{18}=\frac{5}{9}$
As the ball is replaced after first throw,
Hence, Probability of getting a red ball in second draw = $\frac{8}{18}=\frac{4}{9}$
Now, Probability of getting first ball is black and second is red = $\frac{5}{9} \times \frac{4}{9}=\frac{20}{81}$
View full question & answer→Question 291 Mark
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.
Find the probability that both balls are red.
AnswerGiven: A box containing 10 black and 8 red balls.
Total number of balls in box = 18
Probability of getting a red ball in first draw = $\frac{8}{18}=\frac{4}{9}$
As the ball is replaced after first throw,
Hence, Probability of getting a red ball in second draw = $\frac{8}{18}=\frac{4}{9}$
Now, Probability of getting both balls red = $\frac{4}{9} \times \frac{4}{9}=\frac{16}{81}$
View full question & answer→Question 301 Mark
Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find: P(neither A nor B)
AnswerP (neither A nor B) = P [not $\left( {A \cup B} \right)$] $ = 1 - P(A \cup B) $= 1 - 0.72 = 0.28
View full question & answer→Question 311 Mark
Given two independent events A and B such that P (A) = 0.3, P (B) = 0.6. Find P(A or B).
AnswerP (A or B) = P (A) + P (B) – P (A and B) = 0.3 + 0.6 – P(A)P(B) = 0.9 – 0.6 $ \times $ 0.3= 0.9 - 0.18 = 0.72
View full question & answer→Question 321 Mark
Given two independent events A and B such that P (A) = 0.3, P (B) = 0.6. Find P (A and not B).
AnswerP (A and not B) = $P\left( {A \cap \bar B} \right)$ = P (A) – $P\left( {A \cap B} \right)$ = 0.3 – 0.18 = 0.12
View full question & answer→Question 331 Mark
Given two independent events A and B such that P (A) = 0.3, P (B) = 0.6. Find P(A and B).
AnswerP (A and B) = P (A).P(B) = 0.3 $ \times $ 0.6 = 0.18
View full question & answer→Question 341 Mark
Two coins are tossed once, determine P(E|F), where E: no tail appears, F: no head appears.
AnswerSample space of the given experiment, 'S' = {HH, HT, TH, TT}
Here, E: no tail appears F: no head appears
$\Rightarrow$ E = {HH} and F = {TT}
$\Rightarrow \mathrm{E} \cap \mathrm{F}=\phi$
So, P(E) = $\frac{1}{4}$, P(F) = $\frac{1}{4}$, $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{0}{4}=0$
$\text {By definition of conditional probability,}~P(E | F)=\frac{P(E \cap F)}{P(F)}$
$\Rightarrow \mathrm{P}(\mathrm{E} | \mathrm{F})=\frac{0}{1 / 4}$
$\Rightarrow$ P(E|F) = 0
View full question & answer→Question 351 Mark
Two coins are tossed once, determine P(E|F), where E: tail appears on one coin, F: one coin shows head.
AnswerSample space of the experiment, 'S' = {HH, HT, TH, TT}
Here, E: tail appears on one coin and F: head appears on one coin
$\Rightarrow$ E = {HT, TH} and F = {HT,TH}
$\Rightarrow$ E $\cap$ F = {HT, TH}
So, P(E) = $\frac{2}{4}=\frac{1}{2}$, P(F) = $\frac{2}{4}=\frac{1}{2}$, $P(E \cap F)=\frac{2}{4}=\frac{1}{2}$
By the definition of conditional probability, $\mathrm{P}(\mathrm{E} | \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}$
$\Rightarrow \mathrm{P}(\mathrm{E} | \mathrm{F})=\frac{1 / 2}{1 / 2}$
$\Rightarrow$ P(E|F) = 1
View full question & answer→Question 361 Mark
A coin is tossed three times determine P(E|F),
where E: at most two tails, F: at least one tail.
AnswerThe sample space of the given experiment will be:
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Here, E: at most two tails
And F: at least one tail
⇒ E = {HHH, HHT, HTH, THH, HTT, THT, TTH}
And F = {HHT, HTH, THH, HTT, THT, TTH, TTT}
$P(E \cap F)= \text {{HHT, HTH, THH, HTT, THT, TTH}}$
So, P(E) = $\frac{7}{8}$, P(F) = $\frac{7}{8}$, $P(E \cap F)=\frac{6}{8}=\frac{3}{4}$
$\text {Bydefinition of conditional probability,}~ \mathrm{P}(\mathrm{E} | \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}$
$\Rightarrow P(E | F)=\frac{\frac{3}{ 4}}{\frac{7}{8}}=\frac{6}{7}$
View full question & answer→Question 371 Mark
A coin is tossed three times, determine P(E|F),
where E: at least two heads, F: at most two heads.
AnswerThe sample space of the given experiment will be:
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Here, E: at least two heads
And F: at most two heads
⇒ E = {HHH, HHT, HTH, THH}
and F = {HHT, HTH, THH, HTT, THT, TTH, TTT}
⇒ E ∩ F = {HHT, HTH, THH}
So, P(E) = $\frac{4}{8}=\frac{1}{2}$, P(F) = $\frac{7}{8}$, $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{3}{8}$
By the definition of conditional probability $P(E | F)=\frac{P(E \cap F)}{P(F)}$
$\Rightarrow P(E | F)=\frac{\frac{3}{8}}{\frac{7}{ 8}}=\frac{3}{7}$. Which is the required solution.
View full question & answer→Question 381 Mark
A coin is tossed three times, determine P(E|F),
where E: Head on third toss, and F: Head on first two tosses.
AnswerThe sample space of the given experiment will be:
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Here, E: head on third toss
And F: head on first two tosses
⇒ E = {HHH, HTH, THH, TTH} and F = {HHH, HHT}
⇒ E ∩ F = {HHH}
So, P(E) = $\frac{4}{8}=\frac{1}{2}$, P(F) = $\frac{2}{8}=\frac{1}{4}$, $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{8}$
$\text{Now, we know that}~P(E | F)=\frac{P(E \cap F)}{P(F)}$
$\Rightarrow P(E | F)=\frac{\frac{1 }{ 8}}{\frac{1}{ 4}}=\frac{4}{8}=\frac{1}{2}$
$\Rightarrow P(E | F)=\frac{1}{2}$
View full question & answer→Question 391 Mark
If P(A) = $\frac{6}{11}$, P(B) = $\frac{5}{11}$ and $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{7}{11}$, find P(B|A)
AnswerGiven: P(A) = $\frac{6}{11}$, P(B) = $\frac{5}{11}$, $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{7}{11}$
We know that,
$\because$ By definition of conditional probability,
$\mathrm{P}(\mathrm{B} | \mathrm{A})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}$
$\Rightarrow P(B | A)=\frac{4 / 11}{6 / 11}=\frac{4}{6}=\frac{2}{3}$
$\Rightarrow P(B | A)=\frac{2}{3}$
View full question & answer→Question 401 Mark
If P(A) = $\frac{6}{11}$, P(B) = $\frac{5}{11}$ and $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{7}{11}$, find P(A|B)
AnswerGiven: P(A) = $\frac{6}{11}$, P(B) = $\frac{5}{11}$, $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{7}{11}$
we know that $P(A \cap B)=P(A)+P(B)-P(A \cup B) $
$=\frac{6}{11}+\frac{5}{11}-\frac{7}{11} = \frac{4}{11}$
Now, By definition of conditional probability,$\mathrm{P}(\mathrm{A} | \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}$
$\Rightarrow \mathrm{P}(\mathrm{A} | \mathrm{B})=\frac{\frac{4}{11}}{\frac{5 }{ 11}}$
$\Rightarrow \mathrm{P}(\mathrm{A} | \mathrm{B})=\frac{4}{5}$
View full question & answer→Question 411 Mark
If P(A) = $\frac{6}{11}$, P(B) = $\frac{5}{11}$ and $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{7}{11}$, find $\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
AnswerGiven that P(A) = $\frac{6}{11}$, P(B) = $\frac{5}{11}$, $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{7}{11}$
we know that $P(A \cap B)=P(A)+P(B)-P(A \cup B)$
$\Rightarrow P(A \cap B)=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}=\frac{11-7}{11}$
$\Rightarrow P(A \cap B)=\frac{4}{11}$
View full question & answer→Question 421 Mark
If P(A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4, find $\mathrm{P}(\mathrm{A} \cup \mathrm{B})$
AnswerGiven: P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4
$ P(A \cap B) = P(B|A) \cdot P(A) = 0.32$
We know that, $ P(A \cup B)=P(A)+P(B) -P(A \cap B) $
$\Rightarrow$ $P(A \cup B)$ = 0.8 + 0.5 – 0.32 = 1.3 - 0.32
$\Rightarrow P(A \cup B)$ = 0.98
View full question & answer→Question 431 Mark
If P(A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4, find P(A|B)
AnswerGiven: P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4
By definition of conditional probability $P(B | A)=\frac{P(A \cap B)}{P(A)}$
$\Rightarrow P(A \cap B) = P(B | A) ~P(A) = 0.32$
$Now,~ P(A | B)=\frac{0.32}{0.5}=0.64$
$\Rightarrow$ P(A|B) = 0.64
View full question & answer→Question 441 Mark
If P(A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4, find $P(A \cap B)$
AnswerGiven: P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4
We know that
By definition of conditional probability,
${P}({B} | {A})=\frac{{P}({A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}$
$\Rightarrow P(A \cap B)=P(B | A) P(A)$
$\Rightarrow P(A \cap B)=0.4 \times 0.8$
$\Rightarrow P(A \cap B)=0.32$
View full question & answer→Question 451 Mark
If A and B are events such that P(A|B) = P(B|A), then
AnswerIt is given that : P( A | B) = P( B | A)
$ \Rightarrow \frac{{P(A \cap B)}}{{P(B)}} = \frac{{P(B \cap A)}}{{P(A)}} $ $\Rightarrow \frac{1}{{P(B)}} = \frac{1}{{P(A)}} \Rightarrow P(A) = P(B) $
View full question & answer→Question 461 Mark
If P(A) = $\frac{1}{2}$, P(B) = 0, then P(A|B) is
AnswerWe know that :
$ \\ P(A/B) = \frac{{P(A \cap B)}}{{P(B)}} = \frac{{P(A \cap B)}}{0} \\$
which is not defined
View full question & answer→Question 471 Mark
Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that at least one is a girl?
AnswerLet B denote boy and G denote girl.
Sample space of the experiment is, S = {GG, GB, BG, BB}
Let E be the event that ‘both are girls’.
$\Rightarrow$ E = {GG}
$\Rightarrow$ P(E) = $\frac{1}{4}$
Let H be the event that ‘at least one is a girl’.
$\Rightarrow$ H = {GG, GB, BG}
$\Rightarrow \mathrm{P}(\mathrm{H})=\frac{3}{4}$ ......(i)
Now, E $\cap$ H = {GG}
$\Rightarrow \mathrm{P}(\mathrm{E} \cap \mathrm{H})=\frac{1}{4}$ .....(ii)
Now, By definition of conditional probability,
$P(E | F)=\frac{P(E \cap F)}{P(E)}$
$\Rightarrow P(E | H)=\frac{P(E \cap H)}{P(H)}=\frac{1 / 4}{3 / 4}=\frac{1}{3}$ [Using (i) and (ii)]
$\Rightarrow P(E | H)=\frac{1}{3}$
View full question & answer→Question 481 Mark
Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that the youngest is a girl?
AnswerLet B denote boy and G denote girl.
Sample space of the experiment, S = {GG, GB, BG, BB}
Let E be the event that ‘both are girls’.
$\Rightarrow$ E = {GG}
$\Rightarrow$ P(E) = $\frac{1}{4}$
Let F be the event that ‘youngest one is a girl’.
⇒ F = {GG, BG}
$\Rightarrow P(F)=\frac{2}{4}=\frac{1}{2}$ ......(i)
Now, E $\cap$ F = {GG}
$\Rightarrow \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{4}$ .....(ii)
Now, By definition of conditional probability, $\mathrm{P}(\mathrm{E} | \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}$
$\Rightarrow P(E | F)=\frac{1 / 4}{1 / 2}=\frac{2}{4}=\frac{1}{2}$ [Using (i) and (ii)]
$\Rightarrow P(E | F)=\frac{1}{2}$
View full question & answer→Question 491 Mark
A fair die is rolled. Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5} Find $\mathrm{P}((\mathrm{E} \cup \mathrm{F}) | \mathrm{G})$ and $\mathrm{P}((\mathrm{E} \cap \mathrm{F}) | \mathrm{G})$
AnswerSample space of given experiment is,S = {1, 2, 3, 4, 5, 6}
Here, E = {1, 3, 5}, F = {2, 3} and G = {2,3,4,5} ......(i)
$\Rightarrow P(E)=\frac{3}{6}=\frac{1}{2}, P(F)=\frac{2}{6}=\frac{1}{3}, P(G)=\frac{4}{6}=\frac{2}{3}$ .....(ii)
Now, E $\cap$ F = {3}, F $\cap$ G = {2, 3}, E $\cap$ G = {3, 5} ......(iii)
$\Rightarrow \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{6}, \mathrm{P}(\mathrm{F} \cap \mathrm{G})=\frac{2}{6}=\frac{1}{3}, \mathrm{P}(\mathrm{E} \cap \mathrm{G})=\frac{2}{6}=\frac{1}{3}$ .....(iv)
Clearly, from (i), we have
E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}
$\Rightarrow$ E $\cup$ F = {1, 2, 3, 5}
$\Rightarrow$ (E $\cup$ F) $\cap$ G = {2, 3, 5}
$\Rightarrow \mathrm{P}((\mathrm{E} \cup \mathrm{F}) \cap \mathrm{G})=\frac{3}{6}=\frac{1}{2}$.....(v)
Now, By definition of conditional probability, $\mathrm{P}(\mathrm{E} | \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}$
$\Rightarrow P((E \cup F) | G)=\frac{P((E \cup F) \cap G)}{P(G)}=\frac{1 / 2}{2 / 3}=\frac{3}{4}$ [Using (ii) and (v)]
$\Rightarrow \mathrm{P}((\mathrm{E} \cup \mathrm{F}) | \mathrm{G})=\frac{3}{4}$
Similarly, we have E $\cap$ F = {3} [Using (iii)]
And G = {2, 3, 4, 5} [Using (i)]
$\Rightarrow(E \cap F) \cap G=\{3\}$
$\Rightarrow \mathrm{P}((\mathrm{E} \cap \mathrm{F}) \cap \mathrm{G})=\frac{1}{6}$ .....(vi)
So,
$P((E \cap F) | G)=\frac{P((E \cap F) \cap G)}{P(G)}=\frac{1 / 6}{2 / 3}=\frac{1}{4}$ [Using (ii) and (vi)]
$\Rightarrow \mathrm{P}((\mathrm{E} \cap \mathrm{F}) | \mathrm{G})=\frac{1}{4}$
View full question & answer→Question 501 Mark
A fair die is rolled. Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5}.
Find P(E|G) and P(G|E)
AnswerSample space for the given experiment, 'S' = {1, 2, 3, 4, 5, 6}
Here, E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5} .....(i)
$\Rightarrow P(E)=\frac{3}{6}=\frac{1}{2}, P(F)=\frac{2}{6}=\frac{1}{3}, P(G)=\frac{4}{6}=\frac{2}{3}$ ......(ii)
Now, E $\cap$ F = {3}, F $\cap$ G = {2, 3}, E $\cap$ G = {3, 5} .....(iii)
$\Rightarrow \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{6}, \mathrm{P}(\mathrm{F} \cap \mathrm{G})=\frac{2}{6}=\frac{1}{3}, \mathrm{P}(\mathrm{E} \cap \mathrm{G})=\frac{2}{6}=\frac{1}{3}$ ......(iv)
By definition of conditional probability, $P(E | F)=\frac{P(E \cap F)}{P(F)}$
$\Rightarrow P(E | G)=\frac{P(E \cap G)}{P(G)}=\frac{1 / 3}{2 / 3}=\frac{1}{2}$
$\Rightarrow \mathrm{P}(\mathrm{E} | \mathrm{G})=\frac{1}{2}$
Similarly, we have
$P(G | E)=\frac{P(G \cap E)}{P(E)}=\frac{1 / 3}{1 / 2}=\frac{2}{3}$
$\Rightarrow P(G | E)=\frac{2}{3}$
View full question & answer→Question 511 Mark
A fair die is rolled. Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5}.
Find P(E|F) and P(F|E).
AnswerSample space for the given experiment, 'S' = {1, 2, 3, 4, 5, 6}
Here, E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5} ......(i)
$\Rightarrow P(E)=\frac{3}{6}=\frac{1}{2}, P(F)=\frac{2}{6}=\frac{1}{3}, P(G)=\frac{4}{6}=\frac{2}{3}$ ......(ii)
Now, $E \cap F$ = {3}, F $\cap$ G = {2, 3}, E $\cap$ G = {3, 5} ......(iii)
$\Rightarrow \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{6}, \mathrm{P}(\mathrm{F} \cap \mathrm{G})=\frac{2}{6}=\frac{1}{3}, \mathrm{P}(\mathrm{E} \cap \mathrm{G})=\frac{2}{6}=\frac{1}{3}$ ......(iv)
By the definition of conditional probability $\mathrm{P}(\mathrm{E} | \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}$
$\Rightarrow \mathrm{P}(\mathrm{E} | \mathrm{F})=\frac{1 / 6}{1 / 3}=\frac{3}{6}=\frac{1}{2}$ [Using (ii) and (iv)]
$\Rightarrow \mathrm{P}(\mathrm{E} | \mathrm{F})=\frac{1}{2}$
Similarly, we have
$P(F | E)=\frac{P(F \cap E)}{P(E)}=\frac{1 / 6}{1 / 2}=\frac{2}{6}=\frac{1}{3}$ [Using (ii) and (iv)]
$\Rightarrow \mathrm{P}(\mathrm{F} | \mathrm{E})=\frac{1}{3}$
View full question & answer→Question 521 Mark
A black and a red dice are rolled. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
AnswerWhen 2 die are rolled, total number of outcomes = 36.
Let first die is red and second die is black.
Let A be the event of getting sum equal to 8.
$\therefore $ A = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
Let A be the event B that red die results in a number less than 4.
$\therefore $ B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}
P (B) = $\frac{{n\left( B \right)}}{{n\left( S \right)}} = \frac{{18}}{{36}} = \frac{1}{2}$
$A \cap B$ = {(2, 6), (3, 5)}
$ \Rightarrow n\left( {A \cap B} \right) = 2$
$P\left( {A \cap B} \right)$ = $\frac{2}{{36}} = \frac{1}{{18}}$
$P\left( {A\over B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} $
$= \frac{{\frac{1}{{18}}}}{{\frac{1}{2}}} \\= \frac{2}{{18}} \\= \frac{1}{9}.$
View full question & answer→Question 531 Mark
A black and a red dice are rolled. Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
AnswerLet the first observation be from the black die and second from the red die.
When two dice (one black die and another red) are rolled, the sample space S has 6×6 = 36 number of elements.
Let A : obtaining a sum greater than 9 = {(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)}
and B: black die resulted in a 5 = {(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}
$$$\therefore A\cap B$ = {(5,5),(5,6)}
$$The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P(A/B)
$\therefore P(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{2}{36}}{\frac{6}{36}}=\frac{2}{6}=\frac{1}{3}$
View full question & answer→Question 541 Mark
Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards.
What is the probability that first two cards are kings and the third card drawn is an ace?
AnswerLet's define the events
K: card drawn is king and A: card drawn is an ace. Clearly, we have to find P (KKA)
Now $P(K)=\frac{\text {Number of kings}}{\text {Total number of cards }} \Rightarrow \frac{4}{52} = \frac{1}{13}$
Also, P (K|K) is the probability of second king with the condition that one king has already been drawn.
Now there are three kings in (52 - 1) = 51 cards.
Therefore P(K|K) = $\frac{3}{51} = \frac{1}{17}$
Lastly, P(A|KK) is the probability of third drawn card to be an ace, with the condition that two kings have already been drawn. Now there are four aces in left 50 cards.
Therefore P(A|KK) = $\frac{4}{50} = \frac{2}{25}$
By multiplication law of probability, we have
P(KKA) = P(K) P(K|K) P(A|KK)
= $=\frac{1}{13} \times \frac{1}{17} \times \frac{2}{25}=\frac{2}{5525}$
View full question & answer→Question 551 Mark
An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?
AnswerLet the events be
E: First ball drawn is black. and F: Second ball drawn is black.
We need to find the probability that both drawn balls are black. i,e $P(E \cap F)$
$P(E \cap F) = $ Probability first ball drawn is black x Probability second ball is black if first is black
$i,e~P(E \cap F)=P(E) P(F | E)$
Now P(E) = P (black ball in first draw) = $\frac{10}{15} = \frac{2}{3}$
$\mathrm{P}(\mathrm{F} | \mathrm{E})$ is the Probability of F after E has happened i,e. probability of second ball drawn black if first ball was black if first ball drawn was black, we are left with 9 black, 5 white balls
$P(F | E)=\frac{\text {Remaining Black ball}}{\text {remaining balls}}$
i.e. P(F|E} = $\frac{9}{14}$
By multiplication rule of probability, we have
P(E $\cap$ F) = P(E) P(F|E)
= $\frac{10}{15} \times \frac{9}{14}=\frac{3}{7}$
View full question & answer→Question 561 Mark
Consider the experiment of tossing a coin. If the coin shows head, toss it again, but if it shows tail, then throw a die. Find the conditional probability of the event that the die shows a number greater than 4, given that there is atleast one tail.
AnswerThe sample space S of the experiment is given as
S = {(H, H), (H, 1), (T, 1), (T, 2), (T, 3),
(T, 4), (T, 5), (T, 6)}
The probabilities of these elementary events are
$P \{ ( H , T ) \} = \frac { 1 } { 2 } \times \frac { 1 } { 2 } = \frac { 1 } { 4 },$ $P \{ ( H , T ) \} = \frac { 1 } { 2 } \times \frac { 1 } { 2 } = \frac { 1 } { 4 },$
$P \{ ( T , 1 ) \} = \frac { 1 } { 2 } \times \frac { 1 } { 6 } = \frac { 1 } { 12 },$ $P \{ ( T , 2 ) \} = \frac { 1 } { 2 } \times \frac { 1 } { 6 } = \frac { 1 } { 12 },$
$P \left\{ ( T , 3 ) = \frac { 1 } { 2 } \times \frac { 1 } { 6 } = \frac { 1 } { 12 }\right.,$ $P \{ ( T , 4 ) \} = \frac { 1 } { 2 } \times \frac { 1 } { 6 } = \frac { 1 } { 12 },$
$P \left\{ ( T , 5 ) = \frac { 1 } { 2 } \times \frac { 1 } { 6 } = \frac { 1 } { 12 }\right.$ and $P \{ ( T , 6 ) \} = \frac { 1 } { 2 } \times \frac { 1 } { 6 } = \frac { 1 } { 12 }$
The outcomes of the experiment can be represented in the following tree diagram.
Consider the following events:
A = the die shows a number greater than 4 and
B = there is atleast one tail.
We have, A = {(T, 5), (T, 6)},
B = {(H, 1), (T, 1), (T, 2),(T, 3),
(T, 4), (T, 5), (T, 6)}
and A $\cap$ B = {(T, 5), (T, 6)}
$\therefore$ P(B) = P{(H, 1)} + P{(T, l)} + P{(T, 2)}
+ P{(T, 3)} + P{(T, 4)} + P{(T, 5)} + P{(T, 6)}
$\Rightarrow P ( B ) = \frac { 1 } { 4 } + \frac { 1 } { 12 } + \frac { 1 } { 12 } + \frac { 1 } { 12 } + \frac { 1 } { 12 } + \frac { 1 } { 12 } + \frac { 1 } { 12 } = \frac { 3 } { 4 }$
and P(A $\cap$ B) = P{(T, 5)} + P{(T, 6)} =$\frac { 1 } { 12 } + \frac { 1 } { 12 } = \frac { 1 } { 6 }$
$\therefore$ Required probability
$= P \left( \frac { A } { B } \right) = \frac { P ( A \cap B ) } { P ( B ) } = \frac { 1 / 6 } { 3 / 4 } = \frac { 4 } { 18 } = \frac { 2 } { 9 }$
View full question & answer→Question 571 Mark
A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?
AnswerLet's define events;
E : Number 4 appears at least once.
F : Sum of the numbers appearing is 6.
Then, E = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)}
and F = {(1,5), (2,4), (3,3), (4,2), (5,1)}
We have P(E) = $\frac{11}{36}$ and P(F) = $\frac{5}{36}$
Also $\mathrm{E} \cap \mathrm{F}$ = {(2,4), (4,2)}
Therefore $\mathrm{P}(\mathrm{E} \cap \mathrm{F})$ = $\frac{2}{36}$
Hence, the required probability
$\mathrm{P}(\mathrm{E} | \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{\frac{2}{36}}{\frac{5}{36}}=\frac{2}{5}$
View full question & answer→Question 581 Mark
A die is thrown three times. Events A and B are defined as below:
A: 4 on the third throw
B: 6 on the first and 5 on the second throw
Find the probability of A given that B has already occurred.
AnswerTotal sample space = 216
A = $\left\{ \begin{gathered} (1,1,4)(1,2,4)...(1,6,4)(2,1,4)(2,2,4)...(2,6,4) \hfill \\ (3,1,4)(3,2,4)...(3,6,4)(4,1,4)(4,2,4)...(4,6,4) \hfill \\ (5,1,4)(5,2,4)...(5,6,4)(6,1,4)(6,2,4)...(6,6,4) \hfill \\ \end{gathered} \right\}$
B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}
$A \cap B $ = {6, 5, 4}
P(B) = $ \frac{6}{{216}}$, $P(A \cap B) = \frac{1}{{216}}$
$P(\frac {A}{B}) = \frac{{P(A \cap B)}}{{P(B)}} = \frac{{\frac{1}{{216}}}}{{\frac{6}{{216}}}} = \frac{1}{6}$
View full question & answer→Question 591 Mark
In a school there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in class XII given that the chosen student is a girl?
AnswerLet E denotes the event that student chosen randomly studies in class XII, F denotes the event that randomly chosen student is girl.
P (E|F) = ?
$P(F) = \frac{{430}}{{1000}} = 0.43$
$P\left( {E \cap F} \right) = \frac{{43}}{{1000}} = 0.043$
$P\left( {\frac{E}{F}} \right) = \frac{{P\left( {E \cap F} \right)}}{{P(F)}}$
$ = \frac{{0.043}}{{0.43}} = 0.1$
View full question & answer→Question 601 Mark
Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?
AnswerSample space of the experiment is 'S' = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Let's define two events
A : Number on the card drawn is even.
B : Number on the card drawn is greater than 3.
$\Rightarrow$ A = {2, 4, 6, 8, 10}, B = {4, 5, 6, 7, 8, 9, 10}
$A \cap B$ = {4, 6, 8, 10}
Also, P(A) = $\frac{5}{10}$, P(B) = $\frac{7}{10}$ and $P(A \cap B)=\frac{4}{10}$
Then $\mathrm{P}(\mathrm{A} | \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\frac{4}{10}}{\frac{7}{10}}=\frac{4}{7}$
View full question & answer→Question 611 Mark
If a machine is correctly set up, it produces $90\%$ acceptable items. If it is incorrectly set up, it produces only $40\%$ acceptable items. Past experience shows that $80\%$ of the set ups are correctly done. If after a certain set up, the machine produces $2$ acceptable items, find the probability that the machine is correctly setup.
AnswerLet's define events;
$A :$ Machine produces $2$ acceptable items.
$B_1:$ Machine is correctly setup.
$B_2:$ Machie is incorrectly setup.
Now $P(B_1) = 0.8, P(B_2) = 0.2$
$P(A|B_1) = 0.9 \times 0.9$ and $P(A|B_2) = 0.4 \times 0.4$
Therefore $\mathrm{P}\left(\mathrm{B}_{1} | \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{B}_{1}\right) \mathrm{P}\left(\mathrm{A} | \mathrm{B}_{1}\right)}{\mathrm{P}\left(\mathrm{B}_{1}\right) \mathrm{P}\left(\mathrm{A} | \mathrm{B}_{1}\right)+\mathrm{P}\left(\mathrm{B}_{2}\right) \mathrm{P}\left(\mathrm{A} | \mathrm{B}_{2}\right)}$
= $\frac{0.8 \times 0.9 \times 0.9}{0.8 \times 0.9 \times 0.9+0.2 \times 0.4 \times 0.4}$ = $\frac{648}{680}= \frac{81}{85}$
View full question & answer→Question 621 Mark
A and B throw a die alternatively till one of them gets a 6 and wins the game. Find their respective probabilities of winning, if A starts first.
AnswerS denote the success(getting a 6) and F denotes the failure(not getting a 6)
$P(S) = \frac{1}{6},P(F) = \frac{5}{6}$
P(A win in the first throw) = (S) $ = \frac{1}{6}$
P(A win in the 3rd throw) $ = {\left( {\frac{5}{6}} \right)^2} \times \frac{1}{6}$
P(A win in the fifth throw) $ = {\left( {\frac{5}{6}} \right)^4} \times \frac{1}{6}$
P (A win) $ = \frac{1}{6} + {\left( {\frac{5}{6}} \right)^2} \times \frac{1}{6} + {\left( {\frac{5}{6}} \right)^4} \times \frac{1}{6} + ........$
$ = \frac{{\frac{1}{6}}}{{1 - \frac{{25}}{{36}}}}$ $\left[ {u\sin g\;s = \frac{a}{{1 - r}}} \right]$
$ = \frac{6}{{11}}$
View full question & answer→Question 631 Mark
Coloured balls are distributed in four boxes as shown in the following table:
| Box |
Colour |
| Black |
White |
Red |
Blue |
| $I$ |
$3$ |
$4$ |
$5$ |
$6$ |
| $II$ |
$2$ |
$2$ |
$2$ |
$2$ |
| $III$ |
$1$ |
$2$ |
$3$ |
$1$ |
| $IV$ |
$4$ |
$3$ |
$1$ |
$5$ |
A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the box $III?$ AnswerLet $A, E_1, E_2, E_3$ and $E_4$ be the events as defined below :
$A :$ black ball is selected
$E_1 :$ box $I$ is selected
$E_2 :$ box $II$ is selected
$E_3 :$ box $III$ is selected
$E_4 :$ box $IV$ is selected
Since the boxes are chosen at random,
Therefore, $P(E_1) = P(E_2) = P(E_3) = P(E_4)$
Also $P(A|E_1) = \frac{3}{18} , P(A|E_2) = \frac {2}{8} , P(A|E_3) = \frac{1}{7} $ and $P(A|E_4) = \frac{4}{13}$
$P($box III is selected, given that the drawn ball is black$) = P(E_3|A).$
By Bayes' theorem,
$P(E_3|A) = \frac{\mathrm{P}\left(\mathrm{E}_{3}\right) \cdot \mathrm{P}\left(\mathrm{A|E}_{3}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A|E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A|E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \mathrm{P}\left(\mathrm{A|E}_{3}\right)+\mathrm{P}\left(\mathrm{E}_{4}\right) \mathrm{P}\left(\mathrm{A|E}_{4}\right)}$
= $\frac{\frac{1}{4} \times \frac{1}{7}}{\frac{1}{4} \times \frac{3}{18}+\frac{1}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{1}{7}+\frac{1}{4} \times \frac{4}{13}} = 0.165$
View full question & answer→Question 641 Mark
A man is known to speak truth $3$ out of $4$ times. He throws a die and reports that it is a six. Find the probability that it is actually a six.
AnswerLet E be the event that the man reports that six occurs in the throwing of the dice and let $S_1$ be the event that six occurs and $S_2$ be the event six does not occur.
Then $P(S_1)$ =Probability that six occurs = $\frac{1}{6}$
$P(S_2)$ =Probability that six does not occur = $\frac{5}{6}$
$P(E|S_1)$ = Probability that the man reports that six occurs when six has actually occurred on the die
= Probability that the man speaks the truth = $\frac{3}{4}$
$P(E|S_2)$ = Probability that the man reports that six occurs when six hasn't actually occurred on the die
= Probability that the man does not speak the truth $1 - \frac{3}{4} = \frac{1}{4}$
Thus, by Bayes' theorem, we get
$P(S_1|E)$ = Probability that the report of the man that six has occurred is actually a six
$= \frac{{P({S_1})P(E/{S_2})}}{{P({S_1})P(E/{S_1}) + P({S_2})P(E/{S_2})}}$
$=\frac{1/6×3/4}{1/6×3/4+5/6×1/4}$ = $\frac {1}{8} \times \frac {24}{8} = \frac {3}{8}$
View full question & answer→Question 651 Mark
A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively $\frac{3}{{10}},\frac{1}{5},\frac{1}{{10}}$ and $\frac{2}{5}$. The probabilities that he will be late are $\frac{1}{4},\frac{1}{3}$ and $\frac{1}{{12}}$, if he comes by train, bus and scooter respectively, but if he comes by other means of transport, that he will not be late. When he arrives, he is late. What is the probability that he comes by train?
AnswerLet $E$ be the event that the doctor visits the patient late and let $T_1, T_2, T_3, T_4$, be the event that the doctor comes by train, bus, scooter and other means of Transport respectively.
$P\left(T_1\right)=\frac{3}{10}, P\left(T_2\right)=\frac{1}{5}, P\left(T_3\right)=\frac{1}{10}, P\left(T_4\right)=\frac{2}{5}$
$P(\frac {E}{T_1}) =$ probability that the doctor arriving late comes by train $= \frac{1}{4}$
Similarly, $P(\frac {E}{T_2}) = \frac{1}{3},P(\frac {E}{T_3}) = \frac{1}{{12}},P(\frac {E}{T_4}) = 0$
$P(\frac {T_1}{E}) $ = $ \frac{{P({T_1})P(E/{T_1})}}{{P({T_1})P(E/{T_1}) + P({T_2})P(E/{T_2}) + P({T_3})P(E/{T_3}) + P({T_4})P(E/{T_4})}}$
$=\frac{\frac{3}{10}\times \frac{1}{4}}{\frac{3}{10}\times\frac{1}{4}+\frac{1}{5}\times\frac{1}{3}+\frac{1}{10}\times\frac{1}{12}+\frac{2}{5}\times 0}$
$ = \frac{1}{2}$
View full question & answer→Question 661 Mark
A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?
AnswerWe are given that a family has two children.
Let b stand for Boy and g for Girl.
The sample space of the experiment is
S = {(b, b), (g, b), (b, g), (g, g)}
Let E and F denote the following events :
E : Both the children are Boys
F : At least one of the child is a Boy
Then, E = {(b,b)} and F = {(b,b), (g,b), (b,g)}
Now ${E} \cap {F}$ = {(b,b)}
Thus P(F) = $\frac{3}{4}$ and $P(E \cap F)=\frac{1}{4}$
Therefore $\mathrm{P}(\mathrm{E} /\mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$
View full question & answer→Question 671 Mark
In a factory which manufactures bolts, machines. $A, B$ and $C$ manufacture respectively $25\%, 35\%$ and $40\%$ of the bolts. Of their output $5, 4$ and $2$ percent are respectively defective bolts. $A$ bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine $B?$
AnswerLet $B_1 =$ bolt is manufactured by $A$
$B_2 =$ bolt is manufactured by $B$
$B_3 =$ bolt is manufactured by $C$
Let E denote the event that bolt is defective.
The event E occurs with $B_1$ or with $B_2$ or with $B_3.$ Given that,
$P (B_1) = 25\% = 0.25$
$P (B_2) = 0.35$
$P (B_3) = 0.40$
$P ( E|B_1) =$Probability that the bolt drawn is defective given that it is manu-factured by machine $A = 5\% = 0.05.$
Similarly, $P (E|B_2) = 0.04$
$P (E|B_3) = 0.02$
Hence, by Bayes theorm, we have,
$P({B_2}/E) = \frac{{P({B_2})P(E/{B_2})}}{{P({B_1})P(E/{B_1}) + P({B_2})P(E/{B_2}) + P({B_3})P(E/{B_3})}}$
$=\frac{0.35\times 0.04}{0.25\times0.05+0.35\times0.04+0.40\times0.02}$
$=\frac{0.0140}{0.0345}$
$=\frac{28}{69}$
View full question & answer→Question 681 Mark
Suppose that the reliability of a HIV test is specified as follows:
Of people having HIV, 90% of the test detect the disease but 10% go undetected. Of people free of HIV, 99% of the test are Judged HIV-ive but 1% are diagnosed as showing HIV+ive. From a large population of which only 0.1% have HIV, one person is selected at random, given the HIV test, and the pathologist reports him/her is HIV+ive. What is the probability that the person actually has HIV?
AnswerLet E denote the event that the person selected is actually having HIV and A the event that the person’s HIV test is diagnosed as +ve.
Let E' denote the event that person selected is actually not having HIV.
Clearly, {E, E'} is a partition of the sample space of all people in the population.
We are given that
P(E) = 0.1% = $\frac{{0.1}}{{100}} $ = 0.001
P(E') = 1 - P(E) = 0.999
$P(\frac {A}{E}) $ = 90% = $ \frac{{90}}{{100}} $ = 0.9
$P(\frac {A}{E'}) $ = 1% = $ \frac{1}{{100}}$ = 0.01
By Bayes theorm,we have,
$P(\frac {E}{A}) = \frac{{P(E)P(A/E)}}{{P(E)P(A/E) + P(E')P(A/E')}}$
$ = \frac{{0.001 \times 0.9}}{{0.001 \times 0.9 + 0.999 \times 0.01}}$
$ = \frac{9}{{9 + 99.9}}$
$ = \frac{{90}}{{1089}}$
= 0.083
View full question & answer→Question 691 Mark
Given three identical boxes I, II and III each containing two coins. In box I both coins are gold coins, in box II, both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?
AnswerLet $E_1, E_2$ and $E_3$ be the events that boxes I, II and III are chosen.
$P (E_1) = P (E_2) = P (E_3)$ = $\frac{1}{3}$
let A be the event the coin drawn is of gold.
$p(A|{E_1}) = \frac{2}{2} = 1$
$P(A|{E_2}) = 0$
$P(A|{E_3}) = \frac{1}{2}$
$P({E_1}|A) = \frac{{P({E_1})P(A|{E_1})}}{{P({E_1})P(A|{E_1}) + P({E_2}) + P(A|{E_2}) + P({E_3})P(A|{E_3})}}$
$=\frac{\frac{1}{3}×1}{\frac{1}{3}×1+\frac{1}{3}×0+\frac{1}{3}×\frac{1}{2}}$
$ = \frac{2}{3}$
View full question & answer→Question 701 Mark
Bag I contains $3$ red and $4$ black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II.
AnswerLet $E_1 :$ Bag selected is Bag $I$
$E_2 :$ Bag selected is Bag $II$
$A :$ Ball selected is Red
$B :$ Ball selected is Black
Now,
$P($ball was drawn from bag $II,$ if ball is red$) = P(\frac{E_2}{A}$)
$P(\frac{E_2}{A}) = \frac{P(A) P\left(A | E_{2}\right)}{P\left(E_{1}\right) P\left(A | E_{1}\right)+P\left(E_{2}\right) P\left(A | E_{2}\right)}$
$P(E_1) =$ Probability bag selected is Bag $I = \frac{1}{2}$
$P(E_2) =$ Probability bag selected is Bag $II = \frac{1}{2}$
$P(\frac{A}{E_2}) =$ Probability red ball was selected from Bag $II = \frac{5}{5+6}$ = $\frac{5}{11}$
$P(\frac{A}{E_1})= $ Probability red ball was selected from Bag $I = \frac{3}{3+4}$ = $\frac{3}{7}$
Putting values in formula,
$P(\frac{E_{2}} { A} ) = \frac{\frac{1}{2} \times \frac{5}{11}}{\frac{1}{2} \times \frac{3}{7}+\frac{1}{2} \times \frac{5}{11}}$
$= \frac{\frac{1}{2} \times \frac{5}{11}}{\frac{1}{2}\left[\frac{3}{7} + \frac{5}{11}\right]}$
$= \frac{\frac{5}{11}}{\frac{33+35}{77}}$
$= \frac{\frac{5}{11}}{\frac{68}{77}}$ = $\frac{5}{11} \times \frac{77}{68}$ = $\frac{35}{68}$
Therefore, required probability is $\frac{35}{68}$
View full question & answer→Question 711 Mark
A person has undertaken a construction job. The probabilities are 0.65 that there will be strike, 0.80 that the construction job will be completed on time if there is no strike, and 0.32 that the construction job will be completed on time if there is a strike. Determine the probability that the construction job will be completed on time.
AnswerLet, A : Construction job will be completed on time,
and B : There will be a strike.
We have
P(B) = 0.65, P(no strike) = P(B′) = 1 - P(B) = 1 - 0.65 = 0.35
P(A|B) = 0.32, P(A|B′) = 0.80
Since events B and B′ form a partition of the sample space S, therefore, by Theorem on total probability, we have
P(A) = P(B) P(A|B) + P(B′) P(A|B′)
= 0.65 $\times$ 0.32 + 0.35 $\times$ 0.8 = 0.208 + 0.28 = 0.488
Thus, the probability that the construction job will be completed in time is 0.488.
View full question & answer→Question 721 Mark
If A and B are two independent events, then the probability of occurrence of at least one of A and B is given by 1- P(A')(P(B').
AnswerWe have,
P(at least one of A and B) $ = P\left( {A \cup B} \right)$
$ = P(A) + P(B) - P(A \cap B)$
= P(A) + P(B) - P(A) P(B)
=P(A) + P(B) [1−P(A)]
=P(A) + P(B). P(A′)
=1− P(A′) + P(B) P(A′)
=1− P(A′) [1− P(B)]
=1− P(A′) P (B′)
View full question & answer→Question 731 Mark
Prove that if E and F are independent events, then so are the events E and F′.
AnswerSince E and F are independent, we have
P(E $\cap$ F) = P(E) . P(F) ......(i)
From the venn diagram in Figure, it is clear that E $\cap$ F and E $\cap$ F′ are mutually exclusive events and
also E = (E $\cap$ F) $\cup$ (E $\cap$ F′).
Therefore P(E) = P(E $\cap$ F) + P(E $\cap$ F′)
or P(E $\cap$ F′) = P(E) - P(E $\cap$ F)
= P(E) - P(E) . P(F) (by (i))
= P(E) (1 - P(F))
= P(E). P(F′)
Hence, E and F′ are independent.
View full question & answer→Question 741 Mark
Three coins are tossed simultaneously. Consider the event E three heads or three tails, F at least two heads and G at most two heads. Of the pairs (E, F), (E, G) and (F, G), which are independent? which are dependent?
AnswerThe sample space of the experiment is given by
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Clearly, E = {HHH, TTT}, F = {HHH, HHT, HTH, THH}
and G = {HHT, HTH, THH, HTT, THT, TTH, TTT}
Also E $\cap$ F = {HHH}, E $\cap$ G = {TTT}, F $\cap$ G = { HHT, HTH, THH}
Therefore P(E) = $\frac{2}{8}=\frac{1}{4}$, P(F) = $\frac{4}{8}=\frac{1}{2}$, P(G) = $\frac{7}{8}$
and $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{8}$, $\mathrm{P}(\mathrm{E} \cap \mathrm{G})=\frac{1}{8}$, $\mathrm{P}(\mathrm{F} \cap \mathrm{G})=\frac{3}{8}$
Also, P(E).P(F) = $\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}$, P(E).P(G) = $\frac{1}{4} \times \frac{7}{8}=\frac{7}{32}$ and P(F).P(G) = $\frac{1}{2} \times \frac{7}{8}=\frac{7}{16}$
Thus P(E $\cap$ F) = P(E) . P(F)
$\mathrm{P}(\mathrm{E} \cap \mathrm{G}) \neq \mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{G})$
and $\mathrm{P}(\mathrm{F} \cap \mathrm{G}) \neq \mathrm{P}(\mathrm{F}) \cdot \mathrm{P}(\mathrm{G})$
Hence, the events (E and F) are independent, and the events (E and G) and (F and G) are dependent.
View full question & answer→Question 751 Mark
An unbiased die is thrown twice. Let the event A be odd number on the first throw and B the event odd number on the second throw. Check the independence of the events A and B.
AnswerIf all the 36 elementary events of the experiment are considered to be equally likely,
we have P(A) = $\frac{18}{36}=\frac{1}{2}$ and P(B) = $\frac{18}{36}=\frac{1}{2}$
Also P(A $\cap$ B) = P (odd number on both throws)
= $\frac{9}{36}=\frac{1}{4}$
Also P(A) P(B) = $\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$
Clearly P(A $\cap$ B) = P(A) $\times$ P(B)
Thus, A and B are independent events
View full question & answer→Question 761 Mark
A die is thrown. If E is the event the number appearing is a multiple of 3 and F be the event the number appearing is even then find whether E and F are independent?
AnswerTwo event A and B are independent if $ P(A \cap B)=P(A) . P(B)$
Sample space of the experiment is, S = {1, 2, 3, 4, 5, 6}
Now E = {3, 6}, F = { 2, 4, 6} and E $\cap$ F = {6}
Then P(E) = $\frac{2}{6}=\frac{1}{3}$, P(F) = $\frac{3}{6}=\frac{1}{2}$ and $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{6}$
Clearly P(E $\cap$ F) = P(E). P(F) $=\frac{1}{6}$
Hence E and F are independent events.
View full question & answer→Question 771 Mark
If P(A) = $\frac{7}{13}$, P(B) = $\frac{9}{13}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{4}{13}$, evaluate P(A|B).
AnswerWe have P(A|B) = $\frac{P(A \cap B)}{P(B)}=\frac{\frac{4}{13}}{\frac{9}{13}}=\frac{4}{9}$
View full question & answer→