2 Marks Questions - Maths STD 12 Science Questions - Vidyadip

Questions

2 Marks Questions

Take a timed test

25 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks

Suppose that 90% of people are right-handed. What is the probability that at most of 6 of a random sample of 10 people are right-handed?

Answer

p = $\frac{90}{100}=\frac{9}{10}$ and q = 1 - p = 1 - $\frac{9}{10}=\frac{1}{10}$

n = 10

P (at most 6 successes) = 1 – [P (X = 7) + P (X = 8) + P (X = 9) + P (X = 10)]

= 1 – $\left[ {{}^{10}{C_7}{{\left( {\frac{9}{{10}}} \right)}^7}{{\left( {\frac{1}{{10}}} \right)}^3} + {}^{10}{C_8}{{\left( {\frac{9}{{10}}} \right)}^8}{{\left( {\frac{1}{{10}}} \right)}^2} + {}^{10}{C_9}{{\left( {\frac{9}{{10}}} \right)}^9}\left( {\frac{1}{{10}}} \right) + {}^{10}{C_{10}}{{\left( {\frac{9}{{10}}} \right)}^{10}}} \right]$

= 1 - $\sum\limits_{r = 7}^{10} {{}^{10}{C_r}{{(0.9)}^r}{{(0.1)}^{10 - r}}} $

View full question & answer→

Question 22 Marks

A couple has two children, find the probability that both children are females, if it is known that the elder child is a female.

Answer

A Couple has two children,
Let boy be denoted by b & girl be denoted by g
Let's define events;
C : Both children are female.
D : Elder child is female.
$\therefore$ C = {(g, g)}
P(C) = $\frac{1}{4}$
And, D = {(g, b), (g, g)}
P(D) = $\frac{2}{4}$
$P(C \cap D) = \frac14$
Hence, P(C|D) = $\frac{P(C \cap D)}{P(D)}$
= $\frac{\frac{1}{4}}{\frac{2}{4}}$
= $\frac{1}{2}$

View full question & answer→

Question 32 Marks

A couple has two children, find the probability that both children are males, if it is known that at least one of the children is male.

Answer

Given that, a couple has two children. Let boy and girl be denoted by 'b' and 'g' respectively
So, Sample space, S = {(b, b), (b, g), (g, b), (g, g)}
Now, Let's define events;
A : Both children are male.
B : At least one of the children is male.
Thus, we have;
A $\cap$ B = {(b, b)}
P(A $\cap$ B) = $\frac{1}{4}$
P(A) = $\frac{1}{4}$
P(B) = $\frac{3}{4}$
Hence, $P(A | B)=\frac{P(A \cap B)}{P(B)}$
= $\frac{\frac{1}{4}}{\frac{3}{4}}$ = $\frac{1}{3}$

View full question & answer→

Question 42 Marks

A and B are two events such that P (A) $\ne$ 0. Find P(B|A), if A is a subset of B.

Answer

Given two events A and B where P(A) $\ne$ 0 and $A \subset B$
We have, $A \cap B=A$
$\therefore P(A \cap B)=P(B \cap A)=P(A)$
Hence, $P(B | A)=\frac{P(B \cap A)}{P(A)}$
= $\frac{P(A)}{P(A)}$
= 1

View full question & answer→

Question 52 Marks

Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are $0.6$ and $0.4$ respectively. Further, if the first group wins, the probability of introducing a new product is $0.7$ and the corresponding probability if $0.3$, if the second group wins. Find the probability that the new product introduced was by the second group.

Answer

Given: $P (G_1) = 0.6, P (G_2) = 0.4$
Let $P$ denotes the launching of new product.
$\therefore P(P|G_1) = 0.7, P(P|G_2) = 0.3$
$P(G_1|P) =\frac{{P\left( {{G_2}} \right)P\left( {P|{G_2}} \right)}}{{P\left( {{G_1}} \right)P\left( {P|{G_1}} \right) + P\left( {{G_2}} \right)P\left( {P|{G_2}} \right)}}$ = $\frac{{0.4 \times 0.3}}{{0.6 \times 0.7 + 0.4 \times 0.3}} = \frac{{12}}{{54}} = \frac{2}{9}$

View full question & answer→

Question 62 Marks

An insurance company insured $2000$ scooter drivers, $4000$ car drivers and $6000$ truck drivers. The probability of an accidents are $0.01, 0.03$ and $0.15$ respectively. One of the insured persons meet with an accident. What is the probability that he is a scooter driver?

Answer

$E_1 :$ Insured person is a scooter driver
$E_2:$ Insured person is a car driver
$E_3 :$ Insured person is a truck driver
$P(E_1) = \frac{{2000}}{{2000 + 4000 + 6000}} = \frac{{2000}}{{12000}} = \frac{1}{6}$
$P(E_2) = \frac{1}{3}$
$P(E_3) = \frac{1}{2}$
Let A insured per son meets with an accident
$P\left( {\frac{A}{{{E_1}}}} \right) = 0.01 = \frac{1}{{100}}$
$P\left( {\frac{A}{{{E_2}}}} \right) = 0.03 = \frac{3}{{100}}$
$P\left( {\frac{A}{{{E_3}}}} \right) = 0.15 = \frac{{15}}{{100}}$
$P\left( {\frac{A}{{{E_1}}}} \right) = \frac{{P({E_1})P\left( {\frac{A}{{{E_1}}}} \right)}}{{P({E_1})P\left( {\frac{A}{{{E_1}}}} \right) + P({E_2})P\left( {\frac{A}{{{E_2}}}} \right) + P({E_3})P\left( {\frac{A}{{{E_3}}}} \right)}}$
$ = \frac{{\frac{1}{6} \times \frac{1}{{100}}}}{{\frac{1}{6} \times \frac{1}{{100}} + \frac{1}{3} \times \frac{3}{{100}} + \frac{1}{2} \times \frac{{15}}{{100}}}}$
$ = \frac{1}{{52}}$

View full question & answer→

Question 72 Marks

There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads $75\%$ of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Answer

Let $E_1$: Two headed coin,
$E_2$ : Biased coin
and $E_3$ : Unbiased coin.
Let A : Coin shows head.
Then $P (E_1) = P (E_2) = P (E_3) = \frac{1}{3}$
As the two-headed coin has head on both sides so it will show head only.
Also P($\frac{A}{E_1}$) = P (correct answer given that he knows) = 1
And P($\frac{A}{E_2}$) = P (coin shows head given that the coin is biased) = 75% = $\frac{75}{100}=\frac{3}{4}$
And P($\frac{A}{E_3}$) = P (coin shows head given that the coin is unbiased) = $\frac{1}{2}$
Now the probability that the coin is two headed, being given that it shows head, is P($\frac{E_1}{A}$).
By using Bayes’ theorem, we have:
$\mathrm{P}\left(\mathrm{E}_{1} | \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} | \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} | \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{A} | \mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \cdot \mathrm{P}\left(\mathrm{A} | \mathrm{E}_{3}\right)}$
= $\frac{\frac{1}{3} \cdot 1}{\frac{1}{3} \cdot 1+\frac{1}{3} \cdot \frac{3}{4}+\frac{1}{3} \cdot \frac{1}{2}}$
= $\frac{\frac{1}{3}}{\frac{1}{3}\left(1+\frac{3}{4}+\frac{1}{2}\right)}$
= $\frac{\frac{1}{9}}{\frac{9}{4}}=\frac{4}{9}$
$\therefore $ P($\frac{E_1}{A}$) = $\frac{4}{9}$

View full question & answer→

Question 82 Marks

In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac {1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac{1}{4}$. What is the probability that the student knows the answer given that he answered it correctly?

Answer

Let $E_1$ : the student knows the answer, then, $P(E_1) = \frac{3}{4}$
$E_2$ : the student guesses the answer, then, $P(E_2) = \frac{1}{4}$
Let A: the answer is correct.
$P(\frac {A}{{E_1}}) $ = 1, $P(\frac {A}{{E_2}}) = \frac{1}{4}$
Hence, by Baye's theorm, we have,
$P(\frac {E_1}{A}) = \frac{{P({E_1})P(A/{E_1})}}{{P({E_1})P(A/{E_1}) + P({E_2})P(A/{E_2})}}$
$ = \frac{{1 \times \frac{3}{4}}}{{1 \times \frac{3}{4} + \frac{1}{4} \times \frac{1}{4}}} = \frac{{12}}{{13}}$

View full question & answer→

Question 92 Marks

Suppose a girl throws a die. If she gets a $5$ or $6$, she tosses a coin three times and notes the number of heads. If she gets $1, 2, 3$ or $4$, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw $1, 2, 3$ or $4$ with the die?

Answer

$E_1 : 1, 2, 3, 4$ is shown on dice
$E_2 : 5$ or $6$ is shown on dice
$P(E_1) = \frac{4}{6} = \frac{2}{3}, P(E_2) = \frac{2}{6} = \frac{1}{3}$
Let, A exactly one head shown up
$P(\frac {A}{{E_1}}) = \frac{1}{2},P(\frac {A}{{E_2}}) = \frac{3}{8}$
$P(\frac {{E_1}}{A}) = \frac{{P({E_1})P(A/{E_1})}}{{P({E_1})P(A/{E_1}) + P({E_2})P(A/{E_2})}} = \frac{8}{{11}}$

View full question & answer→

Question 102 Marks

If A and B are two events such that P (A) = $\frac{1}{4}$, P (B) = $\frac{1}{2}$ and $P\left( {A \cap B} \right) = \frac{1}{8}$, find P(not A and not B).

Answer

P (A) = $\frac{1}{4}$, P (B) = $\frac{1}{2},P(A \cap B) = \frac{1}{8}$
P (not A) = 1 – P (A) = $1 - \frac{1}{4} = \frac{3}{4}$
P (not B) = 1 – P (B) = $1 - \frac{1}{2} = \frac{1}{2}$
Now, P (A). P (B) $ = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$
$\therefore P(A \cap B)$ = P (A).P (B)
Thus, A and B are independent events.
Therefore, ‘not A’ and ‘not B’ are independent events.
Hence, P (not A and not B) = P (not A). P (not B)
$ = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8}$

View full question & answer→

Question 112 Marks

Let E and F be events with P (E) = $\frac{3}{5}$, P (F) = $\frac{3}{{10}}$ and $P(E \cap F)$ = $\frac{1}{5}$. Are E and F independent?

Answer

Given, P (E) = $\frac{3}{5}$ and P (F) = $\frac{3}{{10}}$
$\therefore P\left( {E \cap F} \right) = \frac{1}{5}$
Now P (E). P (F) = $\frac{3}{5} \times \frac{3}{{10}} = \frac{9}{{50}}$
$ \Rightarrow P\left( {E \cap F} \right) \ne $ P (E).P (F)
Therefore, E and F are not independent events.

View full question & answer→

Question 122 Marks

A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event the number is even, and B be the event the number is red. Are A and B independent?

Answer

S = $\left\{ {\frac{{1,2,3}}{{red}}\frac{{4,5,6}}{{green}}} \right\}$ $ \Rightarrow n(s)$ = 6
The number is even = A = {2, 4, 6} $ \Rightarrow n(A)$ = 3
P (A) = $\frac{{n\left( A \right)}}{{n\left( S \right)}} = \frac{3}{6} = \frac{1}{2}$
The number is red = B = $\left\{ {\frac{{1,2,3}}{{red}}} \right\} \Rightarrow n(B)$ = 3
P (B) = $\frac{{n(B)}}{{n(B)}} = \frac{3}{6} = \frac{1}{2}$
$\left( {A \cap B} \right)$ = {2} $ \Rightarrow $$n\left( {A \cap B} \right)$ = 1
$\therefore $ $\frac{{n(A \cap B)}}{{n(s)}} = \frac{1}{6}$
$\Rightarrow$ P(A $\cap$ B) = $\frac{1}{6}$
Now P (A). P (B) = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
Therefore, $P(A \cap B) \ne $ P (A). P (B), i.e., A and B are not independent.

View full question & answer→

Question 132 Marks

A Box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, other wise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Answer

Let A, B, and C be the respective events that the first, second, and third drawn oranges good.
Therefore,probability that first drawn orange is good = P(A) = $\frac{12}{15}$.
The oranges are not replaced. Therefore,probability of getting second orange good = P(B) = $\frac{11}{14}$ .
Similarly, probability of getting third orange good = P(C) = $\frac{10}{13}$
The box is approved for sale if all the three oranges are good.
$\therefore$ Required probability $ = \frac{{12}}{{15}} \times \frac{{11}}{{14}} \times \frac{{10}}{{13}}$$=\frac{44}{91}$
$$

View full question & answer→

Question 142 Marks

Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Answer

S = 52 cards
$ \Rightarrow $ n(s) = 52
Two cards are drawn without replacement.
A = {x: x is a black card}
$ \Rightarrow $ n(A) = 26
P (A) = $\frac{{26}}{{52}}$
And P (B) i.e., probability that second card is black known that first card is black = $\frac{{25}}{{51}}$
P (A and B) = P (A). P (B) = $\frac{{26}}{{52}} \times \frac{{25}}{{51}} = \frac{1}{2} \times \frac{{25}}{{51}} = \frac{{25}}{{102}}$.

Which is the required solution.

View full question & answer→

Question 152 Marks

A die is tossed thrice. Find the probability of getting an odd number at least once.

Answer

S = {1, 2, 3, 4, 5, 6} $\Rightarrow $ n(s) = 6
Let A represents an odd number.
$\therefore $ A = {1, 3, 5} $\Rightarrow $ n(A) = 3
$P (A) = \frac{{n(A)}}{{n(S)}} = \frac{3}{6} = \frac{1}{2}$
${\text{P}}\left( {\bar A} \right) = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}$
n = 3
Now P (at least one success) = 1 - P (an odd number on none of the three dice)
$= 1 - {\left( {\frac{1}{2}} \right)^3} = 1 - \frac{1}{8} = \frac{7}{8}$

View full question & answer→

Question 162 Marks

Events A and B are such that P(A) = $\frac{1}{2}$, P(B) = $\frac{7}{{12}}$ and P(not A or not B) = $\frac{1}{4}$. State whether A and B are independent?

Answer

$P(\overline A \cup \overline B ) = 1 - P(A \cap B) \Rightarrow \frac{1}{4} = 1 - P(A \cap B)$
$P\left( {\bar A \cap \bar B} \right) = 1 - \frac{1}{4} = \frac{3}{4}$
P (A) = $\frac{1}{2}$ and P (B) = $\frac{1}{{12}}$
P (A). P (B) = $\frac{1}{2} \times \frac{7}{{12}} = \frac{7}{{24}}$
Therefore, $P\left( {A \cap B} \right) \ne $ $P(A) \cdot P(B)$, i.e., A and B are not independent.

View full question & answer→

Question 172 Marks

If P (A) = $\frac{3}{5}$ and P (B) = $\frac{1}{5}$ find P ($A \cap B$) if A and B are independent events.

Answer

As, A and B are independent events.

Therefore, $P(A \cap B)$ = P (A). P (B)= $\frac{3}{5} \times \frac{1}{5} = \frac{3}{{25}}$

View full question & answer→

Question 182 Marks

Determine P(E|F): Mother, father and son line up at random for a family picture E: Son on one end, F: Father in middle

Answer

Let m,f and s denote respectively the mother,the father and the son,then sample space S is
S = {mfs, msf, fms, fsm, smf, sfm}
E = {mfs, fms, smf, sfm}
F = {mfs, sfm}
$E \cap F $ = { mfs,sfm}
$P\left( {\frac{E}{F}} \right) = \frac{{P\left( {E \cap F} \right)}}{{P(F)}} = \frac{{\frac{2}{6}}}{{\frac{2}{6}}} = 1$

View full question & answer→

Question 192 Marks

Determine P(E|F): A die is thrown three times, E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses.

Answer

Since a dice has six faces. Therefore $n\left( S \right)$ $ = 6 \times 6 \times 6 = 216$
E = (1, 2, 3, 4, 5, 6) x (1, 2, 3, 4, 5, 6) x (4)
F = (6) x (5) x (1, 2, 3, 4, 5, 6)
$ \Rightarrow n\left( F \right)$ = 1 x 1 x 6 = 6
P (F) = $\frac{{n\left( F \right)}}{{n\left( S \right)}} = \frac{6}{{216}}$
$\therefore E \cap F$ = (6, 5, 4)
$n\left( {E \cap F} \right) $ = 1
$\therefore P\left( {E \cap F} \right) = \frac{{n\left( {E \cap F} \right)}}{{n\left( S \right)}} = \frac{1}{{216}}$
And $P\left( {E|F} \right) = \frac{{P\left( {E \cap F} \right)}}{{P\left( F \right)}} = \frac{{1/216}}{{6/216}} = \frac{1}{6}$

View full question & answer→

Question 202 Marks

Evaluate ${P\left( {A \cup B} \right)}$ if 2P(A) = P (B) = $\frac{5}{{13}}$ and $P\left( {A|B} \right) = \frac{2}{5}$

Answer

Given: 2 P(A) = P (B) = $\frac{5}{{13}}$, $P\left( {A|B} \right) = \frac{2}{5}$
$\therefore $ P (A) = $\frac{5}{{26}}$
Now, $P\left( {A \cap B} \right) = P\left( {B|A} \right)$ . P (B) = $\frac{2}{5} \times \frac{5}{{13}} = \frac{2}{{13}}$
and $P\left( {A \cup B} \right)$ = P(A) + P (B) – $P\left( {A \cap B} \right)$ = $\frac{5}{{26}} + \frac{5}{{13}} - \frac{2}{{13}} = \frac{{11}}{{26}}$

View full question & answer→

Question 212 Marks

Compute P(A|B), if P (B) = 0.5 and $P\left( {A \cap B} \right)$ = 0.32

Answer

Given: P (B) = 0.5, $P\left( {A \cap B} \right)$ = 0.32
$P\left( {A|B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} = \frac{{0.32}}{{0.50}} = \frac{{32}}{{50}} $ = 0.64

View full question & answer→

Question 222 Marks

Consider the experiment of throwing a die, if a multiple of 3 comes up throw the die again and if any other number comes toss a coin. Find the conditional probability of the event the coin shows a tail, given that at least one die shows a 3.

Answer

S = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
(1, H), (2, H), (3, H), (4, H), (5, H), (1, T), (2, T), (3, T), (4, T), (5, T)}
$\therefore $ n(S) = 20
P (first die shows a multiple of 3) = $\frac{{12}}{{36}} = \frac{1}{3}$
P (first die shows a number which is not a multiple of 3) = $\frac{4}{6} \times \frac{1}{2} + \frac{4}{6} \times \frac{1}{2} = \frac{8}{{12}} = \frac{2}{3}$
Let A = the coin shows a tail = {(1, T), (2, T), (4, T), (5, T)}
B = at least one die shows a 3 = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}
$A \cap B = \phi $
n(A) = 4, n(B) = 6, $n\left( {A \cap B} \right) $ = 0
P(B) = $\frac{6}{{36}} = \frac{1}{6}$ and $P\left( {A \cap B} \right)$ = 0
$P\left( {A|B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} = \frac{0}{{\frac{6}{{36}}}} = 0$

View full question & answer→

Question 232 Marks

Given that the two numbers appearing on throwing two dice are different. Find the probability of the event the sum of numbers on the dice is 4.

Answer

S = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)
(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2)
(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3)
(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4)
(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)
(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)}
$\therefore $ n(S) = 36
Let A represents the event “the sum of numbers on the dice is 4” and B represents the event “the two numbers appearing on throwing two dice are different”.
Therefore, A = {(1, 3), (2, 2), (3, 1)} $ \Rightarrow $ n(A) = 3
P (A) = $\frac{{n(A)}}{{n(S)}} = \frac{3}{{36}}$
Also B = {(2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2),(1, 3),
(2, 3), (4, 3), (5, 3), (6, 3), (1, 4), (2, 4), (3, 4), (5, 4), (6, 4),(1, 5), (2, 5),
(3, 5), (4, 5), (6, 5), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6)}
$\therefore$ n(B) = 30
P (B) = $\frac{{n\left( B \right)}}{{n\left( S \right)}} = \frac{{30}}{{36}}$
Now $A \cap B$ = {(1, 3), (3, 1)}
$ \Rightarrow $ $n\left( {A \cap B} \right)$ = 2
$P(A \cap B)$ = $\frac{{n\left( {A \cap B} \right)}}{{n\left( S \right)}} = \frac{2}{{36}}$
Hence, $P\left( {A|B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$
$= \frac{{\frac{2}{{36}}}}{{\frac{{30}}{{36}}}} = \frac{2}{{30}} = \frac{1}{{15}}$

View full question & answer→

Question 242 Marks

An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Answer

Let E: it is an easy question
F: it is a multiple choice question.
Total number of questions = 300 + 200 + 500 + 400 = 1400
$P(E\cap F)=\frac{500}{1400}$ and P(F) = $\frac{500+400}{1400}=\frac{9}{14} $
Hence, required probability = $ P\left( {\frac{E}{F}} \right)$ =$\frac{P(E\cap F)}{P(F)}=\frac{5/14}{9/14}=\frac{5}{9}$

View full question & answer→

Question 252 Marks

Given that E and F are events such that P (E) = 0.6, P (F) = 0.3 and P$({E \cap F})$ = 0.2, find P(E|F) and P(F|E).

Answer

Given: P (E) = 0.6, P (F) = 0.3 and P$({E \cap F})$ = 0.2

$\therefore $ $P\left( {E|F} \right) = \frac{{P\left( {E \cap F} \right)}}{{P\left( F \right)}} = \frac{{0.2}}{{0.3}} = \frac{2}{3}$

And $P\left( {F|E} \right) = \frac{{P\left( {E \cap F} \right)}}{{P\left( E \right)}} = \frac{{0.2}}{{0.6}} = \frac{1}{3}$

View full question & answer→

Generate Paper Free All Probability topics