MCQ 11 Mark
If A and B are two events such that $\text{P(A)}=\frac{3}{8},\text{P(B)}=\frac{5}{4}.$ and $\text{P}(\text{A}|\text{B})\times\text{P}(\overline{\text{A}}\cap\text{B})$ is equals to.
- A
$\frac{2}{5}$
- B
$\frac{3}{8}$
- C
$\frac{3}{20}$
- ✓
$\frac{6}{25}$
AnswerCorrect option: D. $\frac{6}{25}$
$\text{P(A)}=\frac{3}{8},\text{P(B)}=\frac{5}{8},\text{P}\Big({\text{A}}\cup{\text{B}}\Big)=\frac{3}{4}$
$\text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \frac{3}{4}=\frac{3}{8}+\frac{5}{8}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{1}{4}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}\times\frac{\text{P}(\overline{\text{A}}\cap\text{B})}{\text{P(B)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}\times\frac{\text{P(B)}-\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{\frac{1}{4}}{\frac{5}{8}}\times\frac{\big(\frac{5}{8}-\frac{1}{4}\big)}{\frac{5}{8}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{6}{25}$
View full question & answer→MCQ 21 Mark
A bag contains 5 brown and 4 white socks. A man pulls out two socks. The probability that these are of the sane colour is.
- A
$\frac{5}{108}$
- B
$\frac{18}{108}$
- ✓
$\frac{30}{108}$
- D
$\frac{48}{108}$
AnswerCorrect option: C. $\frac{30}{108}$
Total number of balls = 5brown + 4white = 9
Required probability $=\frac{5}{9}\times\frac{4}{8}+\frac{4}{9}\times\frac{3}{8}=\frac{4}{9}$
$\Rightarrow\ \frac{4\times12}{9\times12}=\frac{48}{108}$
View full question & answer→MCQ 31 Mark
An urn contains 9 balls two of which are red, three blue and four black. Three balls are drawn at random. The probability that they are of the same colour is,
- ✓
$\frac{5}{84}$
- B
$\frac{3}{9}$
- C
$\frac{3}{7}$
- D
$\frac{7}{17}$
AnswerCorrect option: A. $\frac{5}{84}$
Given:
Red balls = 2
Blue balls = 3
Black balls = 4
P(All three balls are of same colour) = P(all three are blue) + P(all three are black)
$=\frac{3}{9}\times\frac{2}{8}\times\frac{1}{7}+\frac{4}{9}\times\frac{3}{8}\times\frac{2}{7}$
$=\frac{1}{84}+\frac{4}{84}$
$=\frac{5}{84}$
View full question & answer→MCQ 41 Mark
A die is thrown and a card is selected ar random from a deck $\text{pf}\ 52$ playing cards. The probability of getting an even number of the die and a spade card is
- A
$\frac{1}{2}$
- B
$\frac{1}{4}$
- ✓
$\frac{1}{8}$
- D
$\frac{3}{4}$
AnswerCorrect option: C. $\frac{1}{8}$
A Sample space when a die is thrown,
$S_1 = \{1, 2, 3, 4, 5, 6\} $
$\Rightarrow n(S_1) = 6$
Let $A$ be the event that getting even number.
$A = \{2, 4, 6\} $
$\Rightarrow n(A) = 3$
$\Rightarrow\ \text{P(A)}=\frac{3}{6}=\frac{1}{2}$
A card is selected from a deck of $52$ cards.
$\text{n}(\text{S}_2)= {^{52}}\text{C}_2=52$
Let $B$ be the event that getting spade card.
$\text{n(B)}= {^{13}}\text{C}_2=13$
$\Rightarrow\ \text{P(B)}=\frac{13}{52}=\frac{1}{4}$
Required probability $= P(A) \times P(B)$
$=\frac{1}{2}\times\frac{1}{4}=\frac{1}{8}$
View full question & answer→MCQ 51 Mark
Five persone entered the lift cabin on the ground floor of an $8$ floor house. Suppose that each of them independently and with equal probability can leave the cabin at any flor beginning with the first, then the probability of all $5$ persons leaving at different floors is,
- ✓
$\frac{^{7}\text{P}_5}{7_5}$
- B
$\frac{7_5}{^{7}\text{P}_5}$
- C
$\frac{6}{^{6}\text{P}_5}$
- D
$\frac{^{5}\text{P}_5}{5}$
AnswerCorrect option: A. $\frac{^{7}\text{P}_5}{7_5}$
Five persons can leave different floors
By $^7P_5$ ways.
Possible ways of leavinf the lift $= 7^5$
Required probability $=\frac{^{7}\text{P}_5}{7^5}$
View full question & answer→MCQ 61 Mark
Three persons, A, B and C fine a target in turn starting with A. Their probability of hitting the target are 0.4, 0.2 and 0.2, respectively. The probability of two hits is
AnswerLet:
A be the event of hitting the target by the person A,
B be the event of hitting the target by the person B and
C be the event of hitting the target by the person C
We have,
P(A) = 0.4, P(B) = 0.3 and P(C) = 0.2
Also,
$\text{P}(\overline{\text{A}})=1-\text{P(A)}=1-0.4=-0.6,$
$\text{P}(\overline{\text{B}})=1-0.3=0.7$ and
$\text{P}(\overline{\text{C}})=1-0.2=0.8$
Now,
$\text{P(Two hits)}=\text{P}(\text{AB}\overline{\text{C}})+\text{P}(\text{A}\overline{\text{B}}\text{C})+\text{P}(\overline{\text{A}}\text{BC})$
$=\text{P(A)}\times\text{P(B)}\times\text{P}(\overline{\text{C}})+\text{P(A)}\times(\overline{\text{B}})\times\text{P(C)}\\+\text{P}(\overline{\text{A}})\times\text{P(B)}\times\text{P(C)}$
$=0.4\times0.3\times0.8+0.4\times0.7\times0.2+0.6\times0.3\times0.2$
$=0.096+0.056+0.036$
$=0.188$
Hence, the correct alternative is option (d).
View full question & answer→MCQ 71 Mark
If two events are independent, then.
- A
They must be mutually exclusive.
- B
The sum of their probabilities must be equal to 1.
- C
(a) and (b) both are correct.
- ✓
None of the above is correctIf two. events are independent, then.
AnswerCorrect option: D. None of the above is correctIf two. events are independent, then.
Let A and B are two independent events, Then,
$\text{P}(\text{A}\cap\text{B})=\text{P(A)}\times\text{P(B)}$
As, $\text{P}(\text{A}\cap\text{B})\neq0\text{ or }\text{P(A)}+\text{P(B)}\neq1$
So, both are neither mutually exclisive nor their sum of probability is 1.
Hence, the correct alternative is option (d).
View full question & answer→MCQ 81 Mark
$A$ and $B$ are two students. Their chances of solving a problem correctly are $\frac{1}{3}$ and $\frac{1}{4}$ respectively. If the probability of their making common error is $\frac{1}{20}$ and they obtain the same answer, then the probability of their answer to be correct is.
- ✓
$\frac{10}{13}$
- B
$\frac{13}{120}$
- C
$\frac{1}{40}$
- D
$\frac{1}{12}$
AnswerCorrect option: A. $\frac{10}{13}$
Let $E_1$ be the event that Both $A$ and $B$ solve the problem.
$A$ and $B$ are independent,
$\Rightarrow\ \text{P}(\text{E}_1)=\text{P(A)}\times\text{P(B)}$
$\Rightarrow\ \text{P}(\text{E}_1)=\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}$
Let $E_2$ both $A$ and $B$ got wrong solution.
$\text{P}(\text{E}_2)=\Big(1-\frac{1}{3}\Big)\times\Big(1-\frac{1}{4}\Big)=\frac{1}{2}$
Let $E$ be the event getting same answer.
$\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=1,\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=\frac{1}{20}$
$\Rightarrow\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)=\frac{\frac{1}{12}\times1}{\frac{1}{12}\times1+\frac{1}{2}\times\frac{1}{20}}=\frac{10}{13}$
View full question & answer→MCQ 91 Mark
If A and B are two events such that $\text{P(A)}\neq0$ and $\text{P(B)}\neq1,$ then $\text{P}(\overline{\text{A}}|\overline{\text{B}})=$
- A
$1-\text{P}(\text{A}|\text{B})$
- B
$1-\text{P}(\overline{\text{A}}|\text{B})$
- ✓
$\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\overline{\text{B}})}$
- D
$=\frac{\text{P}(\overline{\text{A}})}{\text{P}(\overline{\text{B}})}$
AnswerCorrect option: C. $\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\overline{\text{B}})}$
We have,
$\text{P(A)}\neq0$ and $\text{P(B)}\neq1$
Now,
$\text{P}(\overline{\text{A}}|\overline{\text{B}})=\frac{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}{\text{P}(\overline{\text{B}})}$
$=\frac{\text{P}(\overline{\text{A}\cap\text{B}})}{\text{P}(\overline{\text{B}})}$
$=\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\overline{\text{B}})}$
Hence, the correct alternative is option (C).
View full question & answer→MCQ 101 Mark
If A and B are such that $\text{P}(\text{A}\cup\text{B})=\frac{5}{9}$ and $\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3},$ then $\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=$
- A
$\frac{9}{10}$
- ✓
$\frac{10}{9}$
- C
$\frac{8}{9}$
- D
$\frac{9}{8}$
AnswerCorrect option: B. $\frac{10}{9}$
$\text{P}(\text{A}\cup\text{B})=\frac{5}{9},\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3}$
Consider,
$\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$\Rightarrow\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3}$
$\Rightarrow 1-\text{P}(\text{A}\cap\text{B})=\frac{2}{3}$
$\Rightarrow\text{P}(\text{A}\cap\text{B})=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}-\frac{5}{9}=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}=\frac{8}{9}$
$\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=1-\text{P(A)}+1-\text{P(B)}$
$\Rightarrow\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=2-\big[\text{P(A)}+\text{P(B)}\big]$
$\Rightarrow\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=2-\frac{8}{9}$
$\Rightarrow\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=\frac{10}{9}$
View full question & answer→MCQ 111 Mark
In a college 30% students fail in Physics, 25% fail in Mathenatics and 10% fail in both. One student is chosen at random. The probability that she fails in Physics if she failed in Mathematics is.
- A
$\frac{1}{10}$
- B
$\frac{1}{3}$
- ✓
$\frac{2}{5}$
- D
$\frac{9}{20}$
AnswerCorrect option: C. $\frac{2}{5}$
Let A be the event that students failed in Physics. B be the event that students failed in Mathematics.Given that, $\text{P(A)}=30\%=\frac{30}{100}$
$\text{P(B)}=25\%=\frac{25}{100}$
$\text{P}(\text{A}\cap\text{B})=10\%=\frac{10}{100}$
Required probability is given by $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$
$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{\frac{10}{100}}{\frac{25}{100}}=\frac{2}{5}$
View full question & answer→MCQ 121 Mark
If $\text{P(B)}=\frac{3}{5},\text{P}(\text{A}|\text{B})=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then $\text{P}(\overline{\text{A}\cap\text{B}})+\text{P}(\overline{\text{A}}\cap\text{B})=$
- A
$\frac{1}{5}$
- B
$\frac{4}{5}$
- C
$\frac{1}{2}$
- ✓
$1$
Answer$\text{P(B)}=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2},\text{P}\Big({\text{A}}\cup{\text{B}}\Big)=\frac{4}{5}$
Consider,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$
$\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{1}{2}$
$\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\frac{3}{5}}=\frac{1}{2}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{3}{10}$
$\text{P}(\overline{\text{A}\cup\text{B}})+\text{P}(\overline{\text{A}}\cap\text{B})$
$=1-\text{P}(\text{A}\cap\text{B})+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=1-\frac{3}{10}+\frac{3}{5}-\frac{3}{10}$
$=1$
View full question & answer→MCQ 131 Mark
Two persons $A$ and $B$ take turns in throwing a pair of dice.The first person to throw $9$ from both dice will be awarded the prize. If $A$ throws first, then the probability that $B$ wins the game is.
- A
$\frac{9}{17}$
- ✓
$\frac{8}{17}$
- C
$\frac{8}{9}$
- D
$\frac{1}{9}$
AnswerCorrect option: B. $\frac{8}{17}$
$9$ can be obtained from throw of two dice in only $4$ cases as given below:
$[(3, 6), (4, 5), (5, 4), (6, 3)]$
$\Rightarrow\ \text{P(getting }9)=\frac{4}{36}=\frac{1}{9}$
$\text{P(not getting }9)=\frac{32}{36}=\frac{8}{9}$
Now,
$P(B$ is winning$) = P($getting $9$ in $2^{nd}$ throw$) + P($getting $p$ in $4^{th}$ throw$) + P($getting $9$ in $6^{th}$ throw$) + .....$
$=\frac{8}{9}\times\frac{1}{9}+\frac{8}{9}\times\frac{8}{9}\times\frac{8}{9}\times\frac{1}{9}+\ .....$
$=\frac{8}{81}\Big[1+\frac{64}{81}+\Big(\frac{64}{81}\Big)^2+\ ......\Big]$
$=\frac{8}{81}\times\frac{1}{1-\frac{64}{81}}$
$=\frac{8}{81}\times\frac{81}{17}$
$=\frac{8}{17}$
View full question & answer→MCQ 141 Mark
If $\text{P}(\text{A}\cup\text{B})=0.8$ and $\text{P}(\text{A}\cap\text{B})=0.3$ then $\text{P}(\overline{\text{A}})=\text{P}(\overline{\text{B}})=$
AnswerIf $\text{P}(\text{A}\cup\text{B})=0.8\text{ P}(\text{A}\cap\text{B})=0.3,$
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P(A)}+\text{P(B)}=\text{P}(\text{A}\cup\text{B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P(A)}+\text{P(B)}=1.1$
$\Rightarrow\ \text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=1-\text{P(A)}+1-\text{P(B)}$
$\Rightarrow\ \text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=\big[\text{P(A)}+\text{P(B)}\big]$
$\Rightarrow\ \text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=2-1.1$
$\Rightarrow\ \text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=0.9$
View full question & answer→MCQ 151 Mark
A bag $X$ contains $2$ white and $3$ black balls and another bag $Y$ contains $4$ white and $2$ black balls. One bag is selected at random and a ball is drawn from it. Then, the probability chosen to be white is,
- A
$\frac{2}{15}$
- B
$\frac{7}{15}$
- ✓
$\frac{8}{15}$
- D
$\frac{14}{15}$
AnswerCorrect option: C. $\frac{8}{15}$
$\frac{8}{15}$
View full question & answer→MCQ 161 Mark
Let A and B be two events such that P(A) = 0.6, P(B) = 0.2, P(A|B) = 0.5. Then $\text{P}(\overline{\text{A}}|\overline{\text{B}})$ equals.
- A
$\frac{1}{10}$
- B
$\frac{3}{10}$
- ✓
$\frac{3}{8}$
- D
$\frac{6}{7}$
AnswerCorrect option: C. $\frac{3}{8}$
Given that,
$\text{P(A)}=0.6,\text{P(B)}=0.2,\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.5$
Consider,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.5$
$\Rightarrow\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=0.5$
$\Rightarrow\frac{\text{P}(\text{A}\cap\text{B})}{0.2}=0.5$
$\Rightarrow \text{P}(\text{A}\cap\text{B})=0.1$
$\Rightarrow \text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=0.1$
$\Rightarrow \text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-0.1$
$\Rightarrow\text{P}(\text{A}\cup\text{B})=0.7$
Now, $\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\text{P}(\text{A}\cup\text{B})$
$\Rightarrow \text{P}(\overline{\text{A}}\cap\overline{\text{B}})=0.3$
To find
$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)=\frac{\text{P}(\overline{\text{A}\cap\text{B}})}{\text{P}(\overline{\text{B}})}$
$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)=\frac{0.3}{0.8}$
$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)=\frac{3}{8}$
View full question & answer→MCQ 171 Mark
If A and B are two events, then $\text{P}(\overline{\text{A}}\cap\text{B})=$
- A
$\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{B}})$
- B
$1-\text{P}(\text{A})-\text{P}(\text{B})$
- C
$\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
- ✓
$\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
AnswerCorrect option: D. $\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
From the diagram, we get $\text{A}\cap\text{B}$ and $\overline{\text{A}}\cap\text{B}$ are mutually exclusive events such that $(\text{A}\cap\text{B})\cup(\overline{\text{A}}\cap\text{B})=\text{B}.$ therefore by
$\text{P}(\text{A}\cap\text{B})+\text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}$
$\therefore\ \text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}-\text{P}(\text{A}\cap\text{B})$ View full question & answer→MCQ 181 Mark
The probability that in a year of $22^{nd}$ century chosen at random, there will be $53$ Sunday, is
- A
$\frac{3}{28}$
- B
$\frac{2}{28}$
- C
$\frac{7}{28}$
- ✓
$\frac{5}{28}$
AnswerCorrect option: D. $\frac{5}{28}$
We know a leap year is fallen within $4$ years,
So its probability is $\frac{25}{100}=\frac{1}{4}$
$53^{rd}$ Sunday leap year $=\frac{1}{4}\times\frac{2}{7}=\frac{2}{28}$
Similarly probability of $53^{rd}$ Sunday in a non leap year $=\frac{75}{100}\times\frac{1}{7}=\frac{3}{4}\times\frac{1}{7}=\frac{3}{28}$
Required probability $=\frac{2}{28}+\frac{3}{28}=\frac{5}{28}$.
View full question & answer→MCQ 191 Mark
India play two matches each with West indies and Australia. In any match the probability of india getting $0,1$ and $2$ points are $0.45, 0.05$ and $0.50$ respectively. Assuming that the outcomes are indepecdent, the probability of india getting at least $7$ points is.
- ✓
$0.0875$
- B
$\frac{1}{16}$
- C
$0.1125$
- D
AnswerCorrect option: A. $0.0875$
Here, there are total $5$ ways by which India can get at least $7$ points.
- $2$ points $+ 2$ points $+ 2$ points $+ 2$ points $= (0.5 \times 0.5 \times 0.5 \times 0.5)$
- $1$ points $+ 2$ points $+ 2$ points $+ 2$ points $= (0.05 \times 0.5 \times 0.5 \times 0.5)$
- $2$ points $+ 1$ points $+ 2$ points $+ 2$ points $= (0.5 \times 0.05 \times 0.5 \times 0.5)$
- $2$ points $+ 2$ points $+ 1$ points $+ 2$ points $= (0.5 \times 0.5 \times 0.05 \times 0.5)$
- $2$ points $+ 2$ points $+ 2$ points $+ 1$ points $= (0.5 \times 0.5 \times 0.5 \times 0.05)$
View full question & answer→MCQ 201 Mark
A box contains 3 orange balls, 3 green balls and 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 22 green balls and one blue ball is
- A
$\frac{167}{168}$
- B
$\frac{1}{28}$
- C
$\frac{2}{21}$
- ✓
$\frac{3}{28}$
AnswerCorrect option: D. $\frac{3}{28}$
Total balls in a box - 3orange + 3green + 2blue = 8
Three balls are drawn at random from the box then samplw space $\text{n(S)}= {^{8}}\text{C}_3=\frac{8\times7\times6}{3\times2\times1}=56$
Let A be the event that drawing 2 green and one blue ball.
$\text{n(A)}={^{3}}\text{C}_2\times{^{2}}\text{C}_2=6$
$\text{P(A)}=\frac{6}{56}=\frac{3}{28}$
View full question & answer→MCQ 211 Mark
A box contains 6 nails and 10 nuts. Half of the nails and half of the nuts are rusted. If one iten is chosen ar random, the probability that it is rusted or is nail is
- A
$\frac{3}{16}$
- B
$\frac{5}{16}$
- ✓
$\frac{11}{16}$
- D
$\frac{14}{16}$
AnswerCorrect option: C. $\frac{11}{16}$
Rusted items = 3 + 5 = 8
Rusted nails = 3
Total nails = 6
P(getting a rusted item or a nail) = P(getting a rusted item) + P(getting a nail) - P(getting a rusted item and a nail)
$=\frac{8}{16}+\frac{6}{16}-\frac{3}{16}$
$=\frac{8+6-3}{16}$
$=\frac{11}{16}$
View full question & answer→MCQ 221 Mark
If the events A and B are independent, then $\text{P}(\text{A}\cap\text{B})$ is equal to,
Answer$\text{P}(\text{A}\cap\text{B})=\text{P(A)} \text{ P(B)}$ for independent events.
View full question & answer→MCQ 231 Mark
If A and B are two events associated to a random experiment such that $\text{P}(\text{A}\cap\text{B})=\frac{7}{10}$ and $\text{P(B)}=\frac{17}{20}$, then P(A|B) =
- ✓
$\frac{14}{17}$
- B
$\frac{17}{20}$
- C
$\frac{7}{8}$
- D
$\frac{1}{8}$
AnswerCorrect option: A. $\frac{14}{17}$
$\text{P}(\text{A}\cap\text{B})=-\frac{7}{10},\text{P(B)}=\frac{17}{20}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{7}{10}}{\frac{17}{20}}=\frac{14}{17}$
View full question & answer→MCQ 241 Mark
If A and B are two events such that $\text{P}(\text{A}|\text{B})=\text{p},\text{P(A)}=\text{p},\text{P(B)}=\frac{1}{3}$ and $\text{P}(\text{A}\cup\text{B})=\frac{5}{9},$ then p =
- A
$\frac{2}{3}$
- B
$\frac{3}{5}$
- ✓
$\frac{1}{3}$
- D
$\frac{3}{4}$
AnswerCorrect option: C. $\frac{1}{3}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{p},\text{P(A)}=\text{p},\text{P(B)}=\frac{1}{3},\text{P}(\text{A}\cup\text{B})=\frac{5}{9}$
Consider,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{p}$
$\Rightarrow \frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\text{P}$
$\Rightarrow \frac{\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})}{\text{P(B)}}=\text{P}$
$\Rightarrow\frac{\text{p}+\frac{1}{3}-\frac{5}{9}}{\frac{1}{3}}=\text{p}$
$\Rightarrow\text{p}+\frac{1}{3}-\frac{5}{9}=\frac{\text{p}}{3}$
$\Rightarrow\frac{-2}{9}=\frac{\text{p}}{3}-\text{p}$
$=\frac{-2}{3}\text{p}=\frac{-2}{9}$
$\Rightarrow\text{p}=\frac{1}{3}$
View full question & answer→MCQ 251 Mark
A flash light has 8 batteries out of which 3 are dead. IF two batteries are selected without replacement and tested, then the probability that both are dead is,
- ✓
$\frac{3}{28}$
- B
$\frac{1}{14}$
- C
$\frac{9}{64}$
- D
$\frac{33}{56}$
AnswerCorrect option: A. $\frac{3}{28}$
We have,
The total number of batteries = 8
The number of dead batteries = 3
Let A be the event of selecting the first dead battery and B be the event of selecting the second dead battery.
Now,
P(both dead batteries are selected) $=\text{P}(\text{A}\cap\text{B})$
$=\text{P(A)}\times\text{P}(\text{B}|\text{A})$
$=\frac{3}{8}\times\frac{2}{7}$
$=\frac{3}{28}$
Hence, the correct alternative is option (a).
View full question & answer→MCQ 261 Mark
Associated to a random experiment two events A and B are such that $\text{P(B)}=\frac{3}{5},\text{P}(\text{A}|\text{B})=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5}$. The value pf P(A) is
- A
$\frac{3}{10}$
- ✓
$\frac{1}{2}$
- C
$\frac{1}{10}$
- D
$\frac{3}{5}$
AnswerCorrect option: B. $\frac{1}{2}$
$\text{P(B)}=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2},\text{P}(\text{A}\cup\text{B})=\frac{4}{5}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{1}{2}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\frac{3}{5}}=\frac{1}{2}$
$\text{P}(\text{A}\cap\text{B})=\frac{3}{10}$
$\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=\frac{3}{10}$
$\text{P(A)}+\frac{3}{5}-\frac{4}{5}=\frac{3}{10}$
$\text{P(A)}=\frac{1}{2}$
View full question & answer→MCQ 271 Mark
A box contains 10 good articles and 6 with defects. One item is drawn at random. The probability that it is either good or has a defect is,
- ✓
$\frac{64}{64}$
- B
$\frac{49}{64}$
- C
$\frac{40}{64}$
- D
$\frac{24}{64}$
AnswerCorrect option: A. $\frac{64}{64}$
P(good item) $=\frac{10}{16}$
P(defected item) $=\frac{6}{16}$
P(eitherr good or defected item) = P(good item) + P(defected item)
$=\frac{10}{16}+\frac{6}{16}$
$=\frac{16}{16}$
$=1$
$=\frac{64}{64}$
View full question & answer→MCQ 281 Mark
If A and B are two events such that $\text{P(A)}=\frac{1}{2},\text{P(B)}=\frac{1}{3},\text{P}(\text{A}|\text{B})=\frac{1}{4},$ then $\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$ equals.
- ✓
$\frac{1}{12}$
- B
$\frac{3}{4}$
- C
$\frac{1}{4}$
- D
$\frac{3}{16}$
AnswerCorrect option: A. $\frac{1}{12}$
$\text{P(A)}=\frac{1}{2},\text{P(B)}=\frac{1}{3},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{4}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{4}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{1}{4}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\frac{1}{3}}=\frac{1}{4}$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{12}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{1}{12}}{\frac{1}{3}}=\frac{1}{4}$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\text{P}(\text{A}\cup\text{B})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=-1\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\Big[\frac{1}{2}+\frac{1}{3}-\frac{1}{12}\Big]$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\frac{1}{4}$
View full question & answer→MCQ 291 Mark
A and B draw two cards each, one after another, from a pack of well-shuffled pack of 52 cards. The probability that all the four cards drawn are of the same suit is
AnswerCorrect option: A. $\frac{44}{85\times49}$
Total cards = 52 There are four suits of cards in a pack, i.e. diamond, heart, spade and club.
Pall 4 cards are of same suit = Pall 4 cards are of diamond + Pall 4 cards are of heart + Pall 4 cards are of spade + Pall 4 cards are of club.
$=4\times\frac{13}{52}\times\frac{12}{51}\times\frac{11}{50}\times\frac{10}{59}$
$=4\times\frac{11}{85\times49}$
$=\frac{44}{85\times49}$
View full question & answer→MCQ 301 Mark
If A and B are two events such that $\text{P(A)}=\frac{4}{5},$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{10},$ then P(B|A) =
- A
$\frac{1}{10}$
- B
$\frac{1}{8}$
- ✓
$\frac{7}{8}$
- D
$\frac{17}{20}$
AnswerCorrect option: C. $\frac{7}{8}$
We have,
$\text{P(A)}=\frac{4}{5}$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{10}$
Now,
$\text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$=\frac{\Big(\frac{7}{10}\Big)}{\Big(\frac{4}{5}\Big)}$
$=\frac{7\times5}{10\times4}$
$=\frac{7}{8}$
Hence, the correct alternative is option (c).
View full question & answer→MCQ 311 Mark
If A and B are two independent events such that P(A) = 0.3 and $\text{P}(\text{A}\cup\text{B})=0.5,$ then P(A|B) - P(B|A) =
- A
$\frac{2}{7}$
- B
$\frac{3}{35}$
- ✓
$\frac{1}{70}$
- D
$\frac{1}{7}$
AnswerCorrect option: C. $\frac{1}{70}$
We have,
$\text{P(A)}=0.3$ and $\text{P}(\text{A}\cup\text{B})=0.5$
As, A and B are independent events
So, $\text{P}(\text{A}\cap\text{B})=\text{P(A)}\times\text{P(B)}$
$=0.3\times\text{P(B)}$
$=0.3\text{ P(B)}\ .....\text{(i)}$
Also, $\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow 0.5 = 0.3+\text{P(B)}-0.3\text{ P(B)}$ [Using (i)]
$\Rightarrow 0.5-0.3 = 0.7\text{ P(B)}$
$\Rightarrow0.7\text{ P(B)}=0.2$
$\Rightarrow\text{ P(B)}=\frac{0.2}{0.7}$
$\Rightarrow\text{ P(B)}=\frac{2}{7}$
Using (i), we get
$\text{P}(\text{A}\cap\text{B})=0.3\times\frac{2}{7}=\frac{6}{70}$
Now,
$\text{P}(\text{A}|\text{B})-\text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}-\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$=\frac{\Big(\frac{6}{70}\Big)}{\Big(\frac{2}{7}\Big)}-\frac{\Big(\frac{6}{70}\Big)}{0.3}$
$=\frac{6\times7}{70\times2}-\frac{6}{70\times0.3}$
$=\frac{3}{10}-\frac{2}{7}$
$=\frac{21-20}{70}$
$=\frac{1}{70}$
Hence, the correct alternative is option (c).
View full question & answer→MCQ 321 Mark
A and B are two events such that P(A) = 0.25 and P(B) = 0.50. The probability pf both happening together is 0.14. The probability of both A and B hot happening is.
Answer$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=0.25+0.5-0.14$
$0.61$
P(Both A and B not happening) $=\text{P}(\text{A}\cup\text{B})'$
$=1-\text{P}(\text{A}\cup\text{B})$
$=1-0.61$
$=0.39$
View full question & answer→MCQ 331 Mark
If one ball is drawn ar random from each of three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, then the probability that 2 white and 1 black balls will be drawn is.
- ✓
$\frac{13}{32}$
- B
$\frac{1}{4}$
- C
$\frac{1}{32}$
- D
$\frac{3}{16}$
AnswerCorrect option: A. $\frac{13}{32}$
Total balls in first box = 3 white + 1 black = 4
Total balls in second box = 2 white + 2black = 4
Total balls in third box = 1white + 3black = 4
Probability of 2 white and 1 black
= P(WWB) + P(WBW) + P(BWW)
$=\frac{3}{4}\times\frac{2}{4}\times\frac{3}{4}+\frac{3}{4}\times\frac{2}{4}\times\frac{1}{4}+\frac{1}{4}\times\frac{2}{4}\times\frac{1}{4}$
$=\frac{18+6+2}{64}=\frac{13}{32}$
View full question & answer→MCQ 341 Mark
From a set of 100 cards numbered 1 to 100, one card is drawn at randow. The probability number obtained on the card is divisible by 6 or 8 but not by 24 is
- ✓
$\frac{6}{25}$
- B
$\frac{1}{4}$
- C
$\frac{1}{6}$
- D
$\frac{2}{6}$
AnswerCorrect option: A. $\frac{6}{25}$
Number of cards divisible by 6 = 16
$\Rightarrow\ \text{P(A)}=\frac{16}{100}$
Number of cards divisible by 8 = 12
$\Rightarrow\ \text{P(B)}=\frac{12}{100}$
Number of cards divisible by 24 = 4
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{4}{100}$
$\text{P}(\text{A}\cup\text{B})=\frac{16}{100}+\frac{12}{100}-\frac{4}{100}$
$\text{P}(\text{A}\cup\text{B})=\frac{6}{25}$
View full question & answer→MCQ 351 Mark
A bag containe 5 black, 4white balls and 3 red balls. if a ball is selected randomwise, the probability that it is black or red ball is,
- A
$\frac{1}{3}$
- B
$\frac{1}{4}$
- ✓
$\frac{5}{12}$
- D
$\frac{2}{3}$
AnswerCorrect option: C. $\frac{5}{12}$
We know that the bag contains 5B (black), 4W(white) and 3R(red) balls.
Now,
$\text{P(B)}=\frac{5}{12}$
$\text{P(R)}=\frac{3}{12}$
$\text{P}(\text{B or R})=\text{P(B)}+\text{P(R)}$
$=\frac{5}{12}+\frac{3}{12}$
$=\frac{8}{12}=\frac{2}{3}$
View full question & answer→MCQ 361 Mark
A coin is tossed three times. If events A and B are defined as A = Two heads come, B = Last should be head, Then, A and B are
AnswerS = [(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)]
$\text{P(A)}=\text{P}(2\text{heads})=\frac{3}{8}$
$\text{P(B)}=\text{P}(\text{last one is heads})=\frac{4}{8}$
$\text{P}(\text{A}\cap\text{B})=\frac{2}{8}=\frac{1}{4}\neq\text{P(A) P(B)}$
Thus, A and B are dependent.
View full question & answer→MCQ 371 Mark
If A and B are two independent events with $\text{P(A)}=\frac{3}{5}$ and $\text{P(B)}=\frac{4}{9},$ then $\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$ equals,
- A
$\frac{4}{15}$
- B
$\frac{8}{45}$
- C
$\frac{1}{3}$
- ✓
$\frac{2}{9}$
AnswerCorrect option: D. $\frac{2}{9}$
$\text{P(A)}=\frac{3}{5},\text{P(B)}=\frac{4}{9}$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\text{P}(\text{A}\cup\text{B})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\Big[\frac{3}{5}+\frac{4}{9}-\frac{3}{5}\times\frac{4}{9}\Big]$
($\because$ A and B are independent)
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\frac{7}{9}$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\frac{2}{9}$
View full question & answer→MCQ 381 Mark
The probability that a leap year will have 53 fridays or 53 Saturdays is.
- A
$\frac{2}{7}$
- ✓
$\frac{3}{7}$
- C
$\frac{4}{7}$
- D
$\frac{1}{7}$
AnswerCorrect option: B. $\frac{3}{7}$
Non-leap year has 365 days = 52 weeks + 1
366 days in leap year.
We want to find probability of 53 Fridays or 53 Saturday.
Favourable cases = {(Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}
Required probability $=\frac{3}{7}$
View full question & answer→MCQ 391 Mark
Two cards are drawn from a well shuffled deck of $52$ playing cards with replacement. The probability that both cards are queen is
- ✓
$\frac{1}{13}\times\frac{1}{13}$
- B
$\frac{1}{13}+\frac{1}{13}$
- C
$\frac{1}{13}\times\frac{1}{17}$
- D
$\frac{1}{13}\times\frac{4}{5}$
AnswerCorrect option: A. $\frac{1}{13}\times\frac{1}{13}$
Two cards are drawn from $52$ cards.
Let, $E_1$ be the event that getting queen in first draw and $E_2$ be the event that getting queen in second draw,
$\text{P}(\text{E}_1\cap\text{E}_2)=\frac{4}{52}\times\frac{4}{52}=\frac{1}{13}\times\frac{1}{13}$
View full question & answer→MCQ 401 Mark
If $\text{P(B)}=\frac{3}{5},\text{P}(\text{A}|\text{B})=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then $\text{P}(\text{B}|\overline{\text{A}})=$
- A
$\frac{1}{5}$
- B
$\frac{3}{10}$
- C
$\frac{1}{2}$
- ✓
$\frac{1}{2}$
AnswerCorrect option: D. $\frac{1}{2}$
$\text{P(B)}=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2},\text{P}(\text{A}\cup\text{B})=\frac{4}{5}$
Consider,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{1}{2}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\frac{3}{5}}=\frac{1}{2}$
$\text{P}(\text{A}\cap\text{B})=\frac{3}{10}$
$\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=\frac{3}{10}$
$\text{P(A)}+\frac{3}{5}-\frac{4}{5}=\frac{3}{10}$
$\text{P(A)}=\frac{1}{2}$
$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}}\Big)=\frac{\text{P}(\text{B}\cap\overline{\text{A}})}{\text{P}(\overline{\text{A}})}$
$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}}\Big)=\frac{\frac{3}{5}-\frac{3}{10}}{1-\frac{1}{2}}$
$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}}\Big)=\frac{3}{5}$
View full question & answer→MCQ 411 Mark
If S is the samle space and $\text{P(A)}=\frac{1}{3}, \text{P(B)}$ and $\text{S}=\text{A}\cup\text{B,}$ where A and B are tow mutually exclusive events, then P(A) =
- A
$\frac{1}{4}$
- B
$\frac{1}{2}$
- ✓
$\frac{3}{4}$
- D
$\frac{3}{8}$
AnswerCorrect option: C. $\frac{3}{4}$
$\text{P(A)}=\frac{1}{3}\text{P(A)}$
$\Rightarrow\ \text{P(B)}=3\text{P(A)}\ .....(\text{i})$
A and B are mutually exclusive events.
$\Rightarrow\ \text{P}(\text{A}\cap\text{B}) = 0$
Now,
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}=\text{P(S)}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}=1$
$\Rightarrow\ \text{P(A)}+3\text{P(A)}=1$ [From (i)]
$\Rightarrow\ 4\text{P(A)}=1$
$\Rightarrow\ \text{P(A)}=\frac{1}{4}$
View full question & answer→MCQ 421 Mark
Out of 30 consecutive integers, 2 are chosen at random. The probability that their sum is odd, is
- A
$\frac{14}{29}$
- B
$\frac{16}{29}$
- ✓
$\frac{15}{29}$
- D
$\frac{10}{29}$
AnswerCorrect option: C. $\frac{15}{29}$
For sum of two integers to be odd, one integer should be even and the other should be odd. In 30 consecutive integers, 15 are even and 15 are odd.
P(Sum is odd) = P(first integer is odd and second is even) + P(first integer is even and second integer is odd)
$=\frac{15}{30}\times\frac{15}{29}+\frac{15}{30}\times\frac{15}{29}$
$=\frac{450}{30\times29}$
$=\frac{15}{29}$
View full question & answer→MCQ 431 Mark
Assume that in a family, each chold is equally likely to be a boy or a girl. A family with tree cgildren is chosen at random. Tere probability that the eldest child is a girl given that the family has at least oe girl.
- A
$\frac{1}{2}$
- B
$\frac{1}{3}$
- C
$\frac{2}{3}$
- ✓
$\frac{4}{7}$
AnswerCorrect option: D. $\frac{4}{7}$
$S = \{\text{GBB, GGB, GBG, GGG, BGG, BGB, BBG, BBB}\}$
Let $E_1$ be the event that choosing a family with a girl as eldest child.
$E_2$ be the event that choosing a family with at least one girl.
$E_1 = \{\text{GBB, GGB, GBG, GGG}\}$
$E_2 = \{\text{GBB, GGB, GBG, GGG, BGG, BGB, BBG}\}$
$\text{n}(\text{E}_1)=4,\text{n}(\text{E}_2)=7,\text{n}(\text{A}\cap\text{B})=4$
$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n(B)}}=\frac{4}{7}$
View full question & answer→MCQ 441 Mark
Let A and B be two events. If $\text{P(A)}=0.2,\text{P(B)}=0.4,\text{P}(\text{A}\cup\text{B})=0.6$ then P(A|B) is equal to
Answer$\text{P(A)}=0.2,\text{P(B)}=0.4,\text{P}(\text{A}\cup\text{B})=0.6$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{A}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\Rightarrow \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})}{\text{P(B)}}$
$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{0.2+0.4-0.6}{\text{P(B)}}$
$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0$
View full question & answer→MCQ 451 Mark
If A and B are two events such that $\text{A}\neq\phi,\text{B}=\phi,$ then,
- ✓
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
- B
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{P(A)}\text{ P(B)}$
- C
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=1$
- D
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P(A)}}{\text{P(B)}}$
AnswerCorrect option: A. $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
If A and B are two events such that $\text{A}\neq\phi, \text{B}=\phi$ then,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
View full question & answer→MCQ 461 Mark
A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement, then the probability that exactly two of the three balls were red, the first ball being red, is
- A
$\frac{1}{3}$
- ✓
$\frac{4}{7}$
- C
$\frac{15}{28}$
- D
$\frac{5}{28}$
AnswerCorrect option: B. $\frac{4}{7}$
Total number of balls = 5 red + 3 Blue = 8
Probability of getting exacctly two red balls given that first ball should be red
Required probability $=\text{P}\Big(\frac{\text{R}_2\text{B}_2}{\text{R}_1}\Big)+\text{P}\Big(\frac{\text{R}_1\text{B}_2}{\text{R}_1}\Big)$
Required probability $=\frac{4}{7}\times\frac{3}{6}+\frac{3}{7}\times\frac{4}{6}=\frac{4}{7}$
View full question & answer→MCQ 471 Mark
If P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6 then $\text{P}(\text{A}\cup\text{B})=$
AnswerWe have,
P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6
As, P(B|A) = 0.6
$\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}=0.6$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=0.6\times\text{P(A)}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=0.6\times0.4$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=0.24$
Now, $\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=0.4+0.8-0.24$
$=1.2-0.24$
$=0.96$
Hence, the correct alternative is option (d).
View full question & answer→MCQ 481 Mark
Three faces of aj ordinary dice are yellow, two faces are red and one face is blue. The dice is rolled 3 times. The probability that yellow red and blue face appear in the first second and third throws respectively, is
- ✓
$\frac{1}{36}$
- B
$\frac{1}{6}$
- C
$\frac{1}{30}$
- D
AnswerCorrect option: A. $\frac{1}{36}$
P(yellow face) $=\frac{3}{6}=\frac{1}{2}$
P(red face) $=\frac{2}{6}=\frac{1}{3}$
P(one face) $=\frac{1}{6}$
P(yellow face, red face and blue face appear in the required order) $=\frac{1}{2}\times\frac{1}{3}\times\frac{1}{6}=\frac{1}{36}$
View full question & answer→MCQ 491 Mark
Three integers are chosen at random from the first 20 integers. The probability that their product is even is,
- A
$\frac{2}{19}$
- B
$\frac{3}{29}$
- ✓
$\frac{17}{19}$
- D
$\frac{4}{19}$
AnswerCorrect option: C. $\frac{17}{19}$
Required probability that product of two integers should be even.
10 integers are odd out of first 20 integers.
Required probability = 1 - Probability of product is odd
Product of three integers is odd if two numbers are odd
Required probability $=1-\frac{10}{20}\times\frac{9}{19}\times\frac{8}{18}=\frac{17}{19}$
View full question & answer→MCQ 501 Mark
A speaks truth in 75% cases and B seaks truth in 80% cases. Probability that they contradict each other in a statement, is
- ✓
$\frac{7}{20}$
- B
$\frac{13}{20}$
- C
$\frac{3}{5}$
- D
$\frac{2}{5}$
AnswerCorrect option: A. $\frac{7}{20}$
P(A speaks truth) = 0.75
P(A lies) = 1 - 0.75 = 0.25
P(B speaks truth) = 0.8
P(B lies) = 1 - 0.8 = 0.2
P(contradicting each other in a statement) = P(A speaks truth and lies) + P(B speaks truth and A lies)
= 0.75 × 0.2 + 0.8 × 0.25
= 0.15 + 0.2
= 0.35
$=\frac{35}{100}=\frac{7}{20}$
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