M.C.Q (1 Marks)

MCQ 1511 Mark

Two dice are thrown simultaneously. The probability of getting a pair of aces is

✓
$\frac{1}{36}$
B
$\frac{1}{3}$
C
$\frac{1}{6}$
D
None of these.

Answer

Correct option: A.

$\frac{1}{36}$

Required probability $=$ Probability of ace in first throw $+$ Probability of ace in second throw
$=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$

View full question & answer→

MCQ 1521 Mark

The probabilities of a student getting $\text{I, II}$ and $\text{III}$ division in an examination are $\frac{1}{10},\frac{3}{5}$ and $\frac{1}{4}$ respectively. The probability that the student fails in the examination is.

A
$\frac{197}{200}$
✓
$\frac{27}{100}$
C
$\frac{83}{100}$
D
None of these.

Answer

Correct option: B.

$\frac{27}{100}$

$\text{P(A)}=\frac{1}{10},\text{P(B)}=\frac{3}{5},\text{P(C)}=\frac{1}{4}$
Required probability $=\text{P}(\overline{\text{A}}\cap\overline{\text{B}}\cap\overline{\text{C}})$
Required probability $=\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{B}})\text{ P}(\overline{\text{C}})$
Required probability $=(1-\text{P(A)})(1-\text{P(B)})(1-\text{P(C)})$
Required probability $=\Big(1-\frac{1}{10}\Big)\Big(1-\frac{3}{5}\Big)\Big(1-\frac{1}{4}\Big)$
Required probability $=\frac{27}{100}$

View full question & answer→

MCQ 1531 Mark

A fair die is thrown twenty times. The probability that on the tenth throw the fourth six appears is:

A
$\frac{\text{ }^{20}\text{C}_{10}\times5^6}{6^{20}}$
B
$\frac{120\times5^7}{6^{10}}$
✓
$\frac{84\times5^6}{6^{10}}$
D
None of these

Answer

Correct option: C.

$\frac{84\times5^6}{6^{10}}$

A fair die is thrown then probebility of getting $6$ is $p =\frac{1}{6}.$
$\Rightarrow\text{q}=\frac{5}{6}$
To find probability that on tenth throw 4$^{th}$ six appears, in the first nine throw $3$ six should appear.
Required probability $= P(3$ six in first $9$ throw$) × P($a six in tenth throw$)$
Required probability $=\text{ }^9\text{C}_3\big(\frac{1}{6}\big)^3\big(\frac{5}{6}\big)^6\times\frac{1}{6}$
Required probability $=\frac{84\times5^6}{6^{10}}$

View full question & answer→

MCQ 1541 Mark

If $A$ and $B$ are two events such that $P(A) = 0.4, P(B) = 0.3$ and $\text{P}(\text{A}\cup\text{B})=0.5,$ then $\text{P}(\overline{\text{B}}\cap\text{A})$ equals.

A
$\frac{2}{3}$
B
$\frac{1}{2}$
C
$\frac{3}{10}$
✓
$\frac{1}{5}$

Answer

Correct option: D.

$\frac{1}{5}$

$P(A) = 0.4, P(B) = 0.3$ and $\text{P}(\text{A}\cup\text{B})=0.5$
$\text{P}(\overline{\text{B}}\cap\text{A})=\text{P(A)}-\text{P}(\text{A}\cap\text{B})$
$\text{P}(\overline{\text{B}}\cap\text{A})=\text{P(A)}-\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})\big]$
$\text{P}(\overline{\text{B}}\cap\text{A})=\text{P}(\text{A}\cup\text{B})-\text{P(B)}$
$\text{P}(\overline{\text{B}}\cap\text{A})=0.5-0.3$
$\text{P}(\overline{\text{B}}\cap\text{A})=0.2$
$\text{P}(\overline{\text{B}}\cap\text{A})=\frac{1}{5}$

View full question & answer→

MCQ 1551 Mark

A person writes $4$ letters and addresses $4$ envelopes. If the letters are placed in the envelopes at random, then the probability that all letters are not placed in the right envelopes, is

A
$\frac{1}{4}$
B
$\frac{11}{14}$
C
$\frac{15}{24}$
✓
$\frac{23}{24}$

Answer

Correct option: D.

$\frac{23}{24}$

$4$ letter can be placed in $4$ envelopes in $4!$ ways$ = 24$ ways
Now, there is only one method, by which all the letters are placed in the right envelope.
$P($all letters are placed in the envelopes$) =\frac{1}{24}$
$P($all letters are not placed in the right envelopes$) = 1 - P($all letters are placed in the right envelopes$)$
$=1-\frac{1}{24}=\frac{23}{24}$

View full question & answer→

MCQ 1561 Mark

In each of the following, choose the correct answer:
The probability that a student is not a swimmer is $\frac{1}{5}.$ Then the probability that out of five students, four are swimmers is

✓
$\ ^5\text{C}_\text{4}\Big(\frac{4}{5}\Big)^4\frac{1}{5}$
B
$\Big(\frac{4}{5}\Big)^4\frac{1}{5}$
C
$\ ^5\text{C}_1\frac{1}{5}\Big(\frac{4}{5}\Big)^4$
D
None of these

Answer

Correct option: A.

$\ ^5\text{C}_\text{4}\Big(\frac{4}{5}\Big)^4\frac{1}{5}$

The repeated selection of students who are swimmers are Bernoulli trials. Let X denote the number of students, out of 5 students, who are swimmers.
Probability of students who are not swimmers, $\text{q}=\frac{1}{5}$
$\therefore\text{p}=1-\text{q}=1-\frac{1}{5}=\frac{4}{5}$
Clearly, X has a binomial distribution with n = 5 and $\text{p}=\frac{1}{5}$
$\text{P}(\text{X=x})=\ ^\text{n}\text{C}_\text{x}\text{q}^\text{n-x}\text{p}^\text{x}=\ ^5\text{C}_\text{x}\bigg(\frac{1}{5}\bigg)^{5-\text{x}}.\bigg(\frac{4}{5}\bigg)^\text{x}$
P(four students are swimmers) = P(X = 4) $=\ ^5\text{C}_4\bigg(\frac{1}{5}\bigg).\bigg(\frac{4}{5}\bigg)^4$
Therefore, the correct answer is A.

View full question & answer→

MCQ 1571 Mark

Choose the correct answer from the given four options.
The probability distribution of a discrete random variable $X$ is given below:

$\text{X}$	$2$	$3$	$4$	$5$
$\text{P}(\text{X})$	$\frac{5}{\text{k}}$	$\frac{7}{\text{k}}$	$\frac{9}{\text{k}}$	$\frac{11}{\text{k}}$

The value of $k$ is:

A
$8.$
B
$16.$
✓
$32.$
D
$48.$

Answer

Correct option: C.

$32.$

We know that, $\sum\text{P}\text{X}=1$
$\Rightarrow\frac{5}{\text{k}}+\frac{7}{\text{k}}+\frac{9}{\text{k}}+\frac{11}{\text{k}}=1$
$\Rightarrow\frac{32}{\text{k}}=1$
$\therefore\text{k}=32$

View full question & answer→

MCQ 1581 Mark

If $\text{P(A)}=\frac{3}{10},\text{P(B)}=\frac{2}{5}$ and $\text{P}(\text{A}\cap\text{B}=\frac{3}{5,}$ then $P(A|B) + P(B|A)$ equals

A
$\frac{1}{4}$
✓
$\frac{7}{12}$
C
$\frac{5}{12}$
D
$\frac{1}{3}$

Answer

Correct option: B.

$\frac{7}{12}$

$\text{P(B)}=\frac{3}{10},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{2}{5},\text{P}(\text{A}\cup\text{B})=\frac{3}{5}$
$\text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{A})$
$\text{P}(\text{A}\cap\text{B})=\frac{3}{10}+\frac{2}{5}-\frac{3}{5}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)+\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}+\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)+\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\frac{1}{10}}{\frac{2}{5}}+\frac{\frac{1}{10}}{\frac{3}{10}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)+\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{7}{12}$
Note: Option is modified.

View full question & answer→

MCQ 1591 Mark

In each of the following choose the correct answer:If A and B are events such that $\text{P}(\text{A}|\text{B})=\text{P}(\text{B}|\text{A}),\ \text{then}:$

A
$\text{A}\subset\text{B}\ \text{but}\ \text{A}\neq\text{B}$
B
$\text{A}=\text{B}$
C
$\text{A}\cap\text{B}=\phi$
✓
$\text{P}(\text{A})=\text{P}(\text{B})$

Answer

Correct option: D.

$\text{P}(\text{A})=\text{P}(\text{B})$

$\text{P}(\text{A}|\text{B})=\text{P}(\text{B}|\text{A})$ $\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\frac{\text{P}(\text{B}\cap\text{A})}{\text{P}(\text{A})}$$\Rightarrow\ \text{P}(\text{A})=\text{P}(\text{B})$
Therefore, option (D) is correct.

View full question & answer→

MCQ 1601 Mark

A bag contains $5$ red and $3$ blue balls are drawn at random without replacement, then the probability of getting exactly one red ball is.

A
$\frac{15}{29}$
✓
$\frac{15}{56}$
C
$\frac{45}{196}$
D
$\frac{135}{392}$

Answer

Correct option: B.

$\frac{15}{56}$

Total balls $= 5$ red $+ 3$ blue $= 8$
Let $R$ be the event of getting red ball
$B$ be the event of getting a blue ball.
Required probability $= \text{P(BBR) + R(BRB) + P(RBB)}$
$=\frac{3}{8}\times\frac{2}{7}\times\frac{5}{6}+\frac{3}{8}\times\frac{5}{7}\times\frac{2}{6}+\frac{5}{8}\times\frac{3}{7}\times\frac{2}{6}$
$=\frac{15}{56}$

View full question & answer→

MCQ 1611 Mark

If $X$ follows a binomial distribution with parameter $\text{n}=8$ and $\text{p}=\frac{1}{2},$ then $\text{P(|X}-4|\leq2)$ equals:

A
$\frac{118}{128}$
✓
$\frac{119}{128}$
C
$\frac{117}{128}$
D
$\text{None of these}$

Answer

Correct option: B.

$\frac{119}{128}$

$\text{n = 8,}\text{p}=\frac{1}{2}=\text{q}$
$\text{P(|X}-4|)\leq2$
$\Rightarrow-2\leq\text{x}-4\leq2$
$\Rightarrow4-2\leq\text{x}\leq2+4$
$\Rightarrow2\leq\text{x}\leq6$
$\text{P}(2\leq\text{x}\leq6)=\text{P(2)+P(3)+P(4)+P(5)+P(6)}$
$\text{P(2}\leq\text{x}\leq6)=\text{ }^8\text{C}_2\Big(\frac{1}{2^8}\Big)+\text{ }^8\text{C}_3\Big(\frac{1}{2^8}\Big)$
$+\text{ }^8\text{C}_4\Big(\frac{1}{2^8}\Big)\text{ }^8\text{C}_5\Big(\frac{1}{2^8}\Big)+\text{ }^8\text{C}_6\Big(\frac{1}{2^8}\Big)$
$=\frac{119}{128}$

View full question & answer→

MCQ 1621 Mark

Choose the correct answer from the given four options.The probability that a person is not a swimmer is $0.3.$ The probability that out of $5$ persons $4$ are swimmers is:

✓
${^5}\text{C}_4(0.7)^4(0.3)$
B
${^5}\text{C}_1(0.7)^4(0.3)^4$
C
${^5}\text{C}_4(0.7)(0.3)^4$
D
$(0.7)^4(0.3)$

Answer

Correct option: A.

${^5}\text{C}_4(0.7)^4(0.3)$

Here, $\bar{\text{p}}=0.3$
$\Rightarrow\text{p}=0.7$
and $\text{q}=0.3,\text{n}=5$ and $\text{r}=4$
$\therefore$ Required probability $={^5}\text{C}_4(0.7)^4(0.3)$

View full question & answer→

MCQ 1631 Mark

A coin is tossed $10$ times. The probability of getting exactly six heads is:

A
$\frac{512}{513}$
✓
$\frac{105}{512}$
C
$\frac{100}{153}$
D
$\text{ }^{10}\text{C}_6$

Answer

Correct option: B.

$\frac{105}{512}$

$\text{n}=10,\text{x}=6,\text{p = q}=\frac{1}{2}$
$\text{P(X}=6)=\text{ }^{10}\text{C}_6\big(\frac{1}{2}\big)^{10}=\frac{105}{512}$

View full question & answer→

MCQ 1641 Mark

Choose the correct answer from the given four options.Two cards are drawn from a well shuffled deck of $52$ playing cards with replacement. The probability, that both cards are queens, is:

✓
$\frac{1}{13}\times\frac{1}{13}$
B
$\frac{1}{13}\times\frac{1}{13}$
C
$\frac{1}{13}\times\frac{1}{17}$
D
$\frac{1}{13}\times\frac{4}{15}$

Answer

Correct option: A.

$\frac{1}{13}\times\frac{1}{13}$

Required probability $=\frac{4}{52}\cdot\frac{4}{52}$
$=\frac{1}{13}\times\frac{1}{13} [$with replacement$]$

View full question & answer→

MCQ 1651 Mark

Choose the correct answer from the given four options.If $\text{P}(\text{A})=0.4,\text{P}(\text{B})=0.8$ and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.6,$ then $\text{P}(\text{A}\cup\text{B})$ is equal to:

A
$0.24$
B
$0.3$
C
$0.48$
✓
$0.96$

Answer

Correct option: D.

$0.96$

Here$, P(A) = 0.4, P(B) = 0.8$
and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.6,$
$\because\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{B}\cap\text{A})}{\text{P}(\text{A})}$
$\Rightarrow\text{P}(\text{B}\cap\text{A})=\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)\cdot\text{P}(\text{A})$
$=0.6\times0.4=0.24$
$=1.2-0.24=0.96$

View full question & answer→

MCQ 1661 Mark

A bag contains $12$ balls out of which $x$ are white. If one ball is drawn at random, what is the probability it will be a white ball?

A
$\frac{\text{x}}{2}$
✓
$\frac{\text{x}}{12}$
C
$\frac{\text{x}}{10}$
D
$\frac{12}{\text{x}}$

Answer

Correct option: B.

$\frac{\text{x}}{12}$

Total number of balls $= 12$
Number of white balls $= x$
$P ($white ball$) =\frac{\text{x}}{12}$

View full question & answer→

MCQ 1671 Mark

Choose the correct answer from the given four options$.A$ and $B$ are events such that $P(A) = 0.4, P(B) = 0.3$ and $\text{P}(\text{A}\cup\text{B})=0.5,$ Then $\text{P}(\text{B}\ '\cap\text{A})$ equals:

A
$\frac{2}{3}$
B
$\frac{1}{2}$
C
$\frac{3}{10}$
✓
$\frac{1}{5}$

Answer

Correct option: D.

$\frac{1}{5}$

Here$, \text{P}(\text{A}) = 0.4,\text{P}(\text{B}) = 0.3$ and $\text{P}(\text{A}\cup\text{B})=0.5$
$\because\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\text{P}(\text{A}\cap\text{B})=0.4+0.3-0.5=0.2$
$\because\text{P}(\text{B}\ '\cap\text{A})=\text{P}(\text{A})-\text{P}(\text{A}\cap\text{B})$
$=0.4-0.2=0.2=\frac{1}{5}$

View full question & answer→

MCQ 1681 Mark

If A and B are two events such that $\text{A}\subset\text{B}$ and $\text{P}(\text{B})\neq0,$ then which of the following is correct?

A
$\text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{B})}{\text{P}(\text{A})}$
B
$\text{P}(\text{A}|\text{B})\ <\text{P}(\text{A})$
✓
$\text{P}(\text{A}|\text{B})\ \geq\text{P}(\text{A})$
D
None of these.

Answer

Correct option: C.

$\text{P}(\text{A}|\text{B})\ \geq\text{P}(\text{A})$

$\text{A}\subset\text{B}\ \Rightarrow\ \ \text{A}\cap\text{B}=\text{A P}\ \ \text{and}\ \ \text{P}(\text{B})\neq0$
$\Rightarrow\ \text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\frac{\text{P}(\text{A})}{\text{P}(\text{B})}$
Since $\text{P}(\text{B})\neq\ 0$
$\therefore\ \frac{\text{P}(\text{A})}{\text{P}(\text{B})}<1\ \ \ \ \ \Rightarrow\ \text{P}(\text{A})<\text{P}(\text{B})\ \Rightarrow\ \text{P}(\text{A}|\text{B})\geq\text{P}(\text{A})$
Hence, option (C) is correct.

View full question & answer→

MCQ 1691 Mark

If $\text{P(A)}=\frac{7}{13},\text{P(B)}=\frac{9}{13}$ and $\text{P}(\text{A}\cap\text{B})=\frac{4}{13}.$ then, $\text{P}(\overline{\text{A}}|\text{B})=$

A
$\frac{5}{9}$
✓
$\frac{4}{9}$
C
$\frac{4}{13}$
D
$\frac{6}{13}$

Answer

Correct option: B.

$\frac{4}{9}$

We have,
$\text{P(A)}=\frac{7}{13},\text{P(B)}=\frac{9}{13}$ and $\text{P}(\text{A}\cap\text{B})=\frac{4}{13}$
As, $\text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=\frac{9}{13}-\frac{4}{13}$
$=\frac{5}{13}$
Now,
$\text{P}(\overline{\text{A}}|\text{B})=\frac{\text{P}(\overline{\text{A}}\cap\text{B})}{\text{P(B)}}$
$=\frac{\Big(\frac{5}{13}\Big)}{\Big(\frac{9}{13}\Big)}$
$=\frac{5}{9}$
Hence, the correct alternative is option $(a).$

View full question & answer→

MCQ 1701 Mark

Two dice are thrown. If it is known that the sun of the numbers on the dice was less than $6,$ than the probability of gettinga sum $3,$ is

A
$\frac{1}{18}$
B
$\frac{5}{18}$
✓
$\frac{1}{5}$
D
$\frac{2}{5}$

Answer

Correct option: C.

$\frac{1}{5}$

$\text{S}= (1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6),(2, 1),(2, 2),(2, 3),(2, 4),(2, 5),$$(2, 6),(3, 1),(3, 2),(3, 3),(3, 4),(3, 5),(3, 6),(4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6),(5, 1),$$(5, 2),(5, 3),(5, 4),(5, 5),(5, 6),(6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6, 6)$
$\text{n(S)}=36$
Let $A$ be the event that sum of the numbers on dice was less than $6.$
$\text{A} =\{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)\}$
$\text{n(A)} = 10$
Let $B$ be the event that getting sum $3.$
$\text{B}=\{(1, 2), (2, 1)\}$
$\Rightarrow\text{n(B)}=2$
$\text{A}\cap\text{B}=\{(1,2),(2,1)\}$
$\Rightarrow\text{n}(\text{A}\cap\text{B})=2$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{2}{10}=\frac{1}{5}$

View full question & answer→

MCQ 1711 Mark

If $\text{P(A)}=\frac{2}{5},\text{P(B)}=\frac{3}{10}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{5},$ then, $\text{P}(\overline{\text{A}}|\overline{\text{B}}) \text{ P}(\overline{\text{B}}|\overline{\text{A}})$ is equal to

A
$\frac{5}{6}$
B
$\frac{5}{7}$
✓
$\frac{25}{42}$
D
$1$

Answer

Correct option: C.

$\frac{25}{42}$

$\text{P(A)}=\frac{2}{5},\text{P(B)}=\frac{3}{10}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{5}$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{5}$
$\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=\frac{1}{5}$
$\frac{2}{5}+\frac{3}{10}-\text{P}(\text{A}\cup\text{B})=\frac{1}{5}$
$\text{P}(\text{A}\cup\text{B})=\frac{1}{2}$
$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)\text{ P}\Big(\frac{\overline{\text{B}}}{\overline{\text{A}}}\Big)=\frac{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}{\text{P}(\overline{\text{B}})}\frac{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}{\text{P}(\overline{\text{A}})}$
$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)\text{ P}\Big(\frac{\overline{\text{B}}}{\overline{\text{A}}}\Big)=\frac{\big[\text{P}(\overline{\text{A}\cup\text{B}})\big]^2}{\text{P}(\overline{\text{B}})\text{ P}(\overline{\text{A}})}$
$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)\text{ P}\Big(\frac{\overline{\text{B}}}{\overline{\text{A}}}\Big)=\frac{\big[1-\text{P}(\text{A}\cup\text{B})\big]^2}{\text{P}(\overline{\text{B}})\text{ P}(\overline{\text{A}})}$
$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)\text{ P}\Big(\frac{\overline{\text{B}}}{\overline{\text{A}}}\Big)=\frac{\Big[1-\frac{1}{2}\Big]^2}{\frac{7}{10}\times\frac{3}{5}}$
$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)\text{ P}\Big(\frac{\overline{\text{B}}}{\overline{\text{A}}}\Big)=\frac{25}{42}$

View full question & answer→

MCQ 1721 Mark

The probablity of selecting a male or a female is same. If the probability that in an office of $n$ persons $(n - 1)$ males being selected is $\frac{3}{2^{10}},$ the value of n is:

A
$5$
B
$3$
C
$10$
✓
$12$

Answer

Correct option: D.

$12$

$X$ represents number of males.
$\text{p = q}=\frac{1}{2}$
$\text{p(n}-1)=\frac{3}{2^{10}}$
$\text{ }^{\text{n}}\text{C}_{\text{n}-1}\text{p}^{\text{n}-1}\text{q}=\frac{3}{2^{10}}$
$\text{n}\big(\frac{1}{2}\big)^{\text{n}-1}\big(\frac{1}{2}\big)^{\text{n}}=\frac{3}{2^{10}}$
$\text{n}\big(\frac{1}{2}\big)^{\text{n}}=\frac{1}{4}\times\frac{3\times4}{2^{10}}$
$\text{n}\big(\frac{1}{2}\big)^{\text{n}}=12\big(\frac{1}{2}\big)^{12}$
$\Rightarrow\text{n}=12$

View full question & answer→

MCQ 1731 Mark

The probability distribution of a random variable $X$ is:

$X$	$ 0$	$1$	$2$	$3$	$4$
$P(X)$	$ 0.1$	$ k$	$ 2k$	$ k$	$0.1$

where $k$ is some unknown constant.
The probability that the random variable $X$ takes the value $2$ is:

A
$\frac{1}{5}$
✓
$\frac{2}{5}$
C
$\frac{4}{5}$
D
$1$

Answer

Correct option: B.

$\frac{2}{5}$

$\text {As } \sum_{i=1}^n p_i=1$
$\Rightarrow 0.1+k+2 k+k+0.1=1$
$\Rightarrow 4 k=1-0.2=0.8$
$\Rightarrow k=\frac{0.8}{4}=0.2$
$\therefore P(x=2)=2 k=2 \times 0.2=0.4=\frac{4}{10}=\frac{2}{5}$

View full question & answer→

MCQ 1741 Mark

If $A$ and $B$ are events such that $P(A / B)=P(B / A) \neq 0$, then

A
$A \subset B$, but $A \neq B$
B
$A=B$
C
$A \cap B=0$
✓
$P(A)=P(B)$

Answer

Correct option: D.

$P(A)=P(B)$

Given, $P(A / B)=P(B / A)$
$\Rightarrow \frac{P(A \cap B)}{P(B)}=\frac{P(B \cap A)}{P(A)}$
$\Rightarrow P(A)=P(A) [\because P(A \cap B)=P(B \cap A)]$

View full question & answer→

MCQ 1751 Mark

Let $E$ be an event of a sample space $S$ of an experiment then $P(S \mid E)=$

A
$P(S \cap E)$
B
$P(E)$
C
1
D
$0$

Answer

$P(S \mid E)=\frac{P(S \cap E)}{P(E)}=\frac{P(E)}{P(E)}=1$

View full question & answer→

MCQ 1761 Mark

A problem in Mathematics is given to three students whose chances of solving it are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ respectively. If the events of their solving the problem are independent then the probability that the problem will be solved, is

A
$\frac{1}{4}$
B
$\frac{1}{3}$
C
$\frac{1}{2}$
D
$\frac{3}{4}$

Answer

Let $A, B, C$ be the respective events of solving the problem. Then, $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}$ and $P(C)=\frac{1}{4}$. Here, $A, B$, $C$ are independent events. Problem is solved if at least one of them solves the problem.
Required probability is $=P(A \cup B \cup C)=1-P(\bar{A}) P(\bar{B}) P(\bar{C})$
$
=1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)=1-\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4}=1-\frac{1}{4}=\frac{3}{4} \text {. }$

View full question & answer→

MCQ 1771 Mark

$M$ and $N$ are two events such that $P(M \cap N)=0$. Which of the following is equal to $P(M \mid(M \cup N))$ ?

A
$\frac{P(M)}{P(N)}$
B
$\frac{P(M \cup N)}{P(M \cup N)}$
✓
$\frac{P(M)}{P(M)+P(N)}$
D
$\frac{P(M)}{P(M) \times P(N)}$

Answer

Correct option: C.

$\frac{P(M)}{P(M)+P(N)}$

View full question & answer→

MCQ 1781 Mark

If for any two events $A$ and $B$, $P(A)=\frac{4}{5}$ and $P(A \cap B)=\frac{7}{10}$, then $P(B / A)$ is

A
$\frac{1}{10}$
B
$\frac{1}{8}$
C
$\frac{7}{8}$
D
$\frac{17}{20}$

Answer

We know that, $P(B / A)=\frac{P(B \cap A)}{P(A)}=\frac{7 / 10}{4 / 5}=\frac{7}{8}$

View full question & answer→

MCQ 1791 Mark

For two events $A$ and $B$, if $P(A)=0.4, P(B)=0.8$ and $P(B / A)=0.6$, then $P(A \cup B)$ is

A
$0.24$
B
$0.3$
C
$0.48$
D
$0.96$

Answer

We have, $P(A)=0.4, P(B)=0.8$ and $P(B / A)=0.6$
We know that $P(B / A)=\frac{P(A \cap B)}{P(A)}$
$
\Rightarrow 0.6=\frac{P(A \cap B)}{0.4} \Rightarrow P(A \cap B)=0.24
$
Now: $P(A \cup B)=P(A)+P(B)-P(A \cap B)=0.4+0.8-0.24=0.96$
Hence. $P(A \cup B)=0.96$

View full question & answer→

MCQ 1801 Mark

If $P\left(\frac{A}{B}\right)=0.3, P(A)=0.4$ and $P(B)=0.8$, then $P\left(\frac{B}{A}\right)$ is equal to:

A
$0.6$
B
$0.3$
C
$0.06$
D
$0.4$

Answer

We have, $P\left(\frac{A}{B}\right)=0.3, P(A)=0.4$ and $P(B)=0.8$
Now, $P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}$
$
\Rightarrow \quad 0.3=\frac{P(A \cap B)}{0.8} \Rightarrow P(A \cap B)=0.3 \times 0.8=0.24
$
Now, $P\left(\frac{B}{A}\right)=\frac{P(B \cap A)}{P(A)}=\frac{0.24}{0.4}=0.6$

View full question & answer→

MCQ 1811 Mark

A family has 2 children and the elder child is a girl. The probability that both children are girls is:

A
$\frac{1}{4}$
B
$\frac{1}{8}$
C
$\frac{1}{2}$
D
$\frac{3}{4}$

Answer

Let $G_i(i=1,2)$ and $B_i(i=1,2)$ denote the $i^{\text {th }}$ child is a girl or a boy respectively.
Then sample space, $S=\left\{G_1 G_2, G_1 B_2, B_1 G_2, B_1 B_2\right\}$
Let $A$ be the event that both children are girls,
$B$ be the event that the elder child is a girl.
Thus, $A=\left\{G_1 G_2\right\}$ and $B=\left\{G_1 G_2, G_1 B_2\right\}$
$\Rightarrow A \cap B=\left\{G_1 G_2\right\}$
$\therefore \quad$ Required probability $=P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{1 / 4}{2 / 4}=\frac{1}{2}$

View full question & answer→

MCQ 1821 Mark

Five fair coins are tossed simultaneously. The probability of the events that atleast one head comes up is

A
$\frac{27}{32}$
B
$\frac{5}{32}$
C
$\frac{31}{32}$
D
$\frac{1}{32}$

Answer

Since each coin turns up on either a head or tail.
$\therefore \quad$ Total possible outcomes $=2^5=32$
Let $A$ be the event that all tails comes up.
$\therefore \quad n(A)=1\{$ i.e., $(T, T, T, T, T)$
So, required probability $=1-P(A)=1-\frac{1}{32}=\frac{31}{32}$

View full question & answer→

MCQ 1831 Mark

If $A$ and $B$ are two independent events with $P(A)=\frac{1}{3}$ and $P(B)=\frac{1}{4}$, then $P\left(B^\ {\prime} \mid A\right)$ is equal to

A
$\frac{1}{4}$
B
$\frac{1}{3}$
✓
$\frac{3}{4}$
D
$1$

Answer

Correct option: C.

$\frac{3}{4}$

Given, $A$ and $B$ are independent events.
Also, $P(A)=\frac{1}{3}$ and $P(B)=\frac{1}{4}$
$\text { Now, } P\left(B^{\prime} \mid A\right)=\frac{P\left(B^{\prime} \cap A\right)}{P(A)}$
$=\frac{P\left(B^{\prime}\right) P(A)}{P(A)} \quad[\because A, B$ are independent events $]$
$=P\left(B^{\prime}\right)=1-P(B)=1-\frac{1}{4}=\frac{3}{4}$

View full question & answer→

MCQ 1841 Mark

A card is picked at random from a pack of 52 playing cards. Given that the picked card is a queen, the probability of this card to be a card of spade is

A
$\frac{1}{3}$
B
$\frac{4}{13}$
C
$\frac{1}{4}$
D
$\frac{1}{2}$

Answer

Let $A$ be the event that the card is a spade and $B$ be the event that the picked card is a queen.
We have a total of 13 spades and 4 queen cards.
Also only one queen is from spade.
$
\therefore \quad P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{52}}{\frac{4}{52}}=\frac{1}{4}
$

View full question & answer→

MCQ 1851 Mark

A die is thrown once. Let $A$ be the event that the number obtained is greater than $3$ . Let $B$ be the event that the number obtained is less than $5$ . Then $P(A \cup B)$ is

A
$\frac{2}{5}$
B
$\frac{3}{5}$
C
$0$
✓
$1$

Answer

Correct option: D.

$1$

Here$, A=\{4,5,6\}, B=\{1,2,3,4\}$
$A \cap B=\{4\}$
Now$, P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$=\frac{3}{6}+\frac{4}{6}-\frac{1}{6}=1$

View full question & answer→

MCQ 1861 Mark

If $P(A)=0.4, P(B)=0.8$ and $P(B \mid A)=0.6$, then $P(A \cup B)$ is equal to

A
0.24
B
0.3
C
0.48
✓
0.96

Answer

Correct option: D.

0.96

(d) : Given, $P(A)=0.4, P(B)=0.8$ and $P(B \mid A)=0.6$.
Clearly, $P(A \cap B)=P(B \mid A) P(A)=0.6 \times 0.4=0.24$
Now, $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$=0.4+0.8-0.24=0.96$

View full question & answer→

MCQ 1871 Mark

A machine operates if all of its three components function. The probability that the first component fails during the year is $0.14$ , the second component fails is $0.10$ and the third component fails is $0.05 .$ What is the probability that the machine will fail during the year?

A
$0.1542$
✓
$0.2647$
C
$0.3642$
D
$0.4231$

Answer

Correct option: B.

$0.2647$

Consider the following events:
$A=$ First component of the machine fails during the year
$B=$ Second component of the machine fails during the year
$C=$ Third component of the machine fails during the year
We have, $P(A)=0.14, P(B)=0.10$ and $P(C)=0.05$
Clearly, the machine will fail if at least one of its three components fails during the year.
Required probability $= P (A \cup B \cup C)$
$=1-P(\overline{A \cup B \cup C})$
$=1-P(\bar{A} \cap \bar{B} \cap \bar{C})$
$=1-P(\bar{A}) P(\bar{B}) P(\bar{C})[\because A, B, C$ are independent events $]$
$=1-(1-0.14)(1-0.10)(1-0.05)$
$=1-(0.86)(0.90)(0.95)=0.2647 .$

View full question & answer→

MCQ 1881 Mark

If $A$ and $B$ are two events such that $P(A)=0.2, P(B)=0.4$ and $P(A \cup B)=0.5$, then value of $P(A / B)$ is

A
0.1
✓
0.25
C
0.5
D
0.08

Answer

Correct option: B.

0.25

(b) : We have, $P(A)=0.2, P(B)=0.4$ and $P(A \cup B)=0.5$
$\because \quad P(A \cap B)=P(A)+P(B)-P(A \cup B)$
$=0.2+0.4-0.5=0.1$
Now, $P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{0.1}{0.4}=\frac{1}{4}=0.25$

View full question & answer→

MCQ 1891 Mark

If $P(A)=\frac{3}{10}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{3}{5}$, then find the value of $P(B / A)$.

A
$\frac{2}{3}$
✓
$\frac{1}{3}$
C
$\frac{2}{5}$
D
$\frac{1}{4}$

Answer

Correct option: B.

$\frac{1}{3}$

(b) : Given, $P(A)=\frac{3}{10}, P(B)=\frac{2}{5}, P(A \cup B)=\frac{3}{5}$
Now, $P(A \cap B)=P(A)+P(B)-P(A \cup B)$
$
=\frac{3}{10}+\frac{2}{5}-\frac{3}{5}=\frac{3+4-6}{10}=\frac{1}{10}
$
Now, $P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{1 / 10}{3 / 10}=\frac{1}{3}$

View full question & answer→

MCQ 1901 Mark

One card is drawn from a well shuffled pack of 52 cards. If $E$ is the event "the card drawn is a king or queen" and $F$ is the event "the card drawn is a queen or an ace", then find the probability of the conditional event $E \mid F$.

✓
$\frac{1}{2}$
B
$\frac{1}{3}$
C
$\frac{2}{13}$
D
$\frac{3}{13}$

Answer

Correct option: A.

$\frac{1}{2}$

(a) : There are 4 kings and 4 queens in pack of 52 cards.
Let $E=$ the card drawn is a king or queen
$
\therefore \quad P(E)=\frac{8}{52}=\frac{2}{13}
$
There are 4 queens and 4 aces in a pack of 52 cards.
Let $F=$ the card drawn is a queen or an ace
$
\therefore \quad P(F)=\frac{8}{52}=\frac{2}{13}
$
Then, $E \cap F$ contains drawing a queen.
$
\therefore \quad P(E \cap F)=\frac{4}{52}=\frac{1}{13}
$
$\therefore$ Required probability, $P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{1 / 13}{2 / 13}=\frac{1}{2}$

View full question & answer→

MCQ 1911 Mark

If two events $A$ and $B$ are independent, then which of the following is true about events $A$ and $B$ ?

A
$A$ and $B$ are mutually exclusive
✓
$P\left(A^{\prime} \cap B^{\prime}\right)=[1-P(A)][1-P(B)]$
C
$P(A)=P(B)$
D
$P(A)+P(B)$=1

Answer

Correct option: B.

$P\left(A^{\prime} \cap B^{\prime}\right)=[1-P(A)][1-P(B)]$

Two events $A$ and $B$ are independent, if $P(A \cap B)=P(A) P(B)$
$\therefore P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}=1-P(A \cup B)$
$=1-[P(A)+P(B)-P(A \cap B)]$
$=1-P(A)-P(B)+P(A) P(B)$
$=[1-P(A)][1-P(B)]$

View full question & answer→

MCQ 1921 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 2 green balls and one blue ball is

✓
$\frac{3}{28}$
B
$\frac{2}{21}$
C
$\frac{1}{28}$
D
$\frac{167}{168}$

Answer

Correct option: A.

$\frac{3}{28}$

(a) : Possible outcomes $=\{G G B, G B G, B G G\}$
Required probability $=P(G G B)+P(G B G)+P(B G G)$
$
=\frac{3}{8} \times \frac{2}{7} \times \frac{2}{6}+\frac{3}{8} \times \frac{2}{7} \times \frac{2}{6}+\frac{2}{8} \times \frac{3}{7} \times \frac{2}{6}=\frac{3}{28}
$

View full question & answer→

MCQ 1931 Mark

$A$ and $B$ are events such that $P(A)=0.4, P(B)$ $=0.3$ and $P(A \cup B)=0.5$. Then $P\left(B^{\prime} \cap A\right)$ equals

A
$\frac{2}{3}$
B
$\frac{1}{2}$
C
$\frac{3}{10}$
✓
$\frac{1}{5}$

Answer

Correct option: D.

$\frac{1}{5}$

(d) : $P(A)=0.4, P(B)=0.3, P(A \cup B)=0.5$
$P(A \cap B)=P(A)+P(B)-P(A \cup B)=0.4+0.3-0.5=0.2$
Now, $P\left(B^{\prime} \cap A\right)=P(A)-P(A \cap B)=0.4-0.2=0.2=\frac{1}{5}$

View full question & answer→

MCQ 1941 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
both (a) and (b) are correct
✓
None of these

Answer

Correct option: D.

None of these

(d)

View full question & answer→

MCQ 1951 Mark

A flashlight has 8 batteries out of which 3 are dead. If two batteries are selected without replacement and tested, then probability that both are dead is

A
$\frac{33}{56}$
B
$\frac{9}{64}$
C
$\frac{1}{14}$
✓
$\frac{3}{28}$

Answer

Correct option: D.

$\frac{3}{28}$

(d) : Required probability $=P(D D)=\frac{3}{8} \times \frac{2}{7}=\frac{3}{28}$

View full question & answer→

MCQ 1961 Mark

If $A$ and $B$ are events such that $P(A)>0$ and $P(B) \neq 1$, then $P\left(A^{\prime} \mid B^{\prime}\right)$ equals

A
$1-P(A \mid B)$
B
$1-P\left(A^{\prime} \mid B\right)$
✓
$\frac{1-P(A \cup B)}{P\left(B^{\prime}\right)}$
D
$P\left(A^{\prime}\right) \mid P\left(B^{\prime}\right)$

Answer

Correct option: C.

$\frac{1-P(A \cup B)}{P\left(B^{\prime}\right)}$

(c) : By definition, $P\left(A^{\prime} \mid B^{\prime}\right)=\frac{P\left(A^{\prime} \cap B^{\prime}\right)}{P\left(B^{\prime}\right)}$
$
=\frac{P\left((A \cup B)^{\prime}\right)}{P\left(B^{\prime}\right)}=\frac{1-P(A \cup B)}{P\left(B^{\prime}\right)}
$

View full question & answer→

MCQ 1971 Mark

The probability distribution of $X$ is

$X=x$	0	1	2	3	4
$P(X=x)$	$k$	$2k$	$4k$	$2k$	$k$

Then find $P(X \leq 1)$.

A
0.1
B
0.3
C
0.4
D
0.5

Answer

\begin{array}{ll}
& \text { (b) }: \because \sum_{x=0}^4 P(X=x)=1 \\
\Rightarrow & P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)=1 \\
\Rightarrow & k+2 k+4 k+2 k+k=1 \Rightarrow 10 k=1 \Rightarrow k=0.1 \\
\therefore & P(X \leq 1)=P(X=0)+P(X=1) \\
\Rightarrow & P(X \leq 1)=k+2 k=3 k=0.3
\end{array}

View full question & answer→

MCQ 1981 Mark

The probability distribution of a discrete random variable $X$ is given below :

$X$	0	1	2	3
$P(X)$	$\frac{4}{k}$	$\frac{6}{k}$	$\frac{4}{k}$	$\frac{2}{k}$

The value of $k$ is

A
8
✓
16
C
32
D
48

Answer

Correct option: B.

(b) : We have $\Sigma P(X)=1$
$
\Rightarrow \frac{4}{k}+\frac{6}{k}+\frac{4}{k}+\frac{2}{k}=1 \Rightarrow \frac{16}{k}=1 \quad \Rightarrow \quad k=16
$

View full question & answer→

MCQ 1991 Mark

A random variable $X$ has the following probability distribution :

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$a$	$6 a$	$6 a$	$4 a$	$8 a$	$8 a$	$6 a$	$9 a$

Find the value of $a$.

A
$\frac{1}{47}$
✓
$\frac{1}{48}$
C
$\frac{1}{33}$
D
$\frac{1}{29}$

Answer

Correct option: B.

$\frac{1}{48}$

We know that, $\Sigma P_i=1$
$\Rightarrow a+6 a+6 a+4 a+8 a+8 a+6 a+9 a=1$
$\Rightarrow 48 a=1$
$\Rightarrow a=\frac{1}{48}$

View full question & answer→

MCQ 2001 Mark

A problem in mathematics is given to $3$ students whose chances of solving it are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$. What is the probability that the problem is solved?

A
$1 / 5$
B
$1 / 4$
✓
$3 / 4$
D
$2 / 3$

Answer

Correct option: C.

$3 / 4$

Let $A, B, C$ be the respective events of solving the problem.
Then, $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}$ and $P(C)=\frac{1}{4}$.
Clearly $A, B, C$ are independent events and the problem is solved if atleast one student solves it.
$\therefore \text { Required probability }=P(A \cup B \cup C)$
$=1-P(\bar{A}) P(\bar{B}) P(\bar{C})$
$=1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)$
$=1-\frac{1}{4}$
$=\frac{3}{4}$

View full question & answer→

Maths M.C.Q (1 Marks)