M.C.Q (1 Marks)

Question 511 Mark

A bag contains six red four green and eight white balls If a ball is picked at random the probability that it is not white is:

$\frac13$
$\frac49$
$\frac59$
$\frac23$

Answer

$\frac59$

Solution:
Number of balls that are not white = 10
Total $=\frac 18$
$∴ $ P(not white) $= \frac{18}{10} = 95$

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Question 521 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

$\frac{1}{3}$
$\frac{4}{7}$
$\frac{15}{28}$
$\frac{5}{28}$

Answer

$\frac{4}{7}$

Solution:
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 $=\frac{4}{7}\times\frac{3}{6}+\frac{3}{7}\times\frac{4}{6}=\frac{4}{7}$

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Question 531 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

Solution:
$\text{P(A)}=0.2,\text{P(B)}=0.4,\text{P}(\text{A}\cup\text{B})=0.6$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{A}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\Rightarrow \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})}{\text{P(B)}}$
$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{0.2+0.4-0.6}{\text{P(B)}}$
$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0$

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Question 541 Mark

If P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6 then $\text{P}(\text{A}\cup\text{B})=$

0.24
0.3
0.48
0.96

Answer

0.96

Solution:
We 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).

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Question 551 Mark

Choose the correct answer from the given four options.Eight coins are tossed together. The probability of getting exactly 3 heads is:

$\frac{1}{256}$
$\frac{7}{32}$
$\frac{5}{32}$
$\frac{3}{32}$

Answer

$\frac{7}{32}$

Solution:
We know that, probaility distribution $\text{P}(\text{X}=\text{r})={^\text{n}\text{C}}_\text{r}(\text{P})^{\text{r}}\text{q}^{\text{n}-\text{r}}$
Here, $\text{n}=8,\text{r}=3,\text{p}=\frac{1}{2}$ and $\text{q}=\frac{1}{2}$
$\therefore$ Reuaired probability $={^8}\text{C}_3\Big(\frac{1}{2}\Big)^3\Big(\frac{1}{2}\Big)^{8-3}=\frac{8!}{5!3!}\Big(\frac{1}{2}\Big)^8 $
$=\frac{8\cdot7\cdot6}{3\cdot2}\cdot\frac{1}{2^8}=\frac{7}{32}$

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MCQ 561 Mark

The probability that a leap year will have $53$ sundays is:

A
$\frac17$
✓
$\frac27$
C
$\frac57$
D
$\frac67$

Answer

Correct option: B.

$\frac27$

A leap year has $52$ weeks and $2$ days.
The $53^{rd}$ Sunday will be from these extra two days
These $2$ days can be $($Sunday, Monday$)$ or $($Mon, Tue$)$ or $($Tue, Wed$).....($Sat, Sun$)$
There are $7$ possibilities for these $2$ days
Out of which Sunday is coming in $2$ possibilities.
$\therefore P(2$ sundays in leap year$) =\frac27$

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Question 571 Mark

Three integers are chosen at random from the first 20 integers. The probability that their product is even is,

$\frac{2}{19}$
$\frac{3}{29}$
$\frac{17}{19}$
$\frac{4}{19}$

Answer

$\frac{17}{19}$

Soluction:
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 $=1-\frac{10}{20}\times\frac{9}{19}\times\frac{8}{18}=\frac{17}{19}$

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Question 581 Mark

The probability that an automobile will be stolen and found within one week is 0.0006. The probability that an automobile will be stolen is 0.0015. The probability that a stolen automobile will be found in one week is:

Answer

Solution:
Let P(S) be the probability of automobile stolen.
And P(F) be the probability of automobile found.
According to the question,
$\text{P(S∩F) = 0.0006, P(S) = 0.0015}$
We know,
$\text{P}\Big(\frac{\text{F}}{\text{S}}\Big)=\frac{\text{P(F}\cap\text{S)}}{\text{P(S)}}=\frac{0.0006}{0. 0015}=0.4$

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Question 591 Mark

Choose the correct answer from the given four options.If $\text{P}(\text{A}\cap\text{B})=\frac{7}{10},$ and $\text{P}(\text{B})=\frac{17}{20},$ then $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ equas:

$\frac{14}{17}$
$\frac{17}{20}$
$\frac{7}{8}$
$\frac{1}{8}$

Answer

$\frac{14}{17}$

Solution:
Here, $\text{P}(\text{A}\cap\text{B})=\frac{7}{10}$ and $\text{P}(\text{B})=\frac{17}{20}$
$\therefore\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$=\frac{\frac{7}{10}}{\frac{17}{20}}=\frac{14}{17}$

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Question 601 Mark

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

$\frac{1}{36}$
$\frac{1}{3}$
$\frac{1}{6}$
None of these.

Answer

$\frac{1}{36}$

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

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Question 611 Mark

The probabilities of a student getting I, II and 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.

$\frac{197}{200}$
$\frac{27}{100}$
$\frac{83}{100}$
None of these.

Answer

$\frac{27}{100}$

Solution:
$\text{P(A)}=\frac{1}{10},\text{P(B)}=\frac{3}{5},\text{P(C)}=\frac{1}{4}$
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 $=\frac{27}{100}$

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MCQ 621 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
$\text{None of these}$

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$) \times P($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 $=\frac{84\times5^6}{6^{10}}$

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Question 631 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.

$\frac{2}{3}$
$\frac{1}{2}$
$\frac{3}{10}$
$\frac{1}{5}$

Answer

$\frac{1}{5}$

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

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Question 641 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:

Answer

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

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Question 651 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

$\frac{1}{4}$
$\frac{7}{12}$
$\frac{5}{12}$
$\frac{1}{3}$

Answer

$\frac{7}{12}$

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

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Question 661 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:

$\frac{118}{128}$
$\frac{119}{128}$
$\frac{117}{128}$
$\text{None of these}$

Answer

$\frac{119}{128}$

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

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Question 671 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:

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

Answer

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

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

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Question 681 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:

$\frac{1}{13}\times\frac{1}{13}$
$\frac{1}{13}\times\frac{1}{13}$
$\frac{1}{13}\times\frac{1}{17}$
$\frac{1}{13}\times\frac{4}{15}$

Answer

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

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

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Question 691 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:

0.24
0.3
0.48
0.96

Answer

0.96

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

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Question 701 Mark

Choose the correct answer from the given four options: If A and B are such events that $\text{P}(\text{A})>0$ and $\text{P}(\text{B})\neq1,$ then $\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)$ equals to:

$1-\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$
$1-\text{P}\Big(\frac{\text{A}'}{\text{B}}\Big)$
$\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\text{B}')}$
$\frac{\text{P}(\text{A}')}{\text{P}(\text{B}')}$

Answer

$\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\text{B}')}$

Solution:
We have, $\text{P}(\text{A})>0$ and $\text{P}(\text{B})\neq1$
$\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)=\frac{\text{P}(\text{A}'\cap\text{B}')}{\text{P}(\text{B}')}$
$=\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\text{B}')}$

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Question 711 Mark

A fair coin is tossed 99 times. If X is the number of times head appears, then P(X = r) is maximum when r is:

49, 50
50, 51
51,,52
None of these

Answer

49, 50

Solution:
When a coin is tossed $\text{p = q}=\frac{1}{2}$
$\Rightarrow\text{P(X = r})^{\text{ }^{\text{n}}}\text{C}_{\text{r}}\times0.5^{\text{n}}$
Coin is tossed 99 times.
For odd number of n maximum terms at
$\text{r}=\frac{\text{n}-1}{2}$ and $\text{r}=\frac{\text{n}+1}{2}$
$\text{n}=99\Rightarrow\text{r}=49 \text{ or }50$

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Question 721 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, $\text{P}(\overline{\text{A}}|\text{B})=$

$\frac{5}{9}$
$\frac{4}{9}$
$\frac{4}{13}$
$\frac{6}{13}$

Answer

$\frac{5}{9}$

Solution:
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).

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Question 731 Mark

Choose the correct answer from the given four options.
In a college, 30% students fail in physics, 25% fail in mathematics and 10% fail in both. One student is chosen at random. The probability that she fails in physics if she has failed in mathematics is:

$\frac{1}{10}$
$\frac{2}{5}$
$\frac{9}{20}$
$\frac{1}{3}$

Answer

$\frac{2}{5}$

Solution:
Here, $\text{P}_{(\text{Ph})}=\frac{30}{100}=\frac{3}{10}$
$\text{P}_{(\text{M})}=\frac{25}{100}=\frac{1}{4}$
And $\text{P}_{(\text{M}\cap\text{Ph})}=\frac{10}{100}=\frac{1}{10}$
$\therefore\text{P}\Big(\frac{\text{Ph}}{\text{M}}\Big)=\frac{\text{P}(\text{Ph}\cap\text{M})}{\text{P}(\text{M})}$
$=\frac{\frac{1}{10}}{\frac{1}{4}}=\frac{2}{5}$

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Question 741 Mark

If the mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is:

$\frac{2}{3}$
$\frac{4}{5}$
$\frac{7}{8}$
$\frac{15}{16}$

Answer

$\frac{15}{16}$

Solution:
$\text{E(X)}=2, \text{V(X})=1$
$\text{np}=2,\text{npq}=1$
$\Rightarrow\text{q}=\frac{1}2{}=\text{p}$
$\Rightarrow\text{n}=4$
$\text{P(X}\geq1)=1-\text{P(X}=0)$
$\text{P(X}\geq1)=1-\big(\frac{1}{2}\big)^4$
$\text{P(X}\geq1)=1-\frac{1}{16}=\frac{15}{16}$

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Question 751 Mark

$\int\limits^1_0\sqrt{\text{x}(1-\text{x})}\text{ dx}$ equals:

$\frac{\pi}{2}$
$\frac{\pi}{4}$
$\frac{\pi}{6}$
$\frac{\pi}{8}$

Answer

$\frac{\pi}{8}$

Solution:
$\text{I}=\int\limits^1_0\sqrt{\text{x}(1-\text{x})}\text{ dx} $
$\text{I}= \int\limits^1_0\sqrt{\text{x}-\text{x}^2}\text{dx}$
$\text{I}=\int\limits^1_0\sqrt{\frac{1}{4}+\text{x}-\text{x}^2+\frac{1}{4}}\text{dx}$
$\text{I}=\int\limits^1_0\sqrt{\frac{1}{4}-\Big(\text{x}^2-\text{x}+\frac{1}{4}\Big)}\text{dx}$
$\text{I}=\int\limits^1_0\sqrt{\Big(\frac{1}{2}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2}\text{dx}$
$\text{I}=\Bigg[\frac{\text{x}-\frac{1}{2}}{2}\sqrt{\text{x}(1-\text{x})}+\frac{1}{2}\times\frac{1}{4}\sin^{-1}(2\text{x}-1)\Bigg]^1_0$
$\text{I}=0+\frac{1}{8}\big(\sin^{-1}(1)-\sin^{-1}(-1)\big)$
$\text{I}= \frac{1}{8}\Big(\frac{\pi}{2}-\Big(\frac{\pi}{2}\Big)\Big)$
$\text{I}= \frac{\pi}{8}$

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MCQ 761 Mark

A letter is known to have come either from $\text{LONDON}$ or $\text{CLIFTON;}$ on the postmark only the two consecutive letters $\text{ON}$ are ellegible. The probability that it came from $\text{LONDON}$ is:

A
$\frac{5}{17}$
✓
$\frac{12}{17}$
C
$\frac{17}{30}$
D
$\frac{3}{5}$

Answer

Correct option: B.

$\frac{12}{17}$

We define the following events:
$A_1:$ Selecting a pair of consecutive letters from the word $\text{LONDON}$
$A_2:$ Selecting a pair of consecutive letters from the word $\text{CLIFTON}$
$E:$ Selecting a pair of letters $\text{ON}$
Then ${\text{P(A}_1∩\text{E})=\frac52},$ as there are $5$ pairs of consecutive letters out of which $2$ are $\text{ON}$.
$\text{P(A}_2∩\text{E})=\frac61,$ as there are $6$ pairs of consecutive letters of which $1$ is $\text{ON}$.
So, required probability $\text{P}=\Big(\frac{\text{A}_1}{\text{E}}\Big)$
$\Rightarrow\Big(\frac{\text{A}_1}{\text{E}}\Big)=\frac{\text{P}(\text{A}_1\cap\text{E})}{\text{P}(\text{A}_1\cap\text{E}) + \text{P}(\text{A}_1\cap\text{E})}=\frac{\frac25}{\frac25+\frac16}=\frac{12}{17}$

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Question 771 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 $\text{P}(\overline{\text{A}\cap\text{B}})+\text{P}(\overline{\text{A}}\cap\text{B})=$

$\frac{1}{5}$
$\frac{4}{5}$
$\frac{1}{2}$
$1$

Answer

Solution:
$\text{P(B)}=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2},\text{P}\Big({\text{A}}\cup{\text{B}}\Big)=\frac{4}{5}$
Consider,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$
$\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{1}{2}$
$\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\frac{3}{5}}=\frac{1}{2}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{3}{10}$
$\text{P}(\overline{\text{A}\cup\text{B}})+\text{P}(\overline{\text{A}}\cap\text{B})$
$=1-\text{P}(\text{A}\cap\text{B})+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=1-\frac{3}{10}+\frac{3}{5}-\frac{3}{10}$
$=1$

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Question 781 Mark

The least number of times a fair coin must be tossed so that the probability of getting at least one head is at least 0.8, is:

Answer

Solution:
A fair coin is tossed $\Rightarrow\text{p = q}=\frac{1}{2}$
$\text{P(X}\geq1)\geq0.8$
$\Rightarrow1-\text{P}(0)\geq0.8$
$\Rightarrow\text{P}(0)=0.2$
$\Rightarrow\big(\frac{1}{2}\big)^{\text{n}}=0.2$
$\Rightarrow2^{-\text{n}}=0.2$
$\Rightarrow2^{\text{n}}\geq5$
$\Rightarrow\text{n}\geq3$

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Question 791 Mark

A rifleman is firing at a distant target and has only 10% chance of hiting it. the least number of round he must fire in order to have more than 50% chance of hitting it at least once is:

Answer

Solution:
Given $\text{p}=\frac{1}{10}\Rightarrow\text{q}=\frac{9}{10}$
Let n be the number of rounds.
$\text{P(X}\geq1)=1-\text{P(X}=0)$
$\Rightarrow\text{P(X}\geq1)\geq0.5$
$\Rightarrow1-\text{P(X}=0)\geq0.5$
$\Rightarrow\text{P(X}=0)\leq0.5$
$\Rightarrow0.9^{\text{n}}\leq0.5$
Using log table,
$\text{n}\leq6.572\approx7$
He must fire in order to have more than
50% chance of hitting the target at least once.

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MCQ 801 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
$\frac{1}{13}\times\frac{4}{5}$

Answer

Correct 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,
$\text{P}(\text{E}_1\cap\text{E}_2)=\frac{4}{52}\times\frac{4}{52}=\frac{1}{13}\times\frac{1}{13}$

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Question 811 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.

$\frac{15}{29}$
$\frac{15}{56}$
$\frac{45}{196}$
$\frac{135}{392}$

Answer

$\frac{15}{56}$

Solution:
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 = P(BBR) + R(BRB) + P(RBB)
$=\frac{3}{8}\times\frac{2}{7}\times\frac{5}{6}+\frac{3}{8}\times\frac{5}{7}\times\frac{2}{6}+\frac{5}{8}\times\frac{3}{7}\times\frac{2}{6}$
$=\frac{15}{56}$

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Question 821 Mark

In a box containing 100 bulbs, 10 are defective. What is the probability that out of a sample of 5 bulbs, none is defective?

$\big(\frac{9}{10}\big)^5$
$\frac{9}{10}$
$10^{-5}$
$\big(\frac{1}{2}\big)^2$

Answer

$\big(\frac{9}{10}\big)^5$

Solution:
Let X denote the number of defective bulbs.
Hence, the binomial distribution is given by
$\text{n}=5,\text{p}=\frac{10}{100}=\frac{1}{10}$
$\& \text{ q}=\frac{90}{100}=\frac{9}{10}$
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{10}\big)^{\text{r}}\big(\frac{9}{10}\big)^{5-\text{r}}$
$\therefore\text{P(X}=0)=\big(\frac{9}{10}\big)^5$

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Question 831 Mark

A five-digit number is written down at raddom. The probability that the number is divisible by 5, and no two consecutive digits are identical, is:

$\frac{1}{5}$
$\frac{1}{5}\big(\frac{9}{10}\big)^3$
$\big(\frac{3}{5}\big)^4$
$\text{None of these}$

Answer

$\text{None of these}$

Solution:
If last digit is either O or 5 then the number is divisible by 5.
Case : 1
Last digit is 0.
First three places can be selected by 9 × 9 × 9 = 729 ways.
If c = 0 then three places can be selected by 9 × 8 × 1 = 72
If C ≠ 0 then 729 - 72 = 657
Fourth place has 8 choices = 657 × 8 = 5256
Total = 72 + 5256 = 5904
Case : 2
If C = 5
First place other than 5
then first three places can be filled in 8 × 8 × 1 = 64
If first place is 5 then first three places can be filled in 1× 9 × 1 = 9 ways.
If third place is other than 5 then 729 - 64 - 9 = 656 ways.
For fourth place has 8 choices.
As per required condition = (64 + 9) × 9 + 656 × 8 = 5905
required probability $=\frac{5904+5905}{9\times10\times10\times10\times10}=\frac{11809}{90000}$
NOTE: Answer not matching with back answer.

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Question 841 Mark

If X is a binomial variate with parameters n and p, where 0 < p < 1 such that $\frac{\text{P(X = r)}}{\text{P(X = n - r})}$ is independent of n and r, then p equals:

$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{4}$
$\text{None of these}$

Answer

$\frac{1}{2}$

Solution:
Consider,
$\text{P(X = r})=\text{kP(X = n}-\text{r})$
Using $\text{ }^{\text{n}}\text{C}_{\text{r}}=\text{ }^{\text{n}}\text{C}_{\text{n}-\text{r}},\text{q}=1-\text{p}$
$\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}=\text{kp}^{\text{n}-\text{r}}\text{q}^{\text{r}}$
$\text{p}^{\text{r}}(1-\text{p})^{\text{n}-\text{r}}=\text{kp}^{\text{n}-\text{r}}(1-\text{p})^{\text{r}}$
$\text{P}^{2\text{r}-\text{n}}=\text{k}(1-\text{p})^{2\text{r}-\text{n}}$
$\big(\frac{\text{p}}{\text{q}}\big)^{2\text{r}-\text{n}}=\text{k}$
when $\text{p = q}$ then $\text{k}=1$
$\Rightarrow\text{p = q}=\frac{1}{2}$

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Question 851 Mark

If the mean and variance of a binomial distribution are 4 and 3, respectively, the probability of getting exactly six successes in this distribution is:

$\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^{10}\big(\frac{3}{4}\big)^6$
$\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^{6}\big(\frac{3}{4}\big)^{10}$
$\text{ }^{12}\text{C}_6\big(\frac{1}{20}\big)\big(\frac{3}{4}\big)^6$
$\text{ }^{12}\text{C}_6\big(\frac{1}{20}\big)^6\big(\frac{3}{4}\big)^6$

Answer

$\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^6\big(\frac{3}{4}\big)^{10}$

Solution:
$\text{np}=4,\text{npq}=3$
$\Rightarrow\text{q}=\frac{3}4{},\text{p}=\frac{1}{4},\text{n}=16$
$\text{P(X}=6)=\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^6\big(\frac{3}{4}\big)^{10}$

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Question 861 Mark

A coin is tossed 4 times. The probability that at least one head turns up is:

$\frac{1}{16}$
$\frac{2}{16}$
$\frac{14}{16}$
$\frac{15}{16}$

Answer

$\frac{15}{16}$

Solution:
$\text{n}=4,\text{p = q}=\frac{1}{2}$
$\text{P(X}\geq1)=1-\text{P(X}=0)$
$\text{P(X}\geq1)=1-\big(\frac{1}{2}\big)^4$
$\text{P(X}\geq1)=\frac{15}{16}$

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Question 871 Mark

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:

Answer

Solution:
$\sum\limits_2^5\text{P}(\text{x})=1$
$\frac{5}{\text{k}}+\frac{7}{\text{k}}+\frac{9}{\text{k}}+\frac{11}{\text{k}}=1$
$\text{k}=32$
NOTE: Question is modified.

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MCQ 881 Mark

Let $X$ be a discrete random variable. Then the variance of $X$ is$:$

A
$E(X^2)$
B
$E(X^2) + (E(X))^2$
✓
$E(X^2) - (E(X))^2$
D
$\sqrt{\text{E}(\text{X}^2)-(\text{E}(\text{X}))^2}$

Answer

Correct option: C.

$E(X^2) - (E(X))^2$

Since, the variance of a discrete random variable $X$ is given by$:$
Var$(X) = E(X^2) - (E(X))^2$
Hence, the correct alternative is option $(c).$

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Question 891 Mark

A coin is tossed n times. The probability of geting at least once is greater than 0.8. Then, the least value of n, is:

Answer

Solution:
A fair coin is tossed $\Rightarrow\text{p = q}=\frac{1}{2}$
$\text{P(X}\geq1)\geq0.8$
$\Rightarrow1-\text{P}(0)\geq0.8$
$\Rightarrow\text{P(0)}=0.2$
$\Rightarrow\big(\frac{1}{2}\big)^{\text{n}}=0.2$
$\Rightarrow2^{-\text{n}}=0.2$
$\Rightarrow2^{\text{n}}\geq5$
$\Rightarrow\text{n}\geq3$

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Question 901 Mark

For the following probability distribution:

X:	-4	-3	-2	-1	0
P(X):	0.1	0.2	0.3	0.2	0.2

The value of E(X) is:

0
-1
-2
-1.8

Answer

-1.8

Solution:
The probability distribution of X is given below:

X:	-4	-3	-2	-1	0
P(X):	0.1	0.2	0.3	0.2	0.2

E(X) = (-4) × 0.1 + (-3) × 0.2 + (-2) × 0.3 + (-1) × 0.2 + 0 × 0.2
= -0.4 - 0.6 - 0.6 - 0.2
= -1.8
Hence, the correct alternative is option (d).

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Question 911 Mark

Mark the correct alternative in the following question:Which one is not a requirement of a binomial dstribution?

There are 2 outcomes for each trial.
There is a fixed number of trials.
The outcomes must be dependent on each other.
The probability of success must be the same for all the trials.

Answer

The outcomes must be dependent on each other.

Solution:
In binomial distribution trails are independent.

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MCQ 921 Mark

If $X$ is a random variable with probability distribution as given below:

$X = x_i$	$0$	$1$	$2$	$3$
$P(X = X_i)$	$k$	$3k$	$3k$	$k$

The value of $k$ and its variance are:

A
$\frac{1}{8},\frac{22}{27}$
B
$\frac{1}{8},\frac{23}{27}$
C
$\frac{1}{8},\frac{24}{27}$
✓
$\frac{1}{8},\frac{3}{4}$

Answer

Correct option: D.

$\frac{1}{8},\frac{3}{4}$

$\sum\limits_0^3\text{P}(\text{x})=1$
$\text{k}+3\text{k}+3\text{k}+\text{k}=1$
$\text{k}=\frac{1}{8}$

$\text{x}$	$\text{P}(\text{x})$	$\text{x}\text{P}(\text{x})$	$\text{x}^2\text{P}(\text{x})$
$0$	$\frac{1}{8}$	$0$	$0$
$1$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{3}{8}$
$2$	$\frac{3}{8}$	$\frac{6}{8}$	$\frac{12}{8}$
$3$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{9}{8}$
$\text{Total}$		$\text{E(x)}=\frac{12}{8}=1.5$	$\text{E}(\text{x}^2)=3$

$\text{V(x)}=\text{E}(\text{x}^2)-[\text{E}(\text{x})^2]$
$\text{V(x)}=3-(1.5)^2$
$\text{V(x)}=0.75=\frac{3}{4}$

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MCQ 931 Mark

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

$\text{X}:$	$1$	$2$	$3$	$4$
$\text{P}(\text{X}):$	$\frac{1}{10}$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$

The value of $E(X^2)$ is:

A
$3$
B
$5$
C
$7$
✓
$10$

Answer

Correct option: D.

$10$

$\text{X}$	$1$	$2$	$3$	$4$
$\text{P}(\text{X})$	$\frac{1}{10}$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$
$\text{X}^2\text{P(X)}$	$\frac{1}{10}$	$\frac{4}{5}$	$\frac{27}{10}$	$\frac{32}{5}$	$\text{E}(\text{X}^2)=10$

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Question 941 Mark

Mark the correct alternative in the following question:The probability of guessing correctly at least 8 out of 10 answers of a true false type examination is:

$\frac{7}{64}$
$\frac{7}{128}$
$\frac{45}{1024}$
$\frac{7}{41}$

Answer

$\frac{7}{128}$

Solution:
$\text{n}=10,\text{p = q}=\frac{1}{2}$
$\text{P(X}\geq8)=\text{P(8) + P(9) + P(10)}$
$\text{P(X}\geq8)=\text{ }^{10}\text{C}_8\big(\frac{1}{2}\big)^{10}+\text{ }^{10}\text{C}_{9}\big(\frac{1}{2}\big)^{10}+\text{ }^{10}\text{C}_{10}\big(\frac{1}{2}\big)^{10}$
$\text{P(X}\geq8)=\frac{45+10+1}{2^8}$
$\text{P(X}\geq8)=\frac{56}{256}=\frac{7}{128}$

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MCQ 951 Mark

A random variable $X$ takes the values $0, 1, 2, 3$ and its mean is $1.3.$ If $P(X = 3) = 2P(X = 1)$ and $P(X = 2) = 0.3,$ then $P(X = 0)$ is:

A
$0.1$
B
$0.2$
C
$0.3$
✓
$0.4$

Answer

Correct option: D.

$0.4$

Let:
$P(X = 0) = m$
$P(X = 1) = k$
Now,$P(X = 3) = 2k$

$x_i$	$p_i$	$p_ix_i$
$0$	$m$	$0$
$1$	$k$	$k$
$2$	$0.3$	$0.6$
$3$	$2k$	$6k$

Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}$
$\Rightarrow 0 + k + 0.6 + 6k = 1.3$
$\Rightarrow 7k = 1.6 - 0.6$
$\Rightarrow\text{k}=\frac{0.7}{7}$
$\Rightarrow 0.1$
We know that the sum of probabilities in a probability distribution is always $1.$
$\therefore P(X = 0) + P(X = 1) + P(X = 3) = 1$
$\Rightarrow m + 0.1 + 0.3 + 0.2 = 1$
$\Rightarrow m + 0.6 = 1$
$\Rightarrow m = 0.4$

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Question 961 Mark

The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

0
$\frac{1}{3}$
$\frac{1}{12}$
$\frac{1}{36}$

Answer

When two dice are rolled, the number of outcomes is 36. The only even prime number is 2.Let E be the event of getting an even prime number on each die.
$\therefore\text{E}=\left\{\left(2,\ 2\right)\right\}$
$\Rightarrow\text{P}(\text{E})=\frac{1}{36}$
Therefore, the correct answer is D.

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Question 971 Mark

In each of the following choose the correct answer:$\text{If}\ \text{P}(\text{A})=\frac{1}{2},\ \text{P}(\text{B})=0,\ \text{then}\ \text{P}(\text{A}|\text{B})\ \text{is}:$

0
$\frac{1}{2}$
not defined
1

Answer

$\text{P}(\text{A})=\frac{1}{2},\ \text{P}(\text{B})=0$ $\therefore\ \text{P}(\text{A}\cap\text{B})=0$$\therefore\ \text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\frac{0}{0}=\text{not defined}$
Therefore, option (C) is correct.

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MCQ 981 Mark

A random variable has the following probability distribution:

$X = x_i$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X = X_i)$	$0$	$2p$	$2p$	$3p$	$p^2$	$2p^2$	$7p^2$	$2p$

✓
$\frac{1}{10}$
B
$-1$
C
$-\frac{1}{10}$
D
$\frac{1}{5}$

Answer

Correct option: A.

$\frac{1}{10}$

We know that the sum of probabilities in a probability distribution is always $1.$
$\therefore P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) +P(X = 6) + P(X = 7) +P(X = 8) = 1$
$\Rightarrow 0 + 2p + 2p + 3p + p^2 + 2p^2 + 7p^2 + 2p = 1$
$\Rightarrow 10p^{2 }+ 9p - 1 = 0$
$\Rightarrow (10p - 1)(p + 1) = 0$
$\Rightarrow\text{p}=\frac{1}{10}\text{ or }-1 ($Negleting $-1$ as the value of the probability cannot be negative$)$

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Question 991 Mark

Out of 30 consecutive integers, 2 are chosen at random. The probability that their sum is odd, is

$\frac{14}{29}$
$\frac{16}{29}$
$\frac{15}{29}$
$\frac{10}{29}$

Answer

$\frac{15}{29}$

Solution:
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)
$=\frac{15}{30}\times\frac{15}{29}+\frac{15}{30}\times\frac{15}{29}$
$=\frac{450}{30\times29}$
$=\frac{15}{29}$

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Question 1001 Mark

In a binomial distribution, the probability of getting success is $\frac{1}{4}$ and standard deviation is 3. Then, its mean is:

Answer

Solution:
$\text{p}=\frac{1}{4},\sqrt{\text{npq}}=3$
$\Rightarrow\text{q}=\frac{3}{4},\text{npq}=9$
$\Rightarrow\text{Mean = np}=\frac{9}{\text{q}}$
$\Rightarrow\text{Mean}=9\times\frac{4}{3}=12$

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MATHS M.C.Q (1 Marks)