M.C.Q (1 Marks) - Page 3 - MATHS STD 12 Science Questions - Vidyadip

Questions · Page 3 of 4

M.C.Q (1 Marks)

Question 1011 Mark

For a binomial variate X, if $\text{n}=3$ and $\text{P(X}=1)=8\text{ P(X = 3}),$ then p =

$\frac{4}{5}$
$\frac{1}{5}$
$\frac{1}{3}$
$\frac{2}{3}$

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Question 1021 Mark

A random variable X has the following probability distribution:

X:	1	2	3	4	5	6	7	8
P(X):	0.15	0.23	0.12	0.10	0.20	0.08	0.07	0.05

Find the events E = {X : X is a prime number}, F{X : X < 4}, the probability $\text{P}(\text{E}\cup\text{F})$ is:

0.50
0.77
0.35
0.87

Answer

0.77

Solution:
P(E) = P(2) + P(3) + P(5) + P(7)
P(E) = 0.23 + 0.12 + 0.20 + 0.07
P(E) = 0.62
And
P(F) = P(1) + P(2) + P(3)
P(F) = 0.15 + 0.23 + 0.12
P(F) = 0.5
Also,
$\text{P}(\text{E}\cap\text{F})=\text{P}(2)+\text{P}(3)$
$\text{P}(\text{E}\cap\text{F})=0.23+0.12$
$\text{P}(\text{E}\cap\text{F})=0.35$
$\text{P}(\text{E}\cup\text{F})=\text{P}(\text{E})+\text{P(F)}-\text{P}(\text{E}\cap\text{F})$
$\text{P}(\text{E}\cup\text{F})=0.62+0.5-0.35$
$\text{P}(\text{E}\cup\text{F})=0.77$

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Question 1031 Mark

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

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

Answer

$\frac{105}{512}$

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

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Question 1041 Mark

Two events A and B will be independent, if

A and B are mutually exclusive
$\text{P}(\text{A}'\text{B}')=\big[1-\text{P}(\text{A})\big]\big[1-\text{P}(\text{B})\big]$
P(A) = P(B)
P(A) + P(B) = 1

Answer

Two events A and B are said to be independent, if $\text{P}(\text{AB})=\text{P}(\text{A})\times\text{P}(\text{B})$Distracter Rationale.

Let P(A) = m, P(B) = n, 0 < m, n < 1

A and B are mutually exclusive.
$\therefore\text{A}\cap\text{B}=\phi$
$\Rightarrow\text{P}(\text{AB})=0$
$\text{However,}\ \text{P}(\text{A})\cdot\text{P}(\text{B})=mn\neq0$
$\therefore\text{P}(\text{A}).\text{P}(\text{B})\neq\text{P}(\text{AB})$

Consider the result given in alternative.

$\text{P}(\text{A}'\text{B}')=\big[1-\text{P}(\text{A})\big]\big[1-\text{P}(\text{B})\big]$
$\Rightarrow\text{P}(\text{A}'\cap\text{B}')=1-\text{P}(\text{A})-\text{P}(\text{B})+\text{P}(\text{A}).\text{P}(\text{B})$
$\Rightarrow1-\text{P}(\text{A}\cup\text{B})=1-\text{P}(\text{A})-\text{P}(\text{B})+\text{P}(\text{A}).\text{P}(\text{B})$
$\Rightarrow\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A})\cdot\text{P}(\text{B})$
$ \Rightarrow\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{AB})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}).\text{P}(\text{B})$
$\Rightarrow\text{P}(\text{AB})=\text{P}(\text{A}).\text{P}(\text{B})$
This implies that A and B are independent, if $\text{P}(\text{A}'\text{B}')=\big[1-\text{P}(\text{A})\big]\big[1-\text{P}(\text{B})\big]$

Let A: Event of getting an odd number on throw of a die = {1, 3, 5}

$\Rightarrow\text{P}(\text{A})=\frac{3}{6}=\frac{1}{2}$
B: Event of getting an even number on throw of a die = {2, 4, 6}
$\text{P}(\text{B})=\frac{3}{6}=\frac{1}{2}$
Here, $\text{A}\cap\text{B}=\phi$
$\therefore\text{P}(\text{AB})=0 $
$\text{P}(\text{A}).\text{P}(\text{B})=\frac{1}{4}\neq0$
$ \Rightarrow\text{P}(\text{A}).\text{P}(\text{B})\neq\text{P}(\text{AB})$

From the above example, it can be seen that,

$\text{P}(\text{A})+\text{P}(\text{B})=\frac{1}{2}+\frac{1}{2}=1$
However, it cannot be inferred that A and B are independent.
Thus, the correct answer is B.

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Question 1051 Mark

Choose the correct answer in each of the following:
Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is

$\frac{37}{221}$
$\frac{5}{13}$
$\frac{1}{13}$
$\frac{2}{13}$

Answer

$\text{n}(\text{S})=52,\ \text{n}(\text{A})=4$ $\text{P}(\text{X}=0)=\frac{^{48}\text{C}_2}{^{52}\text{C}_2}=\frac{48\times47}{52\times51}=\frac{188}{221}$ $\text{P}(\text{X}=1)=\frac{^{48}\text{C}_2\times^4\text{C}_1}{^{52}\text{C}_2}=\frac{2\times48\times4}{52\times51}=\frac{32}{221}$ $\text{P}(\text{X}=2)=\frac{^4\text{C}_2}{^{52}\text{C}_2}=\frac{4\times3}{52\times51}=\frac{1}{221}$

$\text{x}_i$	$\text{p}_i$	$\text{p}_i\text{x}_i$
$0$ $1$ $2$	$\frac{188}{221}$ $\frac{32}{221}$ $\frac{1}{221}$	$0$ $\frac{32}{221}$ $\frac{2}{221}$
		$\sum\text{p}_i\text{x}_i=\frac{34}{221}=\frac{2}{13}$

$\text{Now}\ \text{E}(\text{X})=\frac{2}{13}$ Therefore, option (D) is correct.

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Question 1061 Mark

Choose the correct answer in each of the following:
If A and B are two events such that P(A) ≠ 0 and P(B | A) = 1, then

$\text{A}\subset\text{B}$
$\text{B}\subset\text{A}$
$\text{B}=\phi$
$\text{A}=\phi$

Answer

$\text{A}\subset\text{B}$
$\text{P}(\text{B|A})=1$
$\Rightarrow\ \frac{\text{P}(\text{B}\cap\text{A})}{\text{P}(\text{A})}=1\ \ \text{P}(\text{B}\cap\text{A})=\text{P}(\text{A})$
$\therefore$ (A) is correct answer.

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MCQ 1071 Mark

Probability that A speaks truth is $\frac{4}{5}.$ A coin is tossed. A reports that a head appears. The probability that actually there was head is

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

Answer

Correct option: A.

$\frac{4}{5}$

Let $A$ be the event that the man reports that head occurs in tossing a coin and let $E_{1 }$ be the event that head occurs and $E_{2 }$ be the event head does not occur.
$\text{P}(\text{E}_1)=\frac{1}{2},\ \text{P}(\text{E}_2)=\frac{1}{2}$
$\text{P}(\text{A}|\text{E}_1) = P(A$ reports that head occurs when head had actually occur red on the coin$) = \frac{4}{5}$
$\text{P}(\text{A}|\text{E}_2)= P(A $ reports that head occurs when head had not occur red on the coin$ )=1-\frac{4}{5}=\frac{1}{5}$
By Bayes’ theorem,
$ \text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)+{\text{P}(\text{E}_2)\text{P}(\text{A}|\text{E}_2)}}=\frac{\frac{1}{2}\times\frac{4}{5}}{\frac{1}{2}\times\frac{4}{5}+\frac{1}{2}\times\frac{1}{5}}=\frac{4}{4+1}=\frac{4}{5}$
Hence, option $(A)$ is correct.

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Question 1081 Mark

In each of the following, choose the correct answer:
In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is

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

Answer

The repeated selections of defective bulbs from a box are Bernoulli trials. Let X denote the number of defective bulbs out of a sample of 5 bulbs.
Probability of getting a defective bulb, p $=\frac{10}{100}=\frac{1}{10}$
$\therefore\text{q}=1-\text{p}=1-\frac{1}{10}=\frac{9}{10}$
Clearly, X has a binomial distribution with n = 5 and $\text{p}=\frac{1}{10}$
$\therefore\ \text{P}(\text{X=x})=\ ^\text{n}\text{C}_\text{x}\text{q}^\text{n-x}\text{p}^\text{x}=\ ^5\text{C}_\text{x}\bigg(\frac{9}{10}\bigg)^{5-\text{x}}\bigg(\frac{1}{10}\bigg)^\text{x}$
P(none of the bulbs is defective) = P(X = 0)
$=\ ^5\text{C}_0\cdot\Big(\frac{9}{10}\Big)^5$
$=1\cdot\Big(\frac{9}{10}\Big)^5$
$=\Big(\frac{9}{10}\Big)^5$
The correct answer is C.

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Question 1091 Mark

Choose the correct answer in each of the following:
If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then

P(B|A) = 1
P(A|B) = 1
P(B|A) = 0
P(A|B) = 0

Answer

$\text{P}(\text{A|B})=1$
$\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A and B})=\text{P}(\text{A})$
$\Rightarrow\ \text{P}(\text{B})=\text{P}(\text{A}\cap\text{B})\ \Rightarrow\ 1=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=1=\text{P}(\text{A|B})$
$\therefore$ (B) is correct answer.

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Question 1101 Mark

Choose the correct answer in each of the following:
If P(A|B) > P(A), then which of the following is correct :

P(B|A) < P(B)
P(A ∩ B) < P(A).P(B)
P(B|A) > P(B)
P(B|A) = P(B)

Answer

$\text{P}(\text{B|A})>\text{P}(\text{B})$
$\text{P}(\text{A|B})>\text{P}(\text{A})$
$\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}>\text{P}(\text{A})\ \ \Rightarrow\ \ \frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{A})}=\text{P}(\text{B})$
$\Rightarrow\ {\text{P}(\text{B}|\text{A})}>{\text{P}(\text{B})}.$
(C) is correct answer.

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Question 1111 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
$\ ^5\text{C}_\text{4}\Big(\frac{4}{5}\Big)^4\frac{1}{5}$
$\Big(\frac{4}{5}\Big)^4\frac{1}{5}$
$\ ^5\text{C}_1\frac{1}{5}\Big(\frac{4}{5}\Big)^4$
None of these

Answer

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.

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Question 1121 Mark

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

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

Answer

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

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Question 1131 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?

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

Answer

$\text{A}\subset\text{B}\ \Rightarrow\ \ \text{A}\cap\text{B}=\text{A P}\ \ \text{and}\ \ \text{P}(\text{B})\neq0$
$\Rightarrow\ \text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\frac{\text{P}(\text{A})}{\text{P}(\text{B})}$
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.

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Question 1141 Mark

Choose the correct answer from the given four options.
Suppose a random variable X follows the binomial distribution with parameters n and p, where 0 < p < 1. If $\frac{\text{P}(\text{x}=\text{r})}{\text{P}(\text{x}=\text{n}–\text{r})}$ is independent of n and r, then p equals:

$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{5}$
$\frac{1}{7}$

Answer

$\frac{1}{2}$

Solution:
$\text{P}(\text{x}=\text{r})={^\text{n}}\text{C}_\text{r}(\text{p})^\text{r}(\text{q})^{\text{n}-\text{r}}$
$=\frac{\text{n}!}{(\text{n}-\text{r})!\text{r}!}(\text{p})^\text{r}(1-\text{p})^{\text{n}-\text{r}}[\therefore\text{q}=1-\text{p}]$
Now, $\frac{\text{P}(\text{x}=\text{r})}{\text{P}(\text{x}=\text{n}-\text{r})}=\frac{{^\text{n}\text{C}_\text{r}\text{p}^\text{r}(1-\text{p})^{\text{n}-\text{r}}}}{{{^\text{n}}\text{C}}_{\text{n}-\text{r}}\text{p}^{\text{n}-\text{r}}(1-\text{p})^{\text{r}}}$
$=\frac{\text{P}^\text{r}(1-\text{P})^{\text{n}-\text{r}}}{\text{p}^{\text{n}-\text{r}}(1-\text{p})^\text{r}}$ $\big[\text{as}{^\text{n}}\text{C}_\text{r}={^\text{n}}\text{C}_{\text{n}-\text{r}}\big]$
$=\Big(\frac{1-\text{p}}{\text{p}}\Big)^{\text{n}-2\text{r}}$
Above expression is independent of n and r, if
$\frac{1-\text{p}}{\text{p}}=1\Rightarrow\text{p}=\frac{1}{2}$

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MCQ 1151 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
$\frac{^{5}\text{P}_5}{5}$

Answer

Correct 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 $=\frac{^{7}\text{P}_5}{7^5}$

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Question 1161 Mark

A fair coin is tossed 100 times. The probability of getting tails an odd nimber of times is:

$\frac{1}{2}$
$\frac{1}{8}$
$\frac{3}{8}$
$\text{None of these}$

Answer

$\frac{1}{2}$

Solution:
Here, $\text{n}=100$
Let X denote the number of times a tail is obtained.
Here, $\text{p = q}=\frac{1}{2}$
$\text{P(X = odd})=\text{P(X}=1,3,5,\dots99)$
$=\big(\text{ }^{100}\text{C}_1+\text{ }^{100}\text{C}_3+\dots+\text{ }^{100}{\text{C}}_{99}\big)\big(\frac{1}{2}\big)^{100}$
= Sum of odd coefficients in binomial expansion in $(1+\text{x})^{100}\big(\frac{1}{2}\big)^{100}$
$=\frac{2^{(100-1)}}{2^{100}}$
$=\frac{1}{2}$

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Question 1171 Mark

If two events are independent, then.

They must be mutually exclusive.
The sum of their probabilities must be equal to 1.
(a) and (b) both are correct.
None of the above is correctIf two. events are independent, then.

Answer

None of the above is correctIf two. events are independent, then.

Solution:
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).

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MCQ 1181 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
$\frac{1}{12}$

Answer

Correct 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.
$\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=1,\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=\frac{1}{20}$
$\Rightarrow\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)=\frac{\frac{1}{12}\times1}{\frac{1}{12}\times1+\frac{1}{2}\times\frac{1}{20}}=\frac{10}{13}$

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Question 1191 Mark

If A and B are two events such that $\text{P(A)}\neq0$ and $\text{P(B)}\neq1,$ then $\text{P}(\overline{\text{A}}|\overline{\text{B}})=$

$1-\text{P}(\text{A}|\text{B})$
$1-\text{P}(\overline{\text{A}}|\text{B})$
$\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\overline{\text{B}})}$
$=\frac{\text{P}(\overline{\text{A}})}{\text{P}(\overline{\text{B}})}$

Answer

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

Solution:
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).

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Question 1201 Mark

Choose the correct answer from the given four options.
The probability of guessing correctly at least 8 out of 10 answers on a true-false type examination is:

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

Answer

$\frac{7}{128}$

Solution:
We know that, $\text{P}(\text{x}=\text{r})={^\text{n}}\text{C}_\text{r}(\text{P}){^\text{r}.(\text{q})6^{\text{n}-\text{r}}}$
Here, $\text{n}=10,\text{p}=\frac{1}{2},\text{q}=\frac{1}{2}$
and $\text{r}\geq8\text{ i.e,}\text{ r}=8,9,10$
$\Rightarrow\text{P}(\text{X}=\text{r})=\text{P}(\text{r}=8)+\text{P}(\text{r}=9)+\text{P}(\text{r}=10) $
$={^{10}}\text{C}_8\Big(\frac{1}{2}\Big)^8\Big(\frac{1}{2}\Big)^{10-8}+{^{10}}\text{C}_9\Big(\frac{1}{2}\Big)^9\Big(\frac{1}{2}\Big)^{10-9}+{^{10}}\text{C}_{10}\Big(\frac{1}{2}\Big)^{10}\Big(\frac{1}{2}\Big)^{10-10}$
$=\Big(\frac{10!}{8!2!}+\frac{10!}{9!1!}+1\Big)\Big(\frac{1}{2}\Big)^{10}$
$=[45+10+1]\Big(\frac{1}{2}\Big)^{10}$
$=56\Big(\frac{1}{2}\Big)^{10}=\frac{7}{128}$

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MCQ 1211 Mark

A die is thrown and a card is selected ar random from a deck 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
$\frac{3}{4}$

Answer

Correct 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 $= P(A) \times P(B)$
$=\frac{1}{2}\times\frac{1}{4}=\frac{1}{8}$

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Question 1221 Mark

Choose the correct answer from the given four options.
Two events E and F are independent. If $\text{P}(\text{E})=0.3,\text{P}(\text{E}\cup\text{F})=0.5,$ then $\text{P}\Big(\frac{\text{E}}{\text{F}}\Big)-\text{P}\Big(\frac{\text{F}}{\text{E}}\Big)$ equal:

$\frac{2}{7}$
$\frac{3}{35}$
$\frac{1}{70}$
$\frac{1}{7}$

Answer

$\frac{1}{70}$

Solution:
We have, $\text{P}(\text{E})=0.3,\text{P}(\text{E}\cup\text{F})=0.5$
Also E and F are independent.
Now $\text{P}(\text{E}\cup\text{F})=\text{P}(\text{E})+\text{P}(\text{F})-\text{P}(\text{E}\cap\text{F})$
$\Rightarrow0.5=0.3+\text{P}(\text{F})-0.3\text{P}(\text{F})$
$\Rightarrow\text{P}(\text{F})=\frac{0.5-0.3}{0.7}=\frac{2}{7}$
$\therefore\text{P}\Big(\frac{\text{E}}{\text{F}}\Big)-\text{P}\Big(\frac{\text{F}}{\text{E}}\Big)$
$=\text{P}(\text{E})-\text{P}(\text{F})$ (as E and F are indepandent)
$=\frac{3}{10}-\frac{2}{7}=\frac{1}{70}$

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Question 1231 Mark

Choose the correct answer from the given four options.If two events are independent, then:

They must be mutually exclusive.
The sum of their probabilities must be equal to 1.
(a) and (b) both are correct.
None of the above is correct.

Answer

None of the above is correct.

Solution:
If two events A and B are independent, then we know that
$\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A})\cdot\text{P}(\text{B}),\text{P}(\text{A})\neq0,\text{P}(\text{B})\neq0$
Since, A and B have a common outcome.
Further, mutually exclusive events never have a common outcome.
In other words, two independents events having non-zero probabilities of occurrence cannot be mutually exclusive and conversely, i.e., two mutually exclusive events having non-zero probabilities of outcome cannot be independent.

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Question 1241 Mark

If A and B are independent events such that P(A) > 0 and P(B) > 0, then.

$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A}).\text{P}(\text{B})$
$\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A}).\text{P}(\text{B})$
$\text{P}\big(\frac{\text{A}}{\text{B}}\big)=\text{P}(\text{A})$
$\text{P}\big(\frac{\text{B}}{\text{A}}\big)=\text{P}(\text{B})$

Answer

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

Solution:
Since, A and B are independent events
$\therefore\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A}).\text{P}(\text{B})$
$\text{P}\big(\frac{\text{A}}{\text{B}}\big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\text{P}(\text{A})$
and $\text{P}\big(\frac{\text{B}}{\text{A}}\big)=\frac{\text{P}(\text{B}\cap\text{A})}{\text{P}(\text{A})}=\text{P}(\text{B})$

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MCQ 1251 Mark

Choose the correct answer from the given four options. A die is thrown and a card is selected at random from a deck of $52$ playing cards. The probability of getting an even number on the die and a spade card is$:$

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

Answer

Correct option: C.

$\frac{1}{8}$

Let $E_{1 }=$ Event for getting an even number on the die
And $E_{2 }=$ Event that a spade card is selected.
$\therefore\text{P}(\text{E}_1)=\frac{3}{6}=\frac{1}{2}$ and $\text{P}(\text{E}_2)=\frac{13}{52}=\frac{1}{4}$
Then, $\text{P}(\text{E}_1\cap\text{E}_2)=\text{P}(\text{E}_1)\cdot\text{P}(\text{E}_2)$
$=\frac{1}{2}\cdot\frac{1}{4}$
$=\frac{1}{8}$

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MCQ 1261 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
$\frac{1}{9}$

Answer

Correct 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$) + .....$
$=\frac{8}{9}\times\frac{1}{9}+\frac{8}{9}\times\frac{8}{9}\times\frac{8}{9}\times\frac{1}{9}+\ .....$
$=\frac{8}{81}\Big[1+\frac{64}{81}+\Big(\frac{64}{81}\Big)^2+\ ......\Big]$
$=\frac{8}{81}\times\frac{1}{1-\frac{64}{81}}$
$=\frac{8}{81}\times\frac{81}{17}$
$=\frac{8}{17}$

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Question 1271 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.

$\frac{1}{10}$
$\frac{3}{10}$
$\frac{3}{8}$
$\frac{6}{7}$

Answer

$\frac{3}{8}$

Solution:
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
$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)=\frac{\text{P}(\overline{\text{A}\cap\text{B}})}{\text{P}(\overline{\text{B}})}$
$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)=\frac{0.3}{0.8}$
$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)=\frac{3}{8}$

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Question 1281 Mark

Choose the correct answer from the given four options.If $\text{P}(\text{A})=\frac{2}{5},\text{P}(\text{B})=\frac{3}{5}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{5},$ then $\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)\cdot\text{P}\Big(\frac{\text{B}'}{\text{A}'}\Big)$ is equas:

$\frac{5}{6}$
$\frac{5}{7}$
$\frac{25}{42}$
$1$

Answer

$\frac{5}{6}$

Solution:
Here, $\text{P}(\text{A})=\frac{2}{5},\text{P}(\text{B})=\frac{3}{5}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{5}$
$\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)=\frac{\text{P}(\text{A}'\cap\text{B}')}{\text{P}(\text{B}')}+\frac{1-\text{P}(\text{A}\cap\text{B})}{1-\text{P}(\text{B})}$
$=\frac{1-\big[\text{P}(\text{A})+\text{P}(\text{B})-\text{A}(\text{A}\cap\text{B})\big]}{1-\text{P}(\text{B})}$
$=\frac{1-\Big(\frac{2}{5}+\frac{3}{10}-\frac{1}{5}\Big)}{1-\frac{3}{10}}$
$=\frac{1-\Big(\frac{4+3-2}{10}\Big)}{\frac{7}{10}}-\frac{1-\frac{1}{2}}{\frac{7}{10}}$
$=\frac{5}{7}$
And $\text{P}\Big(\frac{\text{B}'}{\text{A}'}\Big)=\frac{\text{P}(\text{B}'\cap\text{A}')}{\text{P}(\text{A}')}$
$=\frac{1-\text{P}(\text{A}\cup\text{B})}{1-\text{P}(\text{A})}$
$=\frac{1-\frac{1}{2}}{1-\frac{2}{5}}$ $\Big[\because\text{P}(\text{A}\cup\text{B})=\frac{1}{2}\Big]$
$=\frac{\frac{1}{2}}{\frac{3}{5}}=\frac{5}{6}$
$\therefore\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)\cdot\text{P}\Big(\frac{\text{B}'}{\text{A}'}\Big)=\frac{5}{7}\cdot\frac{5}{6}=\frac{25}{42}$

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Question 1291 Mark

Choose the correct answer from the given four options.Which one is not a requirement of a binomial distribution?

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

Answer

The outcomes must be dependent on each othere.

Solution:
We know that, in a Binomial distribution:

There are 2 outcomes of each trail.
There is a fixed number of trails.
The probability of success must be the same for all the trails.

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Question 1301 Mark

Choose the correct answer from the given four options.
You are given that A and B are two events such that $\text{P}(\text{B})=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then P(A) equals:

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

Answer

$\frac{1}{2}$

Solution:
We have, $\text{P}(\text{B})=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5}$
$\therefore\text{P}(\text{A}\cap\text{B})=\Big(\frac{\text{A}}{\text{B}}\Big)\cdot\text{P}(\text{B})$
$=\frac{1}{2}\cdot\frac{3}{5}=\frac{3}{10}$
Now $\text{P}(\text{A}\cup\text{B})=\text{P}({\text{A}})+\text{P}({\text{B}})\cdot\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\frac{4}{5}=\text{P}(\text{A})+\frac{3}{5}-\frac{3}{10}$
$\therefore\text{P}(\text{A})=\frac{4}{5}-\frac{3}{5}+\frac{3}{10}=\frac{1}{2}$

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Question 1311 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
$\frac{1}{16}$
0.1125
None of these.

Answer

0.0875

Solution:
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 × 0.5 × 0.5 × 0.5)
1 points + 2 points + 2 points + 2 points = (0.05 × 0.5 × 0.5 × 0.5)
2 points + 1 points + 2 points + 2 points = (0.5 × 0.05 × 0.5 × 0.5)
2 points + 2 points + 1 points + 2 points = (0.5 × 0.5 × 0.05 × 0.5)
2 points + 2 points + 2 points + 1 points = (0.5 × 0.5 × 0.5 × 0.05)

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Question 1321 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 $\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=$

$\frac{9}{10}$
$\frac{10}{9}$
$\frac{8}{9}$
$\frac{9}{8}$

Answer

$\frac{10}{9}$

Solution:
$\text{P}(\text{A}\cup\text{B})=\frac{5}{9},\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3}$
Consider,
$\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$\Rightarrow\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3}$
$\Rightarrow 1-\text{P}(\text{A}\cap\text{B})=\frac{2}{3}$
$\Rightarrow\text{P}(\text{A}\cap\text{B})=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}-\frac{5}{9}=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}=\frac{8}{9}$
$\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=1-\text{P(A)}+1-\text{P(B)}$
$\Rightarrow\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=2-\big[\text{P(A)}+\text{P(B)}\big]$
$\Rightarrow\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=2-\frac{8}{9}$
$\Rightarrow\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=\frac{10}{9}$

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MCQ 1331 Mark

A bouquet from $11$ different flowers is to be made so that it contains not less then three flowers. Then the number of the different ways of selecting flowers to from the bouquet.

A
$1972$
B
$1952$
✓
$1981$
D
$1947$

Answer

Correct option: C.

$1981$

No. of ways $=\ ^{11}C_3 +\ ^{11}C_4 +\ ^{11}C_5 +\ ^{11}C_6 +\ ^{11}C_7 +\ ^{11}C_8 +\ ^{11}C_9 +\ ^{11}C_{10} +\ ^{11}C_{11}$
$\Rightarrow 165 + 330 + 462 + 462 + 330 + 165 + 55 + 11 + 1$
$\Rightarrow 1981$

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MCQ 1341 Mark

Choose the correct answer from the given four options.
Two dice are thrown. If it is known that the sum of numbers on the dice was less than $6,$ the probability of getting a sum $3,$ is:

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

Answer

Correct option: C.

$\frac{1}{5}$

Let $E_1 =$ Event that the sum of numbers on the dice was less than $6$
And $E_2=$ Event that the sum of numbers on the dice is 3.
$\therefore E_1 = \{(1, 4), (4, 1), (2, 3), (3, 2), (2, 2), (1, 3), (3, 1), (1, 2), (2, 1), (1, 1)\}$
$\Rightarrow n(E_1) = 10$
And $E_2 = \{(1, 2), (2, 1)\}$
$\Rightarrow n(E_2) = 2$
$\therefore$ Required Probability $=\frac{2}{10}=\frac{1}{5}$

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Question 1351 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

$\frac{167}{168}$
$\frac{1}{28}$
$\frac{2}{21}$
$\frac{3}{28}$

Answer

$\frac{3}{28}$

Solution:
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.
$\text{n(A)}={^{3}}\text{C}_2\times{^{2}}\text{C}_2=6$
$\text{P(A)}=\frac{6}{56}=\frac{3}{28}$

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Question 1361 Mark

A box contain 100 pens of which 10 are defective. What is the probability that out of a sample of 5 pens draws one by one with replacement at most one is defective?

$\big(\frac{9}{10}\big)^5$
$\frac{1}{2}\big(\frac{9}{10}\big)^4$
$\frac{1}{2}\big(\frac{9}{10}\big)^5$
$\big(\frac{9}{10}\big)^5+\frac{1}{2}\big(\frac{9}{10}\big)^4$

Answer

$\big(\frac{9}{10}\big)^5+\frac{1}{2}\big(\frac{9}{10}\big)^4$

Solution:
$\text{p}=\frac{10}{100}=\frac{1}{10},\text{q}=\frac{90}{100}=\frac{9}{10},\text{n}=5$
$\text{P(X}\leq1)=\text{P(0)}+\text{P(1)}$
$\text{P(X}\leq1)=\big(\frac{9}{10}\big)^{5}+\text{ }^5\text{C}_1\big(\frac{1}{10}\big)\big(\frac{9}{10}\big)^{4}$
$\text{P(X}\leq1)=\big(\frac{9}{10}\big)^{5}+\big(\frac{1}{2}\big)\big(\frac{9}{10}\big)^4$

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Question 1371 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

$\frac{3}{16}$
$\frac{5}{16}$
$\frac{11}{16}$
$\frac{14}{16}$

Answer

$\frac{11}{16}$

Solution:
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)
$=\frac{8}{16}+\frac{6}{16}-\frac{3}{16}$
$=\frac{8+6-3}{16}$
$=\frac{11}{16}$

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Question 1381 Mark

Let X denote the number of times heads occur in n tosses of a fair coin. If P(X = 4), P(X = 5) and P(X = 6) are in AP, the value of n is:

7, 14
10, 14
12, 7
14, 12

Answer

7, 14

Solution:
X denotes the number of times heads occurs.
P(X = 4),P(X = 5),P(X = 6) are in AP
$\Rightarrow2\text{P(X = 4),P(X = 5),P(X = 6)}$
$\Rightarrow2\text{ }^{\text{n}}\text{C}_5\big(\frac{1}{2}\big)^5\big(\frac{1}{2}\big)^{\text{n}-5}=\text{ }^{\text{n}}\text{C}_4\big(\frac{1}{2}\big)^4\big(\frac{1}{2}\big)^{\text{n}-4}\times\text{ }^{\text{n}}\text{C}_6\big(\frac{1}{2}\big)^6\big(\frac{1}{2}\big)^{\text{n}-6}$
$\Rightarrow2\text{ }^{\text{n}}\text{C}_5\big(\frac{1}{2}\big)^{\text{n}}=\text{ }^{\text{n}}\text{C}_4\big(\frac{1}{2}\big)^{\text{n}}+\text{ }^{\text{n}}\text{C}_6\big(\frac{1}{2}\big)^{\text{n}}$
$\Rightarrow2\text{ }^{\text{n}}\text{C}_5=\text{ }^{\text{n}}\text{C}_4+\text{ }^{\text{n}}\text{C}_6$
$\Rightarrow\frac{2\text{n}!}{5!(\text{n}-5)!}=\frac{\text{n}!}{4!(\text{n}-4)!}+\frac{\text{n}!}{6!(\text{n}-6)!}$
$\Rightarrow\frac{2}{5\times4!(\text{n}-5)(\text{n}-6)!}=\frac{1}{4!(\text{n}-4)(\text{n}-5)(\text{n}-6)!}+\frac{1}{6\times5\times4!(\text{n}-6)!}$
$\Rightarrow\frac{2}{5(\text{n}-5)}=\frac{1}{(\text{n}-4)(\text{n}-5)}+\frac{1}{6\times5}$
$\Rightarrow\frac{2}{5(\text{n}-5)}=\frac{30+(\text{n}-4)(\text{n}-5)}{30(\text{n}-4)(\text{n}-5)}$
$\Rightarrow12(\text{n}-4)=30+(\text{n}-4)(\text{n}-5)$
$\Rightarrow12(\text{n}-4)-(\text{n}-4)(\text{n}-5)=30$
$\Rightarrow(\text{n}-4)(12-\text{n}+5)=30$
$\Rightarrow(\text{n}-4)(17-\text{n})=30$
Check with options by putting value of n.
$\Rightarrow\text{n}=7,14$

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Question 1391 Mark

If the events A and B are independent, then $\text{P}(\text{A}\cap\text{B})$ is equal to,

P(A) + P(B)
P(A) - P(B)
P(A) P(B)
$\frac{\text{P(A)}}{\text{P(B)}}$

Answer

P(A) P(B)

Solution:
$\text{P}(\text{A}\cap\text{B})=\text{P(A)} \text{ P(B)}$ for independent events.

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Question 1401 Mark

If the random variable X has the following distribution:

X:	0	1	2	3	4	5	6	7	8
P(X):	a	3a	5a	7a	9a	11a	13a	15a	17a

then the value of a is:

$\frac{7}{81}$
$\frac{5}{81}$
$\frac{2}{81}$
$\frac{1}{81}$

Answer

$\frac{1}{81}$

Solution:
We know that the sum of probsabilities in a probability distribution is always 1.
$\therefore$ P(X = 0) + 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
⇒ a + 3a+ 5a+ 7a+ 9a + 11a + 13a + 15a + 17a = 1
⇒ 81a = 1
$\Rightarrow\text{a}=\frac{1}{81}$

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Question 1411 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) =

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

Answer

$\frac{14}{17}$

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

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Question 1421 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 =

$\frac{2}{3}$
$\frac{3}{5}$
$\frac{1}{3}$
$\frac{3}{4}$

Answer

$\frac{1}{3}$

Solution:
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{p},\text{P(A)}=\text{p},\text{P(B)}=\frac{1}{3},\text{P}(\text{A}\cup\text{B})=\frac{5}{9}$
Consider,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{p}$
$\Rightarrow \frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\text{P}$
$\Rightarrow \frac{\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})}{\text{P(B)}}=\text{P}$
$\Rightarrow\frac{\text{p}+\frac{1}{3}-\frac{5}{9}}{\frac{1}{3}}=\text{p}$
$\Rightarrow\text{p}+\frac{1}{3}-\frac{5}{9}=\frac{\text{p}}{3}$
$\Rightarrow\frac{-2}{9}=\frac{\text{p}}{3}-\text{p}$
$=\frac{-2}{3}\text{p}=\frac{-2}{9}$
$\Rightarrow\text{p}=\frac{1}{3}$

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Question 1431 Mark

Choose the correct answer from the given four options:
let $\text{P}(\text{A})=\frac{7}{13},\text{P}(\text{B})=\frac{9}{13}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{13}.$ Then $\text{P}\Big(\frac{\text{A'}}{\text{B}}\Big)$ is equal to:

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

Answer

$\frac{5}{9}$

Solution:
Here, $\text{P}(\text{A})=\frac{7}{13},\text{P}(\text{B})=\frac{9}{13}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{13}$
$\therefore\text{P}\Big(\frac{\text{A}'}{\text{B}}\Big)=\frac{\text{P}(\text{A}'\cap\text{B})}{\text{P}(\text{B})}=\frac{\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$=\frac{\frac{9}{13}-\frac{4}{13}}{\frac{9}{13}}=\frac{\frac{5}{13}}{\frac{9}{13}}=\frac{5}{9}$

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Question 1441 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,

$\frac{3}{28}$
$\frac{1}{14}$
$\frac{9}{64}$
$\frac{33}{56}$

Answer

$\frac{3}{28}$

Solution:
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).

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Question 1451 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

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

Answer

$\frac{1}{2}$

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

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Question 1461 Mark

Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?

9, 7, 4, 0
0, 2, 4, 6
6, 7, 7, 2
6, 4,2, 0

Answer

0, 2, 4, 6

Solution:
A coin is tossed six times and X represents the difference between the number of heads and the number of tails.
$\therefore$ X(6H, 0T)=∣6 - 0∣ = 6
X(5H, 1T) = ∣5 - 1∣ = 4
X(4H, 2T) = ∣4 - 2∣ = 2
X(3H, 3T) = ∣3 - 3∣ = 0
X(2H, 4T) = ∣2 - 4∣ = 2
X(1H, 5T) = ∣1 - 5∣ = 4
X(0H, 6T) = ∣0 - 6∣ = 6
Thus, the possible values of X are 0, 2, 4 and 6.

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MCQ 1471 Mark

Fifteen coupons are numbered $1$ to $15.$ Seven coupons are selected at random one at a time with replacement. The probability that the largest number appearing on a selected coupon is $9$ is:

A
$\big(\frac{3}{7}\big)^7$
B
$\big(\frac{1}{15}\big)^7$
C
$\big(\frac{8}{15}\big)^7$
✓
$\text{None of these}$

Answer

Correct option: D.

$\text{None of these}$

The sample space $= 15^7$ for selecting seven coupons from $15$ coupons.
Maximum number on selected coupon is $9$ can be made by $9^7$ ways.
A number selected on second card is less than $9$ can be made by $8^7$ ways.
Required probability $=\frac{9^7-8^7}{15^7}$

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Question 1481 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,

$\frac{64}{64}$
$\frac{49}{64}$
$\frac{40}{64}$
$\frac{24}{64}$

Answer

$\frac{64}{64}$

Solution:
P(good item) $=\frac{10}{16}$
P(defected item) $=\frac{6}{16}$
P(eitherr good or defected item) = P(good item) + P(defected item)
$=\frac{10}{16}+\frac{6}{16}$
$=\frac{16}{16}$
$=1$
$=\frac{64}{64}$

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Question 1491 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.

$\frac{1}{12}$
$\frac{3}{4}$
$\frac{1}{4}$
$\frac{3}{16}$

Answer

$\frac{1}{4}$

Solution:
$\text{P(A)}=\frac{1}{2},\text{P(B)}=\frac{1}{3},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{4}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{4}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{1}{4}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\frac{1}{3}}=\frac{1}{4}$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{12}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{1}{12}}{\frac{1}{3}}=\frac{1}{4}$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\text{P}(\text{A}\cup\text{B})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=-1\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\Big[\frac{1}{2}+\frac{1}{3}-\frac{1}{12}\Big]$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\frac{1}{4}$

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Question 1501 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

$\frac{44}{85\times49}$
$\frac{11}{85\times49}$
$\frac{13\times24}{17\times25\times49}$
None of these.

Answer

$\frac{44}{85\times49}$

Solution:
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.
$=4\times\frac{13}{52}\times\frac{12}{51}\times\frac{11}{50}\times\frac{10}{59}$
$=4\times\frac{11}{85\times49}$
$=\frac{44}{85\times49}$

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M.C.Q (1 Marks) - Page 3 - MATHS STD 12 Science Questions - Vidyadip