MCQ - Maths STD 11 Science Questions - Vidyadip

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40 questions · 39 auto-graded MCQ + 1 self-marked written.

MCQ 11 Mark

If the probability for $A$ to fail in an examination is $0.2$ and that for $B$ is $0.3,$ then the probability that either $A$ or $B$ fails is :

A
$>0.5$
B
$0.5$
✓
$\leq0.5$
D
$0$

Answer

Correct option: C.

$\leq0.5$

Let $X$ and $Y$ be two events given by,
$X : A$ fails in an examination
$Y : B$ fails in an examination
$P(A$ fails$) = P(X) = 0.2$
$P(B$ fails$) = P(Y) = 0.3$
Now, $P($either $A$ or $B$ fails$) =\text{P}(\text{X}\cup\text{Y})$
We know that,
$=\text{P}(\text{X}\cup\text{Y})\leq\text{P(X)}+\text{P()Y}=0.2+0.3=0.5$
$\Rightarrow\text{P}(\text{X}\cup\text{Y})\leq0.5$
$\therefore\text{P}\text{(either A or B fails)}\leq0.5$
Hence, the correct answer is option $(c)$.

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MCQ 21 Mark

The probabilities of three mutually exclusive events $A, B$ and $C$ are given by $\frac{2}{3}$, $\frac{1}{4}$ and $\frac{1}{6}$respectively. The statement

A
Is true.
✓
Is false.
C
Nothing can be said.
D
Could be either.

Answer

Correct option: B.

Is false.

Since the events $A, B$ and $C$ are mutually exclusive, we have:
$\text{P}(\text{A}\cup\text{B}\cup\text{C})$
$=\frac{2}{3}+\frac{1}{4}+\frac{1}{6}=\frac{13}{12}>1$
which is not possible.
Hence, the given statement is false.

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MCQ 31 Mark

If $S$ is the sample space and $ \text{P(A)} = \frac{1}{3} \text{P(B)}$ and $\text{S} = \text{A}\cup\text{B}$ where $A$ and $B$ are two mutually exclusive events, then $P (A) =$

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

Answer

Correct option: A.

$\frac{1}{4}$

Let $\text{P(B)}=\text{P}$
Than $\text{P(A)}=\frac{1}{3}\text{P}$
Since $A$ and $B$ are two mutually exclusive events, we have :
$\text{A}\cup\text{B}=\text{S}$
$\Rightarrow\text{P}\text{(A}\cup\text{B)}=\text{P}\text{(S)}$
$\Rightarrow\text{P}\text{(A}\cup\text{B)}=1$
$\Rightarrow\text{P}\text{(A)}+\text{(B)}=1$
$\Rightarrow\frac{1}{2}\text{P}+\text{P}=1$
$\Rightarrow\frac{4\text{p}}{3}=1$
$\therefore\text{P(A)}=\frac{1}{3}\text{P}$
$=\frac{1}{3}\times\frac{3}{4}=\frac{1}{4}$

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MCQ 41 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 :

✓
$\frac{16}{16}$
B
$\frac{49}{64}$
C
$\frac{40}{64}$
D
$\frac{24}{64}$

Answer

Correct option: A.

$\frac{16}{16}$

Let $A$ be the event of drawing one good article whereas $B$ be the event of drawing one defected article.
Here,
$\text{P(A)}=\frac{10}{10+6}=\frac{10}{16}$
$\text{P(B)}=\frac{6}{10+6}=\frac{6}{16}$
The events $A$ and $B$ are mutually exclusive.
Thus, the required probability,
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}$
$\Rightarrow\text{P}(\text{A}\cup\text{B})=\frac{10}{16}+\frac{6}{16}=\frac{16}{16}=1$
Hence, the correct option is $(a)$.

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MCQ 51 Mark

Without repetition of the numbers, four digit numbers are formed with the numbers $\{0, 2, 3, 5\}$. The probability of such a number divisible by $5$ is:

A
$\frac{1}{5}$
B
$\frac{4}{5}$
C
$\frac{1}{30}$
✓
$\frac{5}{9}$

Answer

Correct option: D.

$\frac{5}{9}$

The given digits are $\{0, 2, 3\}$ and $5$.

Thousands

Hundreds

Tens

Ones

Now, there are $3$ ways to fill the thousands place $(0$ cannot occupy the thousands place$), 3$ ways to fill the hundreds place, $2$ ways to fill the tens place and $1$ way to fill the ones place.
Total number of four digit numbers formed $= 3 \times 3 \times 2 \times 1 = 18$
We know that a number is divisible by $5$ if it ends in $0$ or $5$.
When $0$ is at the ones place,
Number of four digits numbers divisible by $5$ formed $= 3 \times 2 \times 1 = 6$
When $5$ is at the ones place,
Number of four digits numbers divisible by $5$ formed $= 2 × 2 × 1 = 4 (0$ cannot occupy the thousands place$)$
Total number of four digit numbers divisible by $5 = 6 + 4 = 10$
$\therefore P($four digit number formed is divisible by $5)$
$=\frac{\text{Total Number of four digit numbers divisible by 5}}{\text{Total Number of w4 digit numbers formed}}$
$=\frac{10}{18}=\frac{5}{9}$
Hence, the correct answer is option $(d)$.

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MCQ 61 Mark

Two dice are thrown together. The probability that at least one will show its digit greater than $3$ is :

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

Answer

Correct option: B.

$\frac{3}{4}$

When two dice are thrown, there are $(6 \times 6) = 36$ outcomes.
The set of all these outcomes is the sample space, given by
$S = \{(1, 1) , (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)\}$
$\{(2, 1) , (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)\}$
$\{(3, 1) , (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)\}$
$\{(4, 1) , (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\}$
$\{(5, 1) , (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\}$
$\{(6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$
i.e. $n(S) = 36$
Let $E$ be the event of getting at least one digit greater than $3$.
Then$ E = \{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 1) , (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\},$
$\{(5, 1) , (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}$
$\therefore\ \text{n(E)}=27$
Hence, required probability $=\frac{27}{36}=\frac{3}{4}$

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MCQ 71 Mark

If $A$ and $B$ are mutually exclusive events then :

✓
$\text{P(A)}\leq\text{P}(\overline{\text{B}})$
B
$\text{P(A)}\geq\text{P}(\overline{\text{B}})$
C
$\text{P(A)}<\text{P}(\overline{\text{B}})$
D
None of these

Answer

Correct option: A.

$\text{P(A)}\leq\text{P}(\overline{\text{B}})$

It is given that $A$ and $B$ are mutually exclusive events.
We know that,
$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A)}+\text{(B})-\text{P}(\text{A}\cap\text{B}) \big[$From $(1) \big]$
$\Rightarrow\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A)}+\text{P}\text{(B})$
$\big[\text{P}(\text{A}\cup\text{B})\leq1\big]$
$\Rightarrow\text{P}(\text{A)}+\text{P}\text{(B})\leq1$
$\Rightarrow\text{P}(\text{A)}\leq1-\text{P}\text{(B})=\text{P}(\overline{\text{B}})$
$\therefore\text{P}(\text{A)}\leq\text{P}\text{(B})$
Hence, the correct answer is option $(a)$.
$\therefore\text{P}(\text{A }\cap\text{B})=0\ ...(1)$

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MCQ 81 Mark

If $A, B, C$ are three mutually exclusive and exhaustive events of an experiment such that $3$
$\text{P(A)}=2\text{P(B)}=\text{C},$ then $P(A)$ is equal to :

A
$\frac{1}{11}$
✓
$\frac{2}{11}$
C
$\frac{5}{11}$
D
$\frac{6}{11}$

Answer

Correct option: B.

$\frac{2}{11}$

Let $3 P(A) = 2 P(B) = P(C) = p.$
Then, $\text{P(A)}=\frac{\text{P}}{3},\text{P(B)}=\frac{\text{P}}{2}$ and $\text{P(C)}=\text{P}$
It is given that $A, B, C$ are three mutually exclusive and exhaustive events.
$\therefore\text{P(A)}+\text{P(B)}+\text{P(C)}=1$
$=0 $ and $\big[\text{P}(\text{A}\cup\text{B)=}\text{P}(\text{B}\cap\text{C)}$
$=\text{P}(\text{C}\cap\text{A)=}\text{P}(\text{A}\cup\text{B }\cup\text{C}=1\big]$
$\Rightarrow\frac{\text{P}}{3}+\frac{\text{p}}{2}+\text{P}=1$
$\Rightarrow\frac{\text{11P}}{6}=1$
$\Rightarrow\text{P}=\frac{6}{11}$
$\therefore\text{P(A)}=\frac{\text{P}}{3}=\frac{\frac{6}{11}}{3}=\frac{2}{11}$
Hence, the correct answer is option $(b)$.

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MCQ 91 Mark

Three numbers are chosen from $1$ to $20$. The probability that they are not consecutive is :

A
$\frac{186}{190}$
✓
$\frac{187}{190}$
C
$\frac{188}{190}$
D
$\frac{18}{\ ^{20}\text{C}_3}$

Answer

Correct option: B.

$\frac{187}{190}$

Number of ways to choose three numbers from $1$ to $20 =\ ^{20}\text{C}_3=1140$
Now, the set of three consecutive numbers from $1$ to $20$ are $\{(1, 2, 3), (2, 3, 4), (3, 4, 5), ...., (18, 19, 20)\}$.
So, the number of ways to choose three numbers from $1$ to $20$ such that they are consecutive is $18$.
$P\ ($three numbers choosen are consecutive$)$
$=\frac{\text{Number of ways to choose three consecutive numbers from 1 to 20 }}{\text{Number of ways to choose three numbers from 1 to 20}}$
$=\frac{18}{\ ^{20}\text{c}_3}=\frac{18}{1140}=\frac{3}{190}$
$\therefore P\ ($three numbers choosen are not consecutive$) = 1 - P\ ($three numbers choosen are consecutive$) =1-\frac{3}{190}=\frac{187}{190}$
Hence, the correct answer is option $(b).$

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MCQ 101 Mark

Three digit numbers are formed using the digits $\{0, 2, 4, 6, 8\}$. A number is chosen at random out of these numbers. What is the probability that this number has the same digits?

A
$\frac{1}{16}$
B
$\frac{16}{25}$
C
$\frac{1}{645}$
✓
$\frac{1}{25}$

Answer

Correct option: D.

$\frac{1}{25}$

The given digits are $\{0, 2, 4, 6, 8\}$.

____	____	____
Hundreds	Tens	Ones

Now, there are $4$ ways to fill the hundreds place $(0$ cannot occupy the hundreds place$), 5$ ways to fill the tens place and $5$ ways to fill the ones place.
Total number of $3$ digit numbers formed using the given digits $= 4 \times 5 \times 5 = 100$
The three digit numbers formed using given digits that have the same digits are $222, 444, 666$ and $888.$
Number of $3 $ digit numbers that have the same digits $= 4$
$\therefore P\ ($three digit number formed has the same digits$) \frac{\text{Number of 3 digits numbers that have the same digits}}{\text{Total number of 3 digit numbers formed using the given digits}} =\frac{4}{100}=\frac{1}{25}$
Hence, the correct answer is option $(d)$.

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MCQ 111 Mark

One card is drawn from a pack of $52$ cards. The probability that it is the card of a king or spade is :

A
$\frac{1}{26}$
B
$\frac{3}{26}$
✓
$\frac{4}{13}$
D
$\frac{3}{13}$

Answer

Correct option: C.

$\frac{4}{13}$

If $A$ and $B$ denote the events of drawing a king and a spade card, respectively, then event $A$ consists of four sample points, whereas event $B$ consists of $13$ sample points.
Thus, $\text{P(A)}=\frac{4}{52}$ and $\text{P(B)}=\frac{13}{52}$
The compound event $(A \cap B)$ consists of only one sample point, king of spade.
So, $\text{P}(\text{A}\cap\text{B})=\frac{1}{52}$
By addition theorem , we have:
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=\frac{4}{52}+\frac{13}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}$
Hence, the probability that the card drawn is either a king or a spade is given by $\frac{4}{13}.$

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MCQ 121 Mark

Two dice are thrown simultaneously. The probability of obtaining total score of seven is :

A
$\frac{5}{36}$
✓
$\frac{6}{36}$
C
$\frac{7}{36}$
D
$\frac{8}{36}$

Answer

Correct option: B.

$\frac{6}{36}$

When two dices are thrown, there are $(6 \times 6) = 36$ outcomes.
The set of all these outcomes is the sample space given by
$S = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)\}$
$\{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)\}$
$\{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)\}$
$\{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\}$
$\{(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\}$
$\{(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$
$\therefore n(S) = 36$
Let $E$ be the event of getting a total score of $7$.
Then $E = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\}$
$\therefore n(E) = 6$
Hence, required probability $=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}=\frac{6}{36}$

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MCQ 131 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.

$\frac{1}{36}$

When two dice are thrown simultaneously, the sample space associated with the random experiment is given by :
$S = \{(1, 1), (1, 2), (1, 3), (6, 4), (6, 5), (6, 6)\}$
Clearly, total number of elementary events $= 36$
Let $A$ be the event of getting a pair of aces.
Then $A = \{(1, 1)\}$
$\therefore\text{n(A)}=1$
Hence, required probability $=\frac{\text{n(A)}}{\text{n(S)}}=\frac{1}{36}$

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MCQ 141 Mark

A pack of cards contains $4$ aces, $4$ kings, $4$ queens and $4$ jacks. Two cards are drawn at random. The probability that at least one of them is an ace is

A
$\frac{1}{5}$
B
$\frac{3}{16}$
✓
$\frac{9}{20}$
D
$\frac{1}{9}$

Answer

Correct option: C.

$\frac{9}{20}$

We have :
$\text{P(both are aces)}=\frac{\ ^{4}\text{C}_2}{\ ^{16}\text{C}_2}$
$=\frac{4}{16}\times\frac{3}{15}=\frac{1}{20}$
$\text{P(one are ace)}=\frac{\ ^{4}\text{C}_1\times\ ^{12}\text{C}_1}{\ ^{16}\text{C}_2}=\frac{2}{5}$
$\therefore\text{P(at least one are ace)}=\frac{1}{20}+\frac{2}{5}=\frac{9}{20}$

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MCQ 151 Mark

If $\frac{(1-3\text{P})}{2},\frac{(1+4\text{P})}{3},\frac{(1+\text{P})}{6}$ are the probabilities of three mutually exclusive and exhaustive events, then the set of all values of $p$ is :

A
$(0,1)$
✓
$\Big(\frac{-1}{4},\frac{1}{3}\Big)$
C
$\big(0,\frac{1}{3}\big)$
D
$(0,\infty)$

Answer

Correct option: B.

$\Big(\frac{-1}{4},\frac{1}{3}\Big)$

$\text{P(A)}=\frac{(1-3\text{P})}{2}$
$\text{P(B)}=\frac{(1+4\text{P})}{3}$
$\text{P(C)}=\frac{(1+\text{P})}{6}$
The events are mutually exclusive and exhaustive.
$\therefore\text{P}(\text{A}\cup\text{B }\cup\text{C})=\text{P(A)}+\text{P(B)}+\text{P(C)}=1$
$\Rightarrow0\leq\text{P(A)}\leq1,0\leq\text{P(B)}\leq1,0\leq\text{P(C)}\leq1$
$\Rightarrow0\leq\frac{1-3\text{P}}{2}\leq1$, $0\leq\frac{1-4\text{P}}{3}\leq1, 0\leq\frac{1-\text{P}}{6}\leq1$
$\Rightarrow\frac{-1}{3}\leq\text{P}\leq\frac{1}{3}\ ...(1)$
$\frac{-1}{4}\leq\text{P}\leq\frac{1}{2}\ ...(2)$ and ${-1}\leq\text{P}\leq{5}\ ...(3)$
The common solution of $(1), (2),$ and $(3)$ is $\frac{-1}{4}\leq\text{P}\leq\frac{1}{3}$
$\therefore$ The set values of $P$ are $\Big(\frac{-1}{4},\frac{1}{3}\Big)$

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MCQ 161 Mark

Two dice are thrown together. The probability that neither they show equal digits nor the sum of their digits is $9$ will be :

A
$\frac{13}{15}$
✓
$\frac{13}{18}$
C
$\frac{1}{9}$
D
$\frac{8}{9}$

Answer

Correct option: B.

$\frac{13}{18}$

When two dices are thrown, there are $(6 \times 6) = 36$ outcomes.
The set of all these outcomes is the sample space is given by
$S = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)\}$
$\{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)\}$
$\{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)\}$
$\{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\}$
$\{(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\}$
$\{(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$
$\therefore\text{n}\text{(S)} = 36$
Let $E$ be the event of getting the digits which are neither equal nor give a total of $9$.
Then $E' =$ event of getting either a doublet or a total of $9$
Thus, $E' = \{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (3, 6), (4, 5), (5, 4), (6, 3)\}$
i.e. $n(E') = 10$
$\text{P(E}')=\frac{\text{n(E}')}{\text{n(E)}}=\frac{10}{36}=\frac{5}{18}$
Hence, required probability $P(E) = 1 - P(E')$
$=1-\frac{5}{18}=\frac{13}{18}$

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MCQ 171 Mark

One of the two events must occur. If the chance of one is $\frac{2}{3}$ of the other, then odds in favour of the other are

A
$\text{1 : 3}$
✓
$\text{3 : 1}$
C
$\text{2 : 3}$
D
$\text{3 : 2}$

Answer

Correct option: B.

$\text{3 : 1}$

Let $\text{P(B)}=\text{X}$
Than, $\text{P(A)}=\frac{2\text{x}}{3}$
$\text{P(A)}+\text{P(B)}=\text{x}+\frac{2x}{3}=\frac{5x}{3}$
$\Rightarrow\frac{5\text{x}}{3}=1$
$\big(\therefore$ They are exhaustive events $\big)$
$\Rightarrow\text{x}=\frac{3}{5}$
Now, $\text{P(A)}=\frac{2}{5}$ and $\text{P(B)}=\frac{3}{5}$
$\therefore$ odd in favour of ${B}=\frac{\frac{3}{5}}{\frac{1-3}{5}}=\frac{3}{2}=\text{3 : 1}$

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MCQ 181 Mark

One mapping is selected at random from all mappings of the set $A = \{1, 2, 3, ..., n\}$ into itself. The probability that the mapping selected is one to one is :

A
$\frac{1}{\text{n}^n}$
B
$\frac{1}{\text{n}}$
✓
$\frac{\text{n}-1}{\text{n}^n-1}$
D
None of these

Answer

Correct option: C.

$\frac{\text{n}-1}{\text{n}^n-1}$

Number of ways to map $1^{st}$ element in set $A = n$
Number of ways to map $2^{nd}$ element in set $A = n$ and so on
$\therefore$Total number of mapping from set $A$ to itself $=\text{n}\times\text{n}\times\ ...\ \times\text{n}(\text{n times})=\text{n}^\text{n}$
For one to one mapping,
Number of ways to map $1^{st}$ element in set $A = n$
Number of ways to map $2^{nd}$ element in set $A = n −1$
Number of ways to map nth element in set $A = 1$
Total number of one to one mappings from set $A$ to itself $=\text{n}\times\text{(n -1)}\times\text{(n - 2)}\ \times\ ...\times1=\text{n}$
$\therefore$ Required probability $=\frac{\text{Total number of one mappings from set A to inselp} }{\text{Total number of mappings form set A to itself}}$
$=\frac{\text{n}}{\text{n}^\text{n}}=\frac{(\text{n}-1)}{\text{n}^\text{n-1}}$
Hence, the correct answer is option $(c).$

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MCQ 191 Mark

The probability that a leap year will have $53$ Fridays or $53$ Saturdays is :

A
$\frac{2}{7}$
✓
$\frac{3}{7}$
C
$\frac{4}{7}$
D
$\frac{1}{7}$

Answer

Correct option: B.

$\frac{3}{7}$

We know that a leap year has $366$ days $($i.e. $7 \times 52 + 2) = 52$ weeks and $2$ extra days .
The sample space for these $2$ extra days is given below:
$S = \text{ \{(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday)}\},$
$\text{(Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)} \}$ There are $7$ cases.
$\therefore\text{n(S)}=7$
Let $E$ be the event that the leap year has $53$ Fridays or $53$ Saturdays.
$E = \text{\{(Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}\}$
i.e. $n(E) = 3$
$\therefore\text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}=\frac{3}{7}$
Hence, the probability that a leap year has $53$ Fridays or $53$ Saturdays is $\frac{3}{7}$.

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MCQ 201 Mark

If three dice are throw simultaneously, then the probability of getting a score of $5$ is :

A
$\frac{5}{216}$
B
$\frac{1}{6}$
✓
$\frac{1}{36}$
D
$\text{None of these}$

Answer

Correct option: C.

$\frac{1}{36}$

When three dice are thrown together, the sample space $S$ associated with the random experiment is given by,
$S = \{(1, 1, 1), (1, 1, 2), (1, 1, 3) ...(6, 6, 5), (6, 6, 6)\}$
Clearly, total number of elementary events $n(S) = 216$
Let $A$ be the event of getting a total score of $5$.
Then $A = \{ (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1)\}$
$\therefore$ Favourable number of elementary events $= 6$
i.e. $n(A) = 6$
Hence, required probability $=\frac{6}{216}=\frac{1}{36}$

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MCQ 211 Mark

If the probability of $A$ to fail in an examination is $\frac{1}{5}$ and that of $B$ is $\frac{3}{10}$Then, the probability that either $A$ or $B$ fails is

A
$\frac{1}{2}$
B
$\frac{11}{25}$
✓
$\frac{19}{50}$
D
None of these

Answer

Correct option: C.

$\frac{19}{50}$

Given :
$\text{P(A)}=\frac{1}{5}$
$\therefore\text{P(A}')=1-\frac{1}{5}=\frac{4}{5}$
$\text{P(B)}=\frac{3}{10}$
$\therefore\text{P(B}')=1-\frac{3}{10}=\frac{7}{30}$
Hence, required probability $=\text{P}(\text{A}\cap\text{B}')+\text{P}(\text{A}'\cap\text{B})$
$=\frac{1}{5}\times\frac{7}{10}+\frac{4}{5}\times\frac{3}{10}$
$=\frac{7}{50}+\frac{12}{50}$
$=\frac{19}{50}$

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MCQ 221 Mark

Two dice are thrown simultaneously. The probability of obtaining a total score of $5$ is :

A
$\frac{1}{18}$
B
$\frac{1}{12}$
✓
$\frac{1}{9}$
D
None of these

Answer

Correct option: C.

$\frac{1}{9}$

When two dice are thrown, there are $(6 \times 6) = 36$ outcomes.
The set of all these outcomes is the sample space given by
$S = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)\}$
$\{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)\}$
$\{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)\}$
$\{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\}$
$\{(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\}$
$\{(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$
i.e. $n(S) = 36$
Let $E$ be the event of getting a total score of $5$.
Then $E = \{(1, 4), (2, 3), (3, 2), (4, 1)\}$
$\therefore n(E) = 4$
Hence, required probability $=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}=\frac{4}{36}=\frac{1}{9}$

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MCQ 231 Mark

Five persons 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 floor beginning with the first, then the probability of all $5$ persons leaving at different floor is:

✓
$\frac{\ ^{7}\text{P}_5}{7^5}$
B
$\frac{\ ^{7^5}}{\ ^{7}\text{P}_5}$
C
$\frac{\ ^{6}}{\ ^{6}\text{P}_5}$
D
$\frac{\ ^{5}\text{P}_5}{5^5}$

Answer

Correct option: A.

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

Since, it is an eight $-$ storey building.
So, there are $7$ possible options for them in $7$ floors in total if ground floor is not considered.
Hence, total possible outcomes $=7 \times 7 \times 7 \times 7 \times 7=7^5$
Thus, number of ways in which $5$ persons can leave from seven floors differently $={ }^7 P _5$
Required probability $=\frac{{ }^7 P _5}{7^5}$

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MCQ 241 Mark

A person write $4$ 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
$\frac{11}{24}$
C
$\frac{15}{24}$
✓
$\frac{23}{24}$

Answer

Correct option: D.

$\frac{23}{24}$

Total number of ways of placing four letters in $4$ envelops $= 4 = 24$
All the letters can be dispatched in the right envelops in only one way.
Therefore, the probability that all the letters are placed in the right envelops is $\frac{1}{24}$.
Hence, probability that all the letters are not placed in the right envelops $=1-\frac{1}{24}=\frac{23}{24}$

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MCQ 251 Mark

The probability of getting a total of $10$ in a single throw of two dices is :

A
$\frac{1}{9}$
✓
$\frac{1}{12}$
C
$\frac{1}{6}$
D
$\frac{5}{36}$

Answer

Correct option: B.

$\frac{1}{12}$

When two dices are thrown, there are $(6 \times 6) = 36$ outcomes.
The set of all these outcomes is the sample space, given by
$S = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)\}$
$\{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6\}$
$\{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)\}$
$\{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\}$
$\{(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\}$
$\{(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$
i.e. $n(S) = 36$
Let $E$ be the event of getting a total score of $10.$
Then $ E = \{(4, 6), (5, 5), (6, 4)\}$
$\therefore n(E) = 3$
Hence, required probability $=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}=\frac{3}{36}=\frac{1}{12}$

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MCQ 261 Mark

$6$ boys and $6$ girls sit in a row at random. The probability that all the girls sit together is

A
$\frac{1}{432}$
B
$\frac{12}{431}$
✓
$\frac{1}{132}$
D
$\text{none of these}$

Answer

Correct option: C.

$\frac{1}{132}$

Total number of ways in which $6$ boys and $6$ girls can sit in a row $= 12$
Consider $6$ girls as one group, then $6$ boys and one group can arrange in $7$ ways.
Now, $6$ girls in the group can arrange among themselves in $6$.
So, the number of ways in which all the girls sit together is $7 \times 6$
$\therefore P\ ($all girls sit together$)$
$=\frac{\text{Number of ways in which all girls sit together }}{\text{Total Number of ways in which 6 boys and 6 girls sit in a row}}$
$=\frac{7\times6}{12}=\frac{6\times5\times4\times3\times2\times1}{12\times11\times10\times9\times8}=\frac{1}{132}$
Hence, the correct answer is option $(c).$

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MCQ 271 Mark

An urn contains $9$ balls two of which are red, three blue and four black. Three balls are drawn at random. The probability that they are of the same colour is :

✓
$\frac{5}{84}$
B
$\frac{3}{9}$
C
$\frac{3}{7}$
D
$\frac{7}{17}$

Answer

Correct option: A.

$\frac{5}{84}$

Three balls can be drawn randomly from nine balls in$ \ ^{9}\text{C}_3 = 84$ ways.
Three balls cannot be red as there are only two red balls.
Three balls of the same colour can be drawn in the following ways :
$3$ blue out of a total of $3$ blue balls.
The probability for which is $\frac{\ ^{3}\text{C}_3}{84}=\frac{1}{84}$
$3$ black out of a total of $4$ black balls.
The probability for which is $\frac{\ ^{4}\text{C}_3}{84}=\frac{4}{84}$
Hence, required probability $=\frac{1}{84}+\frac{4}{84}=\frac{5}{84}$

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Question 281 Mark

If $\text{P}\text{(A}\cup\text{B)} = \text{P}\text{(A}\cap\text{B)}$ for any two events A and B, the
$\text{P(A)}=\text{P(B)}$
$\text{P(A)}>\text{P(B)}$
$\text{P(A)}<\text{P(B)}$
None of these.

Answer

We know that,$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A)}+\text{(B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A)}+\text{P}\text{(B})-\text{P}(\text{A}\cap\text{B})$ $\big[\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A}\cup\text{B})\big]$
$\Rightarrow\text{P}(\text{A)}-\text{P}(\text{A}\cup\text{B})+\text{P}(\text{B)}-\text{P}(\text{A}\cap\text{B})=0\ ...(1)$
But,
$\text{P(A)}-\text{P}(\text{A }\cap\text{ B})\geq0$
$\text{P(B)}-\text{P}(\text{A }\cap\text{ B})\geq0$
$\Rightarrow\text{P}(\text{A)}-\text{P}(\text{A}\cup\text{B})+\text{P}(\text{B)}-\text{P}(\text{A}\cap\text{B})\geq0\ ...(2)$
From (1) and (2), we have,
$\Rightarrow\text{P}(\text{A)}-\text{P}(\text{A}\cup\text{B})+\text{P}(\text{B)}-\text{P}(\text{A}\cap\text{B})=0$
$\Rightarrow\text{P}(\text{A)}=\text{P}(\text{A}\cap\text{B})\text{ and }\text{P}(\text{B)}=\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\text{P(A)}=\text{P(B)}$
Hence, the correct answer is option (a).

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MCQ 291 Mark

A die is rolled, then the probability that a number $1$ or $6$ may appear is

A
$\frac{2}{3}$
B
$\frac{5}{6}$
✓
$\frac{1}{3}$
D
$\frac{1}{2}$

Answer

Correct option: C.

$\frac{1}{3}$

Total number of sample space, $S = \{1, 2, 3, 4, 5, 6\}$
$\therefore\text{n}\text{(S)} = 6$
Let A be the event of getting the number $1$ or $6$.
$A = \{1, 6\}$
i.e. $n(A) = 2$
Hence, required probability $=\frac{\text{n(A)}}{\text{n(S)}}=\frac{2}{6}=\frac{1}{3}$

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MCQ 301 Mark

Six boys and six girls sit in a row randomly. The probability that all girls sit together is

A
$\frac{1}{122}$
B
$\frac{1}{112}$
C
$\frac{1}{102}$
✓
$\frac{1}{132}$

Answer

Correct option: D.

$\frac{1}{132}$

Total number of ways in which six boys and six girls can be seated in a row $= (12)$
Taking all the six girls as one person, seven persons can be seated in a row in $7$ ways.
The six girls can be arranged among themselves in $6$ ways.
Then number of ways in which six boys and six girls can be seated in a row so that all
the girls sit together $= 7 \times 6$
$\therefore\text{Required Probability}=\frac{7\times6}{12}$$\frac{720}{12\times11\times10\times9\times8}=\frac{1}{132}$

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MCQ 311 Mark

A card is drawn at random from a pack of $100$ cards numbered $1$ to $100$. The probability of drawing a number which is a square is :

A
$\frac{1}{5}$
B
$\frac{2}{5}$
✓
$\frac{1}{10}$
D
None of these

Answer

Correct option: C.

$\frac{1}{10}$

Clearly, the sample space is given by
$S = \{1, 2, 3, 4, 5 ....97, 98, 99, 100\}$
$\therefore n(S) = 100$
Let $E =$ event of getting a square.
Then $E = \{1, 4, 9, 16, 25, 36, 49, 64, 81, 100\}$
$\therefore n(E) = 10$
Hence, required probability $=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}=\frac{10}{100}=\frac{1}{10}$

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MCQ 321 Mark

$A$ and $B$ are two events such that $P (A) = 0.25$ and $P (B) = 0.50$. The probability of both happening together is $0.14.$ The probability of both $A$ and $B$ not happening is

✓
$0.39$
B
$0.2$
C
$0.11$
D
none of these.

Answer

Correct option: A.

$0.39$

$\text{p(A)}=0.25\text{ and }\text{P(B)=0.50}$
$\text{P}(\text{A}\cap\text{B})=0.14$
$\therefore\text{Required probability}=1-\text{P}(\text{A}\cup\text{B})$
$=1-\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]$
$=1-\big[0.25+0.50-0.14\big]$
$=1-0.61=0.39$

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MCQ 331 Mark

The probabilities of happening of two events $A$ and $B$ are $0.25$ and $0.50$ respectively. If the probability of happening of $A$ and $B$ together is $0.14,$ then probability that neither $A$ nor $B$ happens is :

✓
$\text{0.39}$
B
$0.29$
C
$0.11$
D
None of these.

Answer

Correct option: A.

$\text{0.39}$

$\text{P(A)}=0.25, \text{P(B)=0.50}$ and $\text{P(A}\cap\text{B)}=0 .14$
$\therefore\text{Required Probability}=1-\text{P}(\text{A}\cup\text{B})$
$=1-\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B}\big]$
$=1-\big[0.25+0.50-0.14\big]$
$=1-0.61=0.39$

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MCQ 341 Mark

A box contains $10$ good articles and 6 defective articles. One item is drawn at random. The probability that it is either good or has a defect, is :

✓
$\frac{64}{64}$
B
$\frac{49}{64}$
C
$\frac{40}{69}$
D
$\frac{24}{64}$

Answer

Correct option: A.

$\frac{64}{64}$

The answer is one, because the article would be either good or defective as per the question.
Hence, the only option is $\frac{64}{64}=1$

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MCQ 351 Mark

Four persons are selected at random out of $3$ men, $2$ women and $4$ children. The probability that there are exactly $2$ children in the selection is:

A
$\frac{11}{21}$
B
$\frac{9}{21}$
✓
$\frac {10}{21}$
D
None of these

Answer

Correct option: C.

$\frac {10}{21}$

There are nine persons $($three men, two women and four children$)$ out of which four persons can be selected in $\ ^{9}\text{C}_4 = 126$ ways.
Total number of elementary events $= 126$
Exactly two children means selecting two children and two other people from three men and two women.
This can be done in $\ ^{4}\text{C}_2\times\ ^{ 5}\text{C}_2 $ ways .
Favourable number of elementary events$=\ ^{4}\text{C}_2\times\ ^{ 5}\text{C}_2 = 60$
So, required probability $=\frac{60}{120} =\frac{10}{21}$

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MCQ 361 Mark

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

A
$\frac{14}{29}$
B
$\frac{16}{29}$
✓
$\frac {15}{29}$
D
$\frac{10}{29}$

Answer

Correct option: C.

$\frac {15}{29}$

The total number of ways in which two integers can be chosen from the given $30$ integers is $\ ^{30}\text{C}_2$.
The sum of the selected numbers is odd if exactly one of them is even or odd.
$\therefore$ Favourable number of outcomes $=\ ^{15}\text{C}_1\times\ ^{15}\text{C}_1$
Hence, required probability $=\frac{\ ^{15}\text{C}_1\times\ ^{15}\text{C}_1}{\ ^{30}\text{C}_2}=\frac{15}{29}$

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MCQ 371 Mark

A bag contains $5$ black balls, $4$ white balls and $3$ red balls. If a ball is selected randomwise, the probability that it is black or red ball is :

A
$\frac{1}{3}$
B
$\frac{1}{4}$
C
$\frac{5}{12}$
✓
$\frac{2}{3}$

Answer

Correct option: D.

$\frac{2}{3}$

Out of $12$ balls, one ball can be drawn in $\ ^{12}\text{C}_1$ ways.
Total number of elementary events $= \ ^{12}\text{C}_1 = 12$
Out of fivne black balls, one black ball can be chosen in $\ ^{5}\text{C}_1 = 5$ ways.
Out of three red balls, one red ball can be chosen in $\ ^{3}\text{C}_1 = 3$ ways
Favourable number of events $= 5 + 3 = 8$
Hence, required probability $=\frac{8}{12}=\frac{2}{3}$

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MCQ 381 Mark

A bag contains $3 $ red, $4$ white and $5$ blue balls. All balls are different. Two balls are drawn at random. The probability that they are of different colour is :

✓
$\frac{47}{66}$
B
$\frac{10}{33}$
C
$\frac{1}{3}$
D
$1$

Answer

Correct option: A.

$\frac{47}{66}$

Out of $12$ balls, two balls can be drawn in $^{12}\text{C}_2$ ways.
$\therefore$ Total number of elementary events, $\text{n}(\text{S})=^{12}\text{C}_2=66$
We observe that at least one ball of each colour can be drawn in one of the following mutually exclusive ways :
Red and $1$ white
$1$ red and $1$ blue
$1$ white and $1$ blue
Thus, if we define three events $A, B$ and $C$ as follows:
$A =$ drawing $1$ red and $1$ white
$B =$ drawing $1$ red and $1$ blue
$C =$ drawing $1$ white and $1$ blue
then, $A, B$ and $C$ are mutually exclusive events.
$\therefore$ Required probability $=\text{P}(\text{A}\cup\text{B}\cup\text{C})$
$=\text{P}(\text{A})+\text{P}(\text{B})+\text{P}(\text{C})$
$=\frac{^3\text{C}_1\times^4\text{C}_1}{^{12}\text{C}_2}+\frac{^3\text{C}_1\times^5\text{C}_1}{^{12}\text{C}_2}+\frac{^4\text{C}_1\times^5\text{C}_1}{^{12}\text{C}_2}$
$=\frac{3\times4}{66}+\frac{3\times5}{66}+\frac{4\times5}{66}$
$=\frac{12}{66}+\frac{15}{66}+\frac{20}{56}=\frac{47}{66}$

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MCQ 391 Mark

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

A
$\frac{2}{19}$
B
$\frac{3}{29}$
✓
$\frac{17}{19}$
D
$\frac{4}{19}$

Answer

Correct option: C.

$\frac{17}{19}$

Number of ways in which we can choose three distinct integers from $20$ integers $\ ^{20}\text{C}_3=1140$
We know that, if we take three odd numbers, there product will always be an odd number.
Out of $20$ consecutive integers, $10$ are even and $10$ are odd integers.
Number of ways in which we can choose three distinct odd integers from $10$ odd integers $=\ ^{10}\text{C} _3=120$
$P\ ($product is even$) = 1 - P\ ($product is odd$),$
$=1-\frac{120}{1140}=\frac{1140-120}{1140}=\frac{1020}{1140}=\frac{17}{19}$

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MCQ 401 Mark

A box contains $6$ nails and $10$ nuts. Half of the nails and half of the nuts are rusted. If one item is chosen at random, the probability that it is rusted or is a nail is :

A
$\frac{3}{16}$
B
$\frac{5}{16}$
✓
$\frac{11}{16}$
D
$\frac{14}{16}$

Answer

Correct option: C.

$\frac{11}{16}$

If the numbers of nails and nuts are $6$ and $10,$ respectively, then the numbers of rusted nails and rusted nuts are $3$ and $5,$ respectively.
Total number of items $= 6 + 10 = 16$
Total number of rusted items $= 3 + 5 = 8$
Total number of ways of drawing one item $=\ ^{16}\text{C}_1$
Let $R$ and $N$ be the events where both the items drawn are rusted items and nails, respectively.
$R$ and $N$ are not mutually exclusive events, because there are $3$ rusted nails.
$P\ ($either a rusted item or a nail$)\ = \text{P} \text{(R}\cup\text{N})$
$= \text{P} \text{(R})+\text{P}\text{(N})-\text{P}(\text{R}\cap\text{N})$
$=\frac{\ ^{6}\text{C}_1}{{\ ^{16}\text{C}_1}}+\frac{\ ^{8}\text{C}_1}{{\ ^{16}\text{C}_1}}-\frac{\ ^{3}\text{C}_1}{{\ ^{16}\text{C}_1}}$
$=\frac{6}{16}+\frac{8}{16}-\frac{3}{16}=\frac{11}{16}$

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