MCQ - MATHS STD 11 Science Questions - Vidyadip

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39 questions · auto-graded multiple-choice test.

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:

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

Answer

Correct option: A.

$\frac{64}{64}$

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 × 3 × 2 × 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 × 2 × 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 × 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[\text{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 × 5 × 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 n 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 × 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.

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

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

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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 × 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, $\text{P(A)}=\frac{2}{5}\text{ and }\text{P(B)}=\frac{3}{5}$
$\therefore\text{odd in favour of B}=\frac{\frac{3}{5}}{\frac{1-3}{5}}=\frac{3}{2}=\text{3 : 1}$

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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 1st element in set A = n Number of ways to map 2nd 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 1st element in set A = n
Number of ways to map 2nd 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 × 52 + 2) = 52 weeks and 2 extra days .
The sample space for these 2 extra days is given below:
S = {(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (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 = { (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 × 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.

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

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

Solution:
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 $=\ ^{7}\text{P}_5$
Required probability $=\frac{\ ^{7}\text{P}_5}{7^5}$

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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 × 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 × 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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MCQ 281 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 291 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 × 6
$\therefore\text{Required Probability}=\frac{7\times6}{12}$$\frac{720}{12\times11\times10\times9\times8}=\frac{1}{132}$

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MCQ 301 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 311 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 321 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.

$\text{0.39}$

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

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MCQ 331 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 341 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 $\ ^{9}\text{C}_4 = 126\ \text{ways}.$
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 $\ ^{4}\text{C}_2\times\ ^{ 5}\text{C}_2 \text{ways}.$
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 351 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 361 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 371 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 381 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 391 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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