M.C.Q (1 Marks) - Page 5 - MATHS STD 12 Science Questions - Vidyadip

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M.C.Q (1 Marks)

MCQ 2011 Mark

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

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

Answer

Correct option: D.

$\frac{7}{12}$

(d) : Given, $P(A)=\frac{3}{10}, P(B)=\frac{2}{5}, P(A \cup B)=\frac{3}{5}$
Now, $P(A \cap B)=P(A)+P(B)-P(A \cup B)$
$
=\frac{3}{10}+\frac{2}{5}-\frac{3}{5}=\frac{3+4-6}{10}=\frac{1}{10}
$
Now, $P(B \mid A)+P(A \mid B)=\frac{P(A \cap B)}{P(A)}+\frac{P(A \cap B)}{P(B)}$
$
=\frac{1 / 10}{3 / 10}+\frac{1 / 10}{2 / 5}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}
$

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MCQ 2021 Mark

If six cards are selected at random $($without replacement$)$ from a standard deck of $52$ cards, what is the probability that there will be no pairs? $($two cards of same denomination$)$

A
$0.28$
B
$0.562$
✓
$0.345$
D
$0.832$

Answer

Correct option: C.

$0.345$

Let $E_i$ be the event that the first $i$ cards have no pair among them.
Then we want to compute $P\left(E_6\right)$,
which is actually the same as $P\left(E_1 \cap E_2 \cap \ldots \cap E_6\right)$,
since $E_6 \subset E_5 \subset \ldots \subset E_1$, implying that $E_1 \cap E_2 \cap \ldots \cap E_6=E_6$.
$\therefore P\left(E_1 \cap E_2 \cap \ldots \cap E_6\right)=P\left(E_1\right) P\left(E_2 \mid E_1\right) P\left(E_3 \mid E_1 \cap E_2\right) \ldots$
$=\frac{52}{52} \cdot \frac{48}{51} \cdot \frac{44}{50} \cdot \frac{40}{49} \cdot \frac{36}{48} \cdot \frac{32}{47}=0.345$

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MCQ 2031 Mark

A die is thrown and a card is selected at random from a deck of 52 playing cards, then 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.

$\frac{1}{8}$

(c) : Let $E_1$ be the event for getting an even number on the die and $E_2$ be the event that a spade card is selected.
$
\therefore \quad P\left(E_1\right)=\frac{3}{6}=\frac{1}{2} \text { and } P\left(E_2\right)=\frac{13}{52}=\frac{1}{4}
$
Now, $P\left(E_1 \cap E_2\right)=P\left(E_1\right) \cdot P\left(E_2\right)=\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}$.

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MCQ 2041 Mark

The probability that student entering a university will graduate is 0.4 . Find the probability that out of 3 students of the university none will graduate.

✓
0.216
B
0.36
C
0.6
D
0.1296

Answer

Correct option: A.

0.216

(a) : Let $X$ denote the number of students who graduated.
Now, the probability that a student graduates $=0.4$
$\therefore$ Probability that a student not graduates $=1-0.4=0.6$
$\therefore P($ none will graduate $)=(0.6)^3=0.216$

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MCQ 2051 Mark

An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement, then the probability that both drawn balls are black, is

A
$\frac{2}{7}$
B
$\frac{1}{7}$
C
$\frac{5}{7}$
✓
$\frac{3}{7}$

Answer

Correct option: D.

$\frac{3}{7}$

(d) : Let $E$ and $F$ denote respectively the events that first and second ball drawn are black. We have to find $P(E \cap F)$.
Now, $P(E)=\frac{10}{15}, P(F \mid E)=\frac{9}{14}$
By multiplication rule of probability, we have
$
P(E \cap F)=P(E) \cdot P(F \mid E)=\frac{10}{15} \times \frac{9}{14}=\frac{3}{7}
$

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MCQ 2061 Mark

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

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

The value of $k$ is

A
8
B
16
✓
32
D
48

Answer

Correct option: C.

32

(c) : We have, $\Sigma P(X)=1$
$
\Rightarrow \frac{5}{k}+\frac{7}{k}+\frac{9}{k}+\frac{11}{k}=1 \Rightarrow \frac{32}{k}=1 \Rightarrow k=32
$

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MCQ 2071 Mark

If $A$ and $B$ are two independent events, then the probability of occurrence of atleast one of $A$ and $B$ is given by

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

Answer

Correct option: C.

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

$P($ atleast one of $A$ and $B)$
$=P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$=P(A)+P(B)-P(A) P(B) \quad[\because A, B \text { are independent }]$
$=P(A)+P(B)[1-P(A)]$
$=\left[1-P\left(A^{\prime}\right)\right]+P(B) P\left(A^{\prime}\right)$
$=1-P\left(A^{\prime}\right)+P(B) P\left(A^{\prime}\right)$
$=1-P\left(A^{\prime}\right)[1-P(B)]$
$=1-P\left(A^{\prime}\right) P\left(B^{\prime}\right)$

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MCQ 2081 Mark

A boy tosses fair coin 3 times. If he gets $₹ 2 X$ for $X$ heads, then his expected gain (in ₹) equals to

A
1
B
$\frac{3}{2}$
✓
3
D
4

Answer

Correct option: C.

3

(c) : For $X$ heads, he gets amount $Y=2 X$

$X$	0	1	2	3
$Y$	0	2	4	6
$P(Y)$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

Expected gain, $E(Y)=\Sigma Y P(Y)$
$
=0\left(\frac{1}{8}\right)+2\left(\frac{3}{8}\right)+4\left(\frac{3}{8}\right)+6\left(\frac{1}{8}\right)=\frac{6+12+6}{8}=3.
$

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MCQ 2091 Mark

If $A$ and $B$ are two events and $A \neq \phi, B \neq \phi$, then

A
$P(A \mid B)=P(A)\cdot P(B)$
✓
$P(A \mid B)=\frac{P(A \cap B)}{P(B)}$
C
$P(A \mid B) \cdot P(B \mid A)=1$
D
$P(A \mid B)=P(A) \mid P(B)$

Answer

Correct option: B.

$P(A \mid B)=\frac{P(A \cap B)}{P(B)}$

(b) : By multiplication theorem,
$
\begin{aligned}
& P(A \cap B)=P(A \mid B) \times P(B)=P(B \mid A) \times P(A) \\
\Rightarrow & P(A \mid B)=\frac{P(A \cap B)}{P(B)}
\end{aligned}
$

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MCQ 2101 Mark

A fair die is rolled. Consider the events $A=\{1,3,5\}, B=\{2,3\}$ and $C=\{2,3,4,5\}$. Then the conditional probability $P((A \cup B) \mid C)$ is

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

Answer

Correct option: D.

$\frac{3}{4}$

(d) : $A=\{1,3,5\}, B=\{2,3\}, C=\{2,3,4,5\}$
$\Rightarrow A \cup B=\{1,2,3,5\}$ and $(A \cup B) \cap C=\{2,3,5\}$
Now, $P[(A \cup B) \mid C]$
$
=\frac{P[(A \cup B) \cap C]}{P(C)}=\frac{n[(A \cup B) \cap C] / n(S)}{n(C) / n(S)}=\frac{3}{4} .
$

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MCQ 2111 Mark

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

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

Answer

Correct option: C.

$\frac{3}{4}$

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

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MCQ 2121 Mark

A bag contains 4 identical red balls and 3 identical black balls. The experiment consists of drawing one ball, then putting it into the bag and again drawing a ball. Then the probability of getting both ball red is

A
$\frac{12}{49}$
✓
$\frac{16}{49}$
C
$\frac{2}{7}$
D
None of these

Answer

Correct option: B.

$\frac{16}{49}$

(b) : Required probability $=\frac{4}{7} \times \frac{4}{7}=\frac{16}{49}$

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MCQ 2131 Mark

A random variable $X$ has the following 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

For the event $E=\{X$ is prime number $\}$, find $P(E)$.

A
0.87
✓
0.62
C
0.35
D
0.5

Answer

Correct option: B.

0.62

(b) : $P(E)=P(X=2)+P(X=3)+P(X=5)+P(X=7)$
$=0.23+0.12+0.20+0.07=0.62$

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MCQ 2141 Mark

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

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

Answer

Correct option: C.

$\frac{7}{8}$

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

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MCQ 2151 Mark

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

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

Answer

Correct option: C.

$\frac{15}{56}$

(c) : Possible outcomes $=\{(R B B),(B R B),(B B R)\}$
Required probability $=P(R B B)+P(B R B)+P(B B R)$
$
=\frac{5}{8} \times \frac{3}{7} \times \frac{2}{6}+\frac{3}{8} \times \frac{5}{7} \times \frac{2}{6}+\frac{3}{8} \times \frac{2}{7} \times \frac{5}{6}=3 \times \frac{5}{56}=\frac{15}{56}
$

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MCQ 2161 Mark

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

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

Answer

Correct option: D.

1

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

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MCQ 2171 Mark

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

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

Answer

Correct option: C.

$\frac{31}{32}$

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

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MCQ 2181 Mark

If $P(A)=\frac{7}{13}, P(B)=\frac{9}{13}$ and $P(A \cup B)=\frac{12}{13}$, then evaluate $P(A \mid B)$.

A
$\frac{2}{3}$
✓
$\frac{4}{9}$
C
$\frac{4}{5}$
D
None of these

Answer

Correct option: B.

$\frac{4}{9}$

$\text { (b) : Given, } P(A)=\frac{7}{13}, P(B)=\frac{9}{13} \text { and } P(A \cup B)=\frac{12}{13}$
$\therefore P(A \cap B)=P(A)+P(B)-P(A \cup B)$
$=\frac{7}{13}+\frac{9}{13}-\frac{12}{13} \Rightarrow P(A \cap B)=\frac{4}{13}$
$\therefore P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{4 / 13}{9 / 13}=\frac{4}{9}$

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MCQ 2191 Mark

A bag contains 10 white and 6 black balls. 4 balls are successively drawn out without replacement. What is the probability that they are alternately of different colours?

✓
0.12
B
0.08
C
0.11
D
0.07

Answer

Correct option: A.

0.12

(a) : Required probability $=P(B W B W)+P(W B W B)$
$
=\frac{6}{16} \cdot \frac{10}{15} \cdot \frac{5}{14} \cdot \frac{9}{13}+\frac{10}{16} \cdot \frac{6}{15} \cdot \frac{9}{14} \cdot \frac{5}{13}=\frac{90}{728}=\frac{45}{364} \text {. }
$

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MCQ 2201 Mark

Suppose that five good fuses and two defective ones have been mixed up. To find the defective fuses, we test them one-by-one, at random and without replacement. What is the probability that we are lucky and find both of the defective fuses in the first two tests?

A
$\frac{1}{42}$
B
$\frac{2}{21}$
C
$\frac{1}{18}$
✓
$\frac{1}{21}$

Answer

Correct option: D.

$\frac{1}{21}$

(d) : Let $D_1, D_2$ be the events that we find a defective fuse in the first and second test respectively.
$\therefore \quad$ Required probability $=P\left(D_1 \cap D_2\right)$
$
=P\left(D_1\right) P\left(D_2 \mid D_1\right)=\frac{2}{7} \cdot \frac{1}{6}=\frac{1}{21}
$

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