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 $A$ 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.

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

(c) : $P($ atleast one of $A$ and $B)$
$
\begin{array}{l}
=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)
\end{array}
$

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MCQ 2021 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 2031 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 2041 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 2051 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 2061 Mark

One card is drawn from a well shuffled pack of 52 cards. If $E$ is the event "the card drawn is a king or queen" and $F$ is the event "the card drawn is a queen or an ace", then find the probability of the conditional event $E \mid F$.

✓
$\frac{1}{2}$
B
$\frac{1}{3}$
C
$\frac{2}{13}$
D
$\frac{3}{13}$

Answer

Correct option: A.

$\frac{1}{2}$

(a) : There are 4 kings and 4 queens in pack of 52 cards.
Let $E=$ the card drawn is a king or queen
$
\therefore \quad P(E)=\frac{8}{52}=\frac{2}{13}
$
There are 4 queens and 4 aces in a pack of 52 cards.
Let $F=$ the card drawn is a queen or an ace
$
\therefore \quad P(F)=\frac{8}{52}=\frac{2}{13}
$
Then, $E \cap F$ contains drawing a queen.
$
\therefore \quad P(E \cap F)=\frac{4}{52}=\frac{1}{13}
$
$\therefore$ Required probability, $P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{1 / 13}{2 / 13}=\frac{1}{2}$

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MCQ 2071 Mark

If two events $A$ and $B$ are independent, then which of the following is true about events $A$ and $B$ ?

A
$A$ and $B$ are mutually exclusive
✓
$P\left(A^{\prime} \cap B^{\prime}\right)=[1-P(A)][1-P(B)]$
C
$P(A)=P(B)$
D
$P(A)+P(B)$=1

Answer

Correct option: B.

$P\left(A^{\prime} \cap B^{\prime}\right)=[1-P(A)][1-P(B)]$

(b) : Two events $A$ and $B$ are independent, if $P(A \cap B)=P(A) P(B)$
$\therefore P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}=1-P(A \cup B)$
$
\begin{array}{l}
=1-[P(A)+P(B)-P(A \cap B)]=1-P(A)-P(B)+P(A) P(B) \\
=[1-P(A)][1-P(B)]
\end{array}
$

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MCQ 2081 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 2 green balls and one blue ball is

✓
$\frac{3}{28}$
B
$\frac{2}{21}$
C
$\frac{1}{28}$
D
$\frac{167}{168}$

Answer

Correct option: A.

$\frac{3}{28}$

(a) : Possible outcomes $=\{G G B, G B G, B G G\}$
Required probability $=P(G G B)+P(G B G)+P(B G G)$
$
=\frac{3}{8} \times \frac{2}{7} \times \frac{2}{6}+\frac{3}{8} \times \frac{2}{7} \times \frac{2}{6}+\frac{2}{8} \times \frac{3}{7} \times \frac{2}{6}=\frac{3}{28}
$

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MCQ 2091 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 2101 Mark

A flashlight has 8 batteries out of which 3 are dead. If two batteries are selected without replacement and tested, then probability that both are dead is

A
$\frac{33}{56}$
B
$\frac{9}{64}$
C
$\frac{1}{14}$
✓
$\frac{3}{28}$

Answer

Correct option: D.

$\frac{3}{28}$

(d) : Required probability $=P(D D)=\frac{3}{8} \times \frac{2}{7}=\frac{3}{28}$

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MCQ 2111 Mark

If $A$ and $B$ are events such that $P(A)>0$ and $P(B) \neq 1$, then $P\left(A^{\prime} \mid B^{\prime}\right)$ equals

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

Answer

Correct option: C.

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

(c) : By definition, $P\left(A^{\prime} \mid B^{\prime}\right)=\frac{P\left(A^{\prime} \cap B^{\prime}\right)}{P\left(B^{\prime}\right)}$
$
=\frac{P\left((A \cup B)^{\prime}\right)}{P\left(B^{\prime}\right)}=\frac{1-P(A \cup B)}{P\left(B^{\prime}\right)}
$

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MCQ 2121 Mark

The probability distribution of $X$ is

$X=x$	0	1	2	3	4
$P(X=x)$	$k$	$2k$	$4k$	$2k$	$k$

Then find $P(X \leq 1)$.

A
0.1
B
0.3
C
0.4
D
0.5

Answer

\begin{array}{ll}
& \text { (b) }: \because \sum_{x=0}^4 P(X=x)=1 \\
\Rightarrow & P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)=1 \\
\Rightarrow & k+2 k+4 k+2 k+k=1 \Rightarrow 10 k=1 \Rightarrow k=0.1 \\
\therefore & P(X \leq 1)=P(X=0)+P(X=1) \\
\Rightarrow & P(X \leq 1)=k+2 k=3 k=0.3
\end{array}

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MCQ 2131 Mark

A problem in mathematics is given to 3 students whose chances of solving it are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$. What is the probability that the problem is solved?

A
$1 / 5$
B
$1 / 4$
✓
$3 / 4$
D
$2 / 3$

Answer

Correct option: C.

$3 / 4$

(c) : Let $A, B, C$ be the respective events of solving the problem. Then, $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}$ and $P(C)=\frac{1}{4}$.
Clearly $A, B, C$ are independent events and the problem is solved if atleast one student solves it.
$
\begin{array}{l}
\therefore \quad \text { Required probability }=P(A \cup B \cup C) \\
=1-P(\bar{A}) P(\bar{B}) P(\bar{C}) \\
=1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)=1-\frac{1}{4}=\frac{3}{4}
\end{array}
$

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MCQ 2141 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 2151 Mark

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

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

The value of $k$ is

A
8
✓
16
C
32
D
48

Answer

Correct option: B.

16

(b) : We have $\Sigma P(X)=1$
$
\Rightarrow \frac{4}{k}+\frac{6}{k}+\frac{4}{k}+\frac{2}{k}=1 \Rightarrow \frac{16}{k}=1 \quad \Rightarrow \quad k=16
$

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MCQ 2161 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 2171 Mark

If $A$ and $B$ are two events such that $P(A)=0.2, P(B)=0.4$ and $P(A \cup B)=0.5$, then value of $P(A / B)$ is

A
0.1
✓
0.25
C
0.5
D
0.08

Answer

Correct option: B.

0.25

(b) : We have, $P(A)=0.2, P(B)=0.4$ and $P(A \cup B)=0.5$
$\because \quad P(A \cap B)=P(A)+P(B)-P(A \cup B)$
$=0.2+0.4-0.5=0.1$
Now, $P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{0.1}{0.4}=\frac{1}{4}=0.25$

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MCQ 2181 Mark

A machine operates if all of its three components function. The probability that the first component fails during the year is 0.14 , the second component fails is 0.10 and the third component fails is 0.05 . What is the probability that the machine will fail during the year?

A
0.1542
✓
0.2647
C
0.3642
D
0.4231

Answer

Correct option: B.

0.2647

(b) : Consider the following events:
$A=$ First component of the machine fails during the year
$B=$ Second component of the machine fails during the year
$C=$ Third component of the machine fails during the year
We have, $P(A)=0.14, P(B)=0.10$ and $P(C)=0.05$
Clearly, the machine will fail if at least one of its three components fails during the year.
$
\begin{array}{l}
\text { Required probability }= P (A \cup B \cup C) \\
=1-P(\overline{A \cup B \cup C})=1-P(\bar{A} \cap \bar{B} \cap \bar{C}) \\
=1-P(\bar{A}) P(\bar{B}) P(\bar{C})[\because A, B, C \text { are independent events }] \\
=1-(1-0.14)(1-0.10)(1-0.05) \\
=1-(0.86)(0.90)(0.95)=0.2647 .
\end{array}
$

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MCQ 2191 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 2201 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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