M.C.Q (1 Marks)

Question 11 Mark

If a dice is rolled two times, then what will be the probability of sum of appeared number is 7 ?

Answer

(C)
Total results throwing two dice at a time $=6 \times 6=36$ Sum of number is 7 such that, $\{(1,6),(6,1),(2,5),(5,2),(4,3),(3,4)\}=6$
$
\text { Hence } \quad \begin{aligned}
\text { Such probability } & =\frac{\text { Favourable results }}{\text { Total results }} \\
& =\frac{6}{36}=\frac{1}{6}
\end{aligned}
$
Hence correct option is (C).

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Question 21 Mark

If $P ( A )=0.8, P ( B )=0.5$ and $P ( B / A )=0.4$, then the value of $P ( A \cap B )$ is :

Answer

(A) 0.32

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Question 31 Mark

Two cards are drawn at random and without replacement from a pack of 52 playing cards, then the probability that both the cards are black is :

Answer

(B) $\frac{52}{102}$

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Question 41 Mark

Probability of speaking truth by$ A$ is $\frac{4}{5}$ where as by $B$ is $\frac{3}{4}$. When they speak on a topic, then probability of contradiction:

Answer

Given
$P(A)=\frac{4}{5} $
$\therefore P(\overline{A})=1-\frac{4}{5}=\frac{1}{5}$
$P(B)=\frac{3}{4} $
$\therefore P(\overline{B})=1-\frac{3}{4}=\frac{1}{4}$
Probability of contradiction to speak $A$ and $B$
$=P(A) \cdot P(\overline{B})+P(B) \ P(\overline{A})$
$=\frac{4}{5} \times \frac{1}{4}+\frac{3}{4} \times \frac{1}{5}$
$=\frac{1}{5}+\frac{3}{20}=\frac{7}{20}$
Hence correct option is $(A).$

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Question 51 Mark

A die is rolled. Let $A$ be event of getting number more than 3 and B be event of getting number less than 5 . Then value of $P ( A \cup B )$ :

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Question 61 Mark

Two events A and B such that $P ( A )=\frac{1}{4}, P ( A \mid B )=\frac{1}{2}$, and $P(B \mid A)=\frac{2}{3}$, then value of $P(B)$ :

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Question 71 Mark

In a pack of playing cards there are $4$ Ace $, 4$ King $, 4$ Queen and $4$ Jack. Two cards are drawn. What is the probability getting at least one Ace?

Answer

Total cards in a pack of playing cards $=4+4+4+4$
$=16$
Methods of two cards are drawn $={ }^{16} C _2$
Probability of getting at least one Ace
$=\frac{{ }^4 C_1 \times{ }^{12} C_1}{{ }^{16} C_2}+\frac{{ }^4 C_2}{{ }^{16} C_2}$
$=\frac{4 \times 12 \times 2}{16 \times 15}+\frac{4 \times 3}{16 \times 15}$
$=\frac{96+12}{16 \times 15}$
$=\frac{108}{240}$
$=\frac{9}{20}$
Hence correct option is $(C).$

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Question 81 Mark

4 numbers are taken from 1, 2, 3, 4, 5, 6, 7. Probability of getting sum of 4 numbers is less than 12 .

Answer

(C)
Total methods of taking 4 numbers $=7 \times 6 \times 5 \times 4$ Total methods of taking numbers having sum less than $4\{1,2,3,4\}$ and $\{1,2,3,5\}$
$
=\lfloor 4+\lfloor 4=24+24=48
$
Hence, $\quad$ Such probability $=\frac{48}{7 \times 6 \times 5 \times 4}=\frac{2}{35}$
Hence correct option is (C).

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Question 91 Mark

If one pair of two unbiased dice are rolled then what is the probability of getting sum 5 ?

Answer

(D)
Total number of results when two dice are rolled $=6 \times$ $6=36$
Sum is 5 such as $\{(1,4),(4,1),(2,3),(3,2)\}=4$
$
\begin{aligned}
\text { Hence, }\text { Such probability } & =\frac{\text { Favoured conditions }}{\text { Total conditions }} \\
& =\frac{4}{36}=\frac{1}{9}
\end{aligned}
$
Hence correct option is (D).

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Question 101 Mark

4 white and 3 black balls are in a bag. 3 white and 4 black balls are in other bag. If a ball is drawn, that is black then find the probability of ball is black drawn from second bag.

Answer

(D)
Let $E_1=$ Ball drawn from first bag
$E _2=$ Ball drawn from second bag
$\therefore \quad P\left(E_1\right)=\frac{1}{2}$ and $P\left(E_2\right)=\frac{1}{2}$
Total balls in first bag $=4+3=7$
and no. of black balls $=3$
Hence probability of a black ball drawn from first bag
$
P\left(\frac{A}{E_1}\right)=\frac{{ }^3 C_1}{{ }^7 C_1}=\frac{3}{7}
$
Probability of a black ball drawn from second bag
$
P\left(\frac{A}{E_2}\right)=\frac{{ }^4 C_1}{{ }^7 C_1}=\frac{4}{7}
$
Hence such probability
$
\begin{aligned}
P\left(\frac{E_2}{A}\right) & =\frac{P\left(E_2\right) \cdot P\left(A \mid E_2\right)}{P\left(E_1\right) \cdot P\left(A \mid E_1\right)+P\left(E_2\right) \cdot P\left(A \mid E_2\right)} \\
& =\frac{\frac{1}{2} \times \frac{4}{7}}{\frac{1}{2} \cdot \frac{3}{7}+\frac{1}{2} \cdot \frac{4}{7}}=\frac{4 / 7}{\frac{3}{7}+4 / 7} \\
& =\frac{4}{7}
\end{aligned}
$
Hence correct option is (D).

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MATHS M.C.Q (1 Marks)

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