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Maths 2 Marks Questions

PART - 2 CH - 14 Probability · CBSE STD 11 Science (CBSE - English Medium). 3 questions.

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Question 12 Marks
There are 10 white and 15 red balls in a bag and 16 white and 9 red balls in another bag. One - one ball is drawn from each of the two bags. Find the probability of both balls to be of same colour.
Answer
Total number of balls in first bag $=15+10=25$
Total number of balls in second bag $=16+9=25$
The above event can occur in two ways :
(i) when both balls are white or
(ii) when both balls are red
Probability of drawing of white ball from first bag
$=\frac{10}{25}$
Probability of drawing of white ball from second bag
$=\frac{16}{25}$
Both the events are mutually independent.
Probability of both balls being white
$=\frac{10}{25} \times \frac{16}{25}=\frac{32}{125}$
Similarly, probability of both balls being red
$=\frac{15}{25} \times \frac{9}{25}=\frac{27}{125}$
The probability of both balls being of same colour
$=\frac{32}{25}+\frac{27}{125}=\frac{59}{125}$
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Question 22 Marks
Four persons are chosen at random from a group of 4 males, 3 females and 5 children. Find the probability that exactly two of them are children.
Answer
Total number of people $=4+3+5=12$
4 people from 12 people can be chosen in ${ }^{12} C _4$ ways
$=\frac{12 \times 11 \times 10 \times 9}{1 \times 2 \times 3 \times 4}=495$
There must be two children in every selection which can be done in ${ }^5 C _2$ ways. Remaining two persons will be chosen from $3$ males $+$ $4$ females $=7$ persons which can be done in ${ }^7 C _2$ ways. Therefore, favourable outcomes
$\begin{array}{l}={ }^5 C_2 \times{ }^7 C_2 \\=\frac{5 \times 4}{1 \times 2} \times \frac{7 \times 6}{1 \times 2}=210\end{array}$
Hence, required probability $=\frac{210}{495}=\frac{14}{33}$
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Question 32 Marks
The probability of winning of $A$ in a race is $\frac{1}{3}$ and the probability of winning of $B$ is $\frac{1}{4}$. Find the probability of not winning of either of them.
Answer
Probability of winning of $A , P ( A )=\frac{1}{3}$
Probability of winning of $B , P ( B )=\frac{1}{4}$
Only one can win the race, so these are mutually exclusive events.
Probability of winning of at least any one of them
$\begin{array}{l}P(A \cup B)=P(A)+P(B) \\=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\end{array}$
Probability of not winning of either of them
$=1-\frac{7}{12}=\frac{12-7}{12}=\frac{5}{12}$
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