Question

A bag contains tickets number 1 to 30. Three tickets are drawn at random from the bag. What is the probability that the maximum number of the selected tickets exceeds 25?

Or

Out of 9 outstanding students in a college, there are 4 boys and 5 girls. A team of four students is to be selected for a quiz programme. Find the probability that:

(i) two are boys and two are girls.

(ii) one is boy and three are girls.

Answer 1

Let A be the event that none of the selected tickets bear number exceeding 25.

Then, required probability = $P\left(\overline{A}\right)=1-P\left(A\right)$

The total number of ways of drawing three tickets out of 30 = $30{=}^{30}{C}_3=\frac{30!}{3!27!}=4060$

$\therefore$ Total number of outcomes = 4060

Since none of the selected tickets bear number exceeding 25. Therefore, three tickets are drawn tickets bearing number 1 to 25. This can be done in ${}^{25}{C}_3$ ways.

$\therefore$ Favourable number of outcomes = ${}^{25}{C}_3=\frac{25!}{3!22!}=2300$.

So, $P\left(A\right)=\frac{2300}{4060}=\frac{115}{203}$

Hence, required probability = $P\left(\overline{A}\right)=1-\frac{115}{203}=\frac{88}{203}$

Or

(i) Out of 9 students, 4 students can be selected in ${}^9{C}_4$ ways.

So, total number of outcomes = ${}^9{C}_4=\frac{9!}{4!5!}=126$

There are 4 boys and 5 girls out of which 2 boys and 2 girls can be selected in ${}^4{C}_2{\times }^5{C}_2$ ways.

So, favourable number of outcomes = ${}^4{C}_2{\times }^5{C}_2=6\times 10=60$

$\therefore$ Required probability = $\frac{60}{126}=\frac{10}{21}$

(ii) There are 4 boys and 5 girls out of which 1 boy and 3 girls can be selected in ${}^4{C}_1{\times }^5{C}_3$ ways.

So, favourable number of outcomes = ${}^4{C}_1{\times }^5{C}_3=4\times 10=40$

$\therefore$ Required probability = $\frac{40}{126}=\frac{20}{63}$