Question

A box contains 2 fifty paise coins, 5 twenty-five paise coins and a certain number $n(\ge 2)$ of ten and five paise coins. Five coins are taken out of the box at random. The probability that the total value of these 5 coins is less than one rupee and fifty paise is

• A. $\frac{10(n+2)}{{}^{n+7}{C}_5}$
• B. $1-\frac{10(n+2)}{{}^{n+7}{C}_5}$
• C. $\frac{5(n+7)}{{}^{n+2}{C}_5}$
• D. $1-\frac{5(n+7)}{{}^n{C}_5}$

Answer 1

The correct option is B $1-\frac{10(n+2)}{{}^{n+7}{C}_5}$

The total number of coins in the box is $N+7$.
The total value of selected coins will be
greater that or equal to one rupee and fifty paise if the selected coins are
$-$ 1 fifty paise coin, 4 twenty five paise coins
$-$ 2 fifty paise coin, 3 twenty five paise coins
$-$ 2 fifty paise coin, 2 twenty five paise coins
and 1 coin out of the $N$ five paise and ten paise coins
If $E=$ total value of the $5$ selected coins is graeter than equal to one rupee and fifty paise, then
$n\left(E\right){=}^2{C}_1{.}^5{C}_4{.}^N{C}_0{+}^2{C}_2{.}^5{C}_3{.}^N{C}_0{+}^2{C_2}^5{C}_2{.}^N{C}_1{.}^{10}{C}_1=10(N+2)$
Hence, required probability $=1-P\left(E\right)=1-\frac{n\left(E\right)}{n\left(S\right)}=1-\frac{10(N+2)}{N{+}^7{C}_5}$