Question

A box contains $2$ fifty paise coins, $5$ twenty five paise coins & a certain fixed number $N($$\ge$$2)$ of ten and five paise coins. Five coins are taken out of the box at random. Find the probability that the total value of these five coins is less than $Re.1\&50$ paise.

• A. $1-\left\{\frac{20+10N}{{}^{N+7}{C}_5}\right\}$
• B. $\left\{\frac{20+10N}{{}^{N+7}{C}_5}\right\}$
• C. $\left\{\frac{20+10N}{{}^{N+5}{C}_7}\right\}$
• D. $1-\left\{\frac{20+10N}{{}^{N+5}{C}_7}\right\}$

Answer 1

Answer: (A)

$1-\left\{\frac{20+10N}{{}^{N+7}{C}_5}\right\}$

There are $(n+7)$ coins in the box out of which five coins can be taken out in ${}^{n+7}{C}_5$ ways.
The total value of $5$ coins can be equal to or more than one rupee and fifty paise in the following ways.
1) when one $50$ paise coin and four $25$ paise coins are chosen.
2) when two $50$ paise coins and three $25$ paise coins are chosen.
3) when two $50$ paise, two $25$ paise coins and one from $n$ coins of ten and five paise.
Therefore the total number of ways of selecting five coins so that the total value of the coins is not less than one rupee and fifty paise is,
$\left({}^2{C}_1{.}^5{C}_5{.}^n{C}_0\right)+\left({}^2{C}_2{.}^5{C}_3{.}^n{C}_0\right)+\left({}^2{C}_2{.}^5{c}_2{.}^n{C}_1\right)=10+10+10n=10(n+2)$
So, the number of ways of selecting five coins, so that the total value of the coins is less than one rupee and fifty paise is
${}^{n+7}{C}_5-10(n+2)$
Therefore required probability $=\frac{{}^{n+7}{C}_5-10(n+2)}{{}^{n+7}{C}_5}=1-\frac{10(n+2)}{{}^{n+7}{C}_5}$