Probability Ques 21
A box contains $2$ fifty paise coins, $5$ twenty five paise coins and a certain fixed number $n(\geq 2)$ 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 $5$ coins is less than one rupee and fifty paise.
$(1988,3$ M)
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Answer:
Correct Answer: 21.$(1-\frac{10(n+2)}{{ }^{n+7} C _5})$
Solution:
Formula:
- 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.
(i) When one $50$ paise coin and four $25$ paise coins are chosen.
(ii) When two $50$ paise coins and three $25$ paise coins are chosen.
(iii) When two $50$ paise coins, $2$ twenty five 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
$ \begin{gathered} \left({ }^{2} C _1 \cdot{ }^{5} C _5 \cdot{ }^{n} C _0\right)+\left({ }^{2} C _2 \cdot{ }^{5} C _3 \cdot{ }^{n} C _0\right)+\left({ }^{2} C _2 \cdot{ }^{5} C _2 \cdot{ }^{n} C _1\right) \\ =10+10+10 n=10(n+2) \end{gathered} $
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} $
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