Question

A box contains 2 fifty paise coins ,5 twenty five paise coins and a certain fixed number N $(\ge 2)$of 10 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 $10(N+k)$. Find the value of $k$.

Answer 1

Since there are in all N+7 coins, the total number of ways of drawing 5 coins ${=}^{N+7}{C}_5$ .We have to find the probability that the
total value of 5 coins is $<$ one rupee and fifty paise.We first find the probability that the total value is $\ge$ one rupee
and fifty paisa.Now the total value will be equal or more than one rupee and fifty paisa in three ways only;(i) one fifty paisa coin and 4 twenty five paisa coins.(ii)2 fifty paise coins and 3 twenty five paise coins (iii) 2 fifty paise coins, 2 twenty five paise coins and one ten paise or five paise coin.So the favourable ways ${}^2{C}_2{,}^5{C}_3{,}^N{C}_0,{+}^2{C}_2{,}^5{C}_2{,}^N{C}_1{,}^2{C}_1{,}^5{C}_4,=10N+20$