A box contains two 50 paise coins, five 25 paise coins and a certain fixed number `N(geq2)` of 10 and 5-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 1 rupee and 50 paise.

Explanation

Here the total number of coins is N + 7, Therefore, the total number of ways of choosing 5 coins out of N + 7 is `.^(n + 7)C_(5)`.
Let E denote the event that the sum of the values of the coins is less than 1 rupee and 50 paise. Then E' denotes the event that the total values of the five coins is equal to or more than 1 rupee and 50 paise. The number of cases favourable to E' is
`.^(2)C_(1) xx .^(5)C_(4) xx .^(N)C_(0) + .^(2)C_(2) xx .^(5)C_(3) xx .^(N)C_(0) + .^(2)C_(2) xx .^(5)C_(2) xx .^(N)C_(1) = 2 xx 5 + 10 + 10N = 10(N + 2)`
`therefore P(E') = (10(N + 2))/(.^(N + 7)C_(5))`
`implies P(E) = 1- P(E') = 1 - (10 (N + 2))/(.^(N+7)C_(5))`