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# A box contains 2 fifty paise coins, 5 twenty five paise coins and a certain fixed number N (≥ 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.

Jitender Pal
7 years ago
Hello Student,
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 + 7C5 Let E denotes the event that the sum of the values of the coins is less than one rupee and fifty paisa.
Then E’ denotes the event that the total value of the five coins is equal to or more than one rupee and fifty paisa.
NOTE THIS STEP:
The number of cases favorable to E’ is
= 2 C1 x 5 C4 x N C0 + 2 C2 x 5 C3 x N C0 + 2 C2 x 5 C2 x N C1
= 2 x 5 + 10 + 10 N = 10 (N + 2)
∴ P (E) = 10 (N + 2)/n + 1 C5
⇒ P (E) = 1 – P (E) = 1 – 10 (N + 2)/N + 7 C5

Thanks
Jitender Pal