A box contains 2 fifty paisa coins, 5 twenty five paisa coins and a ce

1.	A box contains 2 fifty paisa coins, 5 twenty five paisa coins and a certain fixed number N (≥ 2) of ten and five paisa 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 paisa.
Answer» 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 = ²C₁ x ⁵C₄ x ^NC₀ + ²C₂ x ⁵C₃ x ^NC₀ + ²C₂ x ⁵C₂x NC₁ = 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

A box contains 2 fifty paisa coins, 5 twenty five paisa coins and a certain fixed number N (≥ 2) of ten and five paisa 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 paisa.

Answer»

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

= ²C₁ x ⁵C₄ x ^NC₀ + ²C₂ x ⁵C₃ x ^NC₀ + ²C₂ x ⁵C₂x NC₁

= 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

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