Fredrick's coins

Question

fredrick has 8 coins in his pocket, consisting of 5 quarters and 3 nickels. If fredrick pulls 2 coins out without replacement, what is the probability that they will both be quarters or they will both be nickels?

Answer 1

n = 8 coins (5 quarters and 3 nickels)

A = picking a quarter

B = picking a nickel

P(A) = 5/8 or 0.625

P(B) = 3/8 or 0.375

We want to find the probability of picking 2 quarters and the probability of picking 2 nickels, separately. Since the coins are not replaced, the events will be dependent. Let C be both quarters and D be both nickels.

P(A₁ and A₂) = P(C) = (5/8)*(4/7) = 20/56 = 5/14 ≈ 0.3571

P(B₁ and B₂) = P(D) = (3/8)*(2/7) = 6/56 = 3/28 ≈ 0.1071

Now, we want to find the probability of both coins will be quarters or both coins will be nickels. Use the Addition Rule. The event (C and D) is mutually exclusive because you can't have both coins be quarters and nickels at the same time.

P(C or D) = P(C) + P(D) - P(C and D) = (20/56) + (6/56) - 0 = 26/56 = 13/28 ≈ 0.4643

The probability that either both coins be quarters or both coins be nickels is about 0.4643.