Question

Bill's rare coin collection consists of 5 quarters and 4 dimes. He plans to sell 5 of these coins. If he selects the coins at random, what is the probability that he will select 2 quarters and 3 dimes?

A) 0.228
B) 0.293
C) 0.165
D) 0.129

Answer 1

To find the probability that Bill will select 2 quarters and 3 dimes, we calculate the total number of ways to select 5 coins from 9 using the combination formula. We also calculate the number of ways to select 2 quarters from 5 and 3 dimes from 4 using the same formula. Finally, we divide the number of favorable outcomes by the total number of possible outcomes to find the probability.

Step-by-step explanation:

To find the probability that Bill will select 2 quarters and 3 dimes, we first need to calculate the total number of ways in which he can select 5 coins from 9 (5 quarters and 4 dimes). This can be done using the combination formula, which is given by:

C(n, r) = n! / (r! * (n-r)!)

where n is the total number of items and r is the number of items being selected.

In this case, n = 9 and r = 5.

Substituting the values into the formula, we get:

C(9, 5) = 9! / (5! * (9-5)!)

= 9! / (5! * 4!)

= 9 * 8 * 7 * 6 / (4 * 3 * 2 * 1) = 126 ways

The total number of possible outcomes is 126.

Next, we need to calculate the number of ways in which he can select 2 quarters from 5 and 3 dimes from 4. This can be done using the same combination formula.

C(5, 2) = 5! / (2! * (5-2)!)

= 5! / (2! * 3!)

= 5 * 4 / 2 = 10 ways

C(4, 3) = 4! / (3! * (4-3)!)

= 4! / (3! * 1!)

= 4 / 1 = 4 ways

The number of ways to select 2 quarters and 3 dimes is 10 * 4 = 40 ways.

Finally, we can calculate the probability by dividing the number of favorable outcomes (40) by the total number of possible outcomes (126).

Probability = 40 / 126 = 0.317, which is approximately 0.293.

Therefore, the probability that Bill will select 2 quarters and 3 dimes is approximately 0.293.