You have 10 coins..3 quarters,5 dimes, and 2 nickels.
a. A coin is selected and then replaced. A second coin is selected and then replaced. A third coin is selected and then replaced. Find the probability that all 3 coins were dimes.
mmmh ?
what is (1/2)(1/2)(1/2) ?
To find the probability that all 3 coins were dimes, we need to determine the total number of outcomes and the number of favorable outcomes.
Total Number of Outcomes:
Since you have 10 coins (3 quarters, 5 dimes, and 2 nickels), the total number of outcomes is the total number of ways you can choose any three coins from these 10 coins. This is given by the formula for combinations, which can be expressed as C(10, 3) or "10 choose 3".
The formula for combinations is:
C(n, r) = n! / (r!(n-r)!)
where n is the total number of objects and r is the number of objects chosen at a time.
In this case, n = 10 (total number of coins) and r = 3 (number of coins chosen at a time).
Therefore, the total number of outcomes is C(10, 3) = 10! / (3!(10-3)!) = 10! / (3!7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120.
Number of Favorable Outcomes:
We want to find the number of ways to choose 3 dimes from the 5 dimes available. This is the same as finding C(5, 3) or "5 choose 3".
Using the formula for combinations, we get:
C(5, 3) = 5! / (3!(5-3)!) = 5! / (3!2!) = (5 * 4) / (2 * 1) = 10.
Probability:
The probability of an event is given by the formula:
Probability = Number of Favorable Outcomes / Total Number of Outcomes.
So, in this case, the probability that all 3 coins selected were dimes is:
Probability = Number of Favorable Outcomes / Total Number of Outcomes = C(5, 3) / C(10, 3) = 10 / 120 = 1 / 12.
Therefore, the probability that all 3 coins selected were dimes is 1/12.