If Mike has 8 dimes and 6 pennies in his pocket. If he randomly selected one coin and then a second coin without replacing the first, what is the probability that both coins were dimes?

prob(2dimes) = (8/14)(7/13)

= 4/13

Probability of choosing the first coin (dime) = 8/14

Probability of choosing the second without replacement = 7/13
Probability of choosing the first AND second = (8/14)(7/13)=4/13

To find the probability that both coins were dimes, we need to calculate two things: the total number of possible outcomes and the number of favorable outcomes.

First, let's calculate the total number of possible outcomes. In the given scenario, Mike can select any one of the 14 coins (8 dimes + 6 pennies) as the first coin. Once the first coin is selected, there are only 13 coins left to choose from for the second coin since we are not replacing the first coin.

Therefore, the total number of possible outcomes is 14 * 13 = 182.

Next, let's calculate the number of favorable outcomes, which is the number of ways to select two dimes. Since Mike has 8 dimes, for the first coin he can select any one of the 8 dimes, and then for the second coin he will have 7 remaining dimes to choose from.

Thus, the number of favorable outcomes is 8 * 7 = 56.

Finally, the probability that both coins were dimes is given by the number of favorable outcomes divided by the total number of possible outcomes:

Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = 56 / 182
Probability = 0.3077 (rounded to four decimal places)

Therefore, the probability that both coins were dimes is approximately 0.3077, or 30.77%.