Question

michael has 3 quarters, 2 dimes, and 3 nickels in his pocket. he randomly draws two coins from his pocket, one at a time, and they are both dimes. he says the probability of that occurring is $\frac{1}{4}$ because 2 of the 8 coins are dimes. is he correct? explain.

Explanation:

Step1: Calculate total number of coins

There are $3 + 2+3=8$ coins initially.

Step2: Calculate probability of first - draw being a dime

The probability of drawing a dime on the first draw is $\frac{2}{8}$ since there are 2 dimes out of 8 coins.

Step3: Calculate probability of second - draw being a dime

After drawing one dime on the first draw, there is 1 dime left and 7 coins left. So the probability of drawing a dime on the second draw given that the first - draw was a dime is $\frac{1}{7}$.

Step4: Calculate probability of both events

By the multiplication rule for independent events (in the case of sequential draws without replacement), the probability of drawing two dimes is $\frac{2}{8}\times\frac{1}{7}=\frac{2}{56}=\frac{1}{28}$.

Answer:

No, he is not correct. The probability of drawing two dimes is $\frac{1}{28}$, not $\frac{1}{4}$.