michael has 3 quarters, 2 dimes, and 3 nickels in his pocket. he randomly draws two coins from his pocket…

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 1/4 because 2 of the 8 coins are dimes. is he correct? explain.

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 1/4 because 2 of the 8 coins are dimes. is he correct? explain.

Answer

Explanation:

Step1: Calculate first - draw probability

The total number of coins is $3 + 2+3=8$. The probability of drawing a dime on the first draw is $\frac{2}{8}$ since there are 2 dimes out of 8 coins.

Step2: Calculate second - draw probability

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 is $\frac{1}{7}$.

Step3: Calculate combined probability

By the multiplication rule for independent events (in the case of sequential non - replacement draws), the probability of both events occurring is the product of their individual probabilities. So 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 $\frac{1}{4}$.