Question

A jar of coins contains 8 pennies, 5 nickels, 12 dimes, 7 quarters. you choose a coin at random, do not replace it, and then choose another coin. what is probability of choosing a dime first and quarter second

Answer 1

The probability of choosing a dime first and quarter second is approximately 0.081.

Step-by-step explanation:

The probability of choosing a dime first and a quarter second can be found by multiplying the probability of choosing a dime first by the probability of choosing a quarter second.

There are a total of 8 + 5 + 12 + 7 = 32 coins in the jar. The probability of choosing a dime first is 12/32, since there are 12 dimes in the jar. Once a dime has been chosen, there are 31 coins remaining, so the probability of choosing a quarter second is 7/31, since there are 7 quarters remaining.

Therefore, the probability of choosing a dime first and a quarter second is (12/32) * (7/31) = 0.08064516129, or approximately 0.081.

Answer 2

`|\Omega|=32\cdot31=992\\ |A|=12\cdot7=84\\\\ P(A)=(84)/(992)=(21)/(248)\approx8.5\%`