Question

A jar of coins has 8 pennies, 5 nickels, 1 dime, and 7 quarters. What is the probability of drawng a quarter, replacing it, and drawing another quarter.

Answer 1

Given:

A jar of coins has 8 pennies, 5 nickels, 1 dime, and 7 quarters.

To find:

The probability of drawing a quarter, replacing it, and drawing another quarter.

Solution:

A jar of coins has 8 pennies, 5 nickels, 1 dime, and 7 quarters.

Total number of coins = 8+5+1+7 = 21

Probability of getting a quarter is

$P(Quarter)=\dfrac{\text{Number of quarters}}{\text{Total number of coins}}$

$P(Quarter)=\dfrac{7}{21}$

$P(Quarter)=\dfrac{1}{3}$

After drawing a quarter, we replaced it. So, the probability of getting quarter in second draw is the same and first one, .i.e., $P(Quarter)=\dfrac{1}{3}$.

The probability of drawing a quarter, replacing it, and drawing another quarter is

$P=\dfrac{1}{3}\times \dfrac{1}{3}$

$P=\dfrac{1}{9}$

Therefore, the required probability is $\dfrac{1}{9}$.