Question

A jar contains 6 pennies, 7 nickels and 8 dimes. A child selects 2 coins at random without replacement from the jar. Let X represent the amount in cents of the selected coins.Round your answers to 3 decimal places.Find the probability X = 10. Find the probability X = 11

Answer 1

Answer:

(a)0.1 (b)0.229

Explanation:

First, briefly recall the following:

• 1 penny = 1 cent

,

• 1 nickel = 5 cents

,

• 1 dime = 10 cents

The child is to select two coins.

There are 6 pennies, 7 nickels and 8 dimes.

The total number of coins = 6 + 7 + 8 = 21.

Let X represent the amount in cents of the selected coins.

(a)X=10

For the child to pick two coins worth 10 cents, the child must pick two nickels.

$\begin{gathered} P(1st\text{ nickel)}=\frac{7}{21} \\ P(2nd\text{ nickel)}=\frac{6}{20} \end{gathered}$

Therefore:

$P(X=10)=\frac{7}{21}\times\frac{6}{20}=0.1$

The probability that X=10 is 0.1.

(b)X=11

For the child to pick two coins worth 11 cents, the child must pick 1 dime and 1 penny in any of the two orders below.

• Penny then dime

,

• Dime then penny

$\begin{gathered} P(\text{penny 1st)}\times P(\text{dime 2nd)}=\frac{6}{21}\times\frac{8}{20}=\frac{4}{35} \\ P(\text{dime 1st)}\times P(\text{penny 2nd)}=\frac{8}{21}\times\frac{6}{20}=\frac{4}{35} \end{gathered}$

Therefore:

$P(X=11)=\frac{4}{35}+\frac{4}{35}=\frac{8}{35}\approx0.229$

The probability that X=11 is 0.229.