Question

A jar contains 8 pennies, 5 nickels and 7 dimes. A child selects 2 coins at random without replacement from the jar. Let X represent the amount in cents of the selected coins. Be very precise with your answers.

a. Find the probability x = 2 cents.

b. Find the probability x = 6 cents.

c. Find the probability x = 10 cents.

d. Find the probability x = 11 cents.

e. Find the probability x = 15 cents.

f. Find the probability x = 20 cents.

g. Find the expected value of x.

Answer 1

a. The probability x = 2 cents = 7/22

b. The probability x = 6 cents = 35/66

c. The probability x = 10 cents = 5/33

d. The probability x = 11 cents= 28/33

e. The probability x = 15 cents = 20/33

f. The probability x = 20 cents = 14/33

g. The expected value of x = 5.9

Explanation:

This is a binomial probability distribution. The number of trials is known .

a. The probability x = 2 cents.

Probability ( X=2) P( selecting 2 dimes)= 7C2 / 12c2

= 21 / 66 = 7/22

b. The probability x = 6 cents.

Probability ( X=6) P( selecting a nickel and a dime)= 5C1 * 7C1/ 12c2

= 5*7 / 66 = 35/66

c. The probability x = 10 cents.

Probability ( X=10) P( selecting two nickels )= 5C2 / 12c2)

= 10/ 66 = 5/33

d. The probability x = 11 cents.

Probability ( X=11) P( selecting a penny and a dime)= 8C1 * 7C1/ 12c2)

= 8*7 / 66 = 56/66= 28/33

e. The probability x = 15 cents.

Probability ( X=15) P( selecting a penny and a nickel)= 8C1 * 5C1/ 12c2)

= 8*5 / 66 = 40/66= 20/33

f. The probability x = 20 cents.

Probability ( X=20) P( selecting 2 pennies )= 8C2 / 12c2)

= 28 / 66 = 14/33

g. The expected value of x.

E(X) = np

E(X) = 2 * (8C2+ 5C2+ 7C2)/(8+5+7) = 2( 28+10+21)/20

=2(59)/20= 5.9