Question

From a box containing 4 dimes and 2 nickels,
3 coins are selected at random without replacement.
Find the probability distribution for the total T of the
3 coins. Express the probability distribution graphically as a probability histogram

Answer 1

The objective of the study is to determine the probability distribution for the total T of the 3 coins.

The provided information is that a box containing 4 dimes and 2 nickels, 3 coins are selected at random without replacement.

Let D denote the dimes and N denote the Nickels.

Here, One dime equals to 10 cents and 1 Nickle is equal to 5 cents.

The total number of three coins with three Dimes (DDD) and no Nickel (N) is,

$T= 3 \times 10 + 0 \times 5 \\= 30+0 \\= 30 \;cents$

The probability that the total number of three coins with three Dimes (DDD) and no Nickel (N) is,

$P(T=30) = \frac{\binom{4}{3} \binom{2}{0}}{\binom{6}{3}} \\ = \frac{4 \times 1}{20} \\ = \frac{1}{5}$

The total number of three coins with two Dimes (DD) and 1 nickel (N) is,

$T= 2 \times 10 + 1 \times 5 \\ = 20+5 \\ = 25 \;cents$

The probability that the total number of three coins with two Dimes (DD) and 1 nickel (N) is,

$P(T=25) = \frac{\binom{4}{2} \binom{2}{1}}{\binom{6}{3}} \\ = \frac{6 \times 2}{20} \\ = \frac{3}{5}$

The total number of three coins with one Dime (D) and 2 nickel (N) is,

$T=1 \times 10 + 2 \times 5 \\ = 10 +10 \\ =20$

The probability that the total number of three coins with one Dime (D) and 2 nickel (N) is,

$P(T=20) = \frac{\binom{4}{1} \binom{2}{2}}{\binom{6}{3} } \\ = \frac{4 \times 1}{20} \\ = \frac{1}{5}$

The probability distribution for the total T of the 3 coins is,

Construct the histogram for the obtained probability distribution for the total T of the 3 coins.

From the histogram, it can be conclude that the distribution for the total T of the 3 coins is symmetric.