Question

A supplier delivers an order for 20 electric toothbrushes to a store. By accident, three of the electric toothbrushes are defective. What is the probability that the first two electric toothbrushes sold are defective? Group of answer choices

Answer 1

Answer:

The probability that the first two electric toothbrushes sold are defective is 0.016.

Step-by-step explanation:

The probability of an event, say E occurring is:

$P (E)=\frac{n(E)}{N}$

Here,

n (E) = favorable outcomes

N = total number of outcomes

Let X = number of defective electric toothbrushes sold.

The number of electric toothbrushes that were delivered to a store is, n = 20.

Number of defective electric toothbrushes is, x = 3.

The number of ways to select two toothbrushes to sell from the 20 toothbrushes is:

${20\choose 2}=\frac{20!}{2!(20-2)!}=\frac{20!}{2!\times 18!}=\frac{20\times 19\times 18!}{2!\times 18!}=190$

The number of ways to select two defective toothbrushes to sell from the 3 defective toothbrushes is:

${3\choose 2}=\frac{3!}{2!(3-2)!}=\frac{3!}{2!\times 1!}=3$

Compute the probability that the first two electric toothbrushes sold are defective as follows:

P (Selling 2 defective toothbrushes) = Favorable outcomes ÷ Total no. of outcomes

                                                            $=\frac{3}{190}$

                                                            $=0.01579\\\approx0.016$

Thus, the probability that the first two electric toothbrushes sold are defective is 0.016.