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

Answer 1

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

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 = the number of defective electric toothbrushes sold.

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

The number of defective electric toothbrushes is x = 3.

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

$(\left {{20} \atop {2}} \right. )$$=\frac{20!}{2!(20-2)!} =\frac{20!}{2!18!} =\frac{20*19*18!}{2!*18!} = 190$

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

$(\left {{3} \atop {2}} \right. )$$=\frac{3!}{2!(3-2)!} =\frac{3!}{2!1!} =\frac{3*2!}{1!2!} =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

$\frac{3}{190}\\ =0.01579\\=0.016$

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

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