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Harlamova29_29 [7]
3 years ago
11

A supplier delivers an order for 20 electric toothbrushes to a store. By accident, three of the electric toothbrushes are defect

ive. What is the probability that the first two electric toothbrushes sold are defective? Group of answer choices
Mathematics
1 answer:
krek1111 [17]3 years ago
8 0

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 <em>E </em>occurring is:

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

Here,

n (E) = favorable outcomes

N = total number of outcomes

Let <em>X</em> = number of defective electric toothbrushes sold.

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

Number of defective electric toothbrushes is, <em>x</em> = 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.

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