Respuesta :
The probability that the first two electric toothbrushes sold are defective is 0.016.
The probability of an event, say E occurring is:
[tex]P(E)=\frac{n(E)}{N}[/tex]
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:
[tex](\left {{20} \atop {2}} \right. )[/tex][tex]=\frac{20!}{2!(20-2)!} =\frac{20!}{2!18!} =\frac{20*19*18!}{2!*18!} = 190[/tex]
The number of ways to select two defective toothbrushes to sell from the 3 defective toothbrushes is:
[tex](\left {{3} \atop {2}} \right. )[/tex][tex]=\frac{3!}{2!(3-2)!} =\frac{3!}{2!1!} =\frac{3*2!}{1!2!} =3[/tex]
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
[tex]\frac{3}{190}\\ =0.01579\\=0.016[/tex]
Thus, the probability that the first two electric toothbrushes sold are defective is 0.016.
Learn more about probability here https://brainly.com/question/27474070
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