Question

a flashlight has 6 batteries, 2 of which are defective, If 2 are selected at radom without replacement, find the proability that both are defective..........help

Answer 1

P(Both Defective) = P(First defective AND second defective)

P(Both Defective) = P(First defective) * P(Second defective)

P(Both Defective) = (2/6) * (1/5)

P(Both Defective) = 2/30

P(Both Defective) = 1/15

Answer 2

Hopefully that all makes sense

Answer 3

It sure does thank you for your help 5910

Answer 4

yw