Question

out of 8 items that have arrived,one is defective.A worker picks these parts one by one.what is the probability that the third is defective given that the first two are not (use condotional probability)​

Answer 1

Given :- Out of 8 items that have arrived, one is defective. A worker picks these parts one by one. What is the probability that the third is defective given that the first two are not ?

Solution :-

we have,

→ Total items arrived = 8 .

→ No. of defective item = 1 .

now, given that,

first 2 packed items are not defective .

then,

→ Items left to be packed = 8 - 2 = 6 items.

and,

→ Defective item is = 1 .

therefore,

→ Probability that the 3rd item is defective = (No. of Defective item) / (Items left to be packed) = (1/6) (Ans.)

Hence, probability that the third is defective will be (1/6).

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