A box contains 20 electric bulbs, out of which 4 are defective. Two

Your answer is almost right. When you pick one defective bulb and one non-defective bulb, you need to count the number of ways to pick the non-defective bulb as well. So you can pick 1 bulb of each type in (4C1)(16C1) ways, and the answer is

[(4C1)(16C1) + 4C2 ] / 20C2

which is equal to 7/19.




This actually makes no difference at all. After all, if you put your two hands in the box and take hold of two lightbulbs, it won't be any more or less likely you get two defective bulbs if you take your hands out of the box one at a time than if you take them both out at once.

I find it much easier in these types of problems to imagine making the selections one at a time. You can then avoid using expressions like '20C2' altogether. If I pick one bulb at a time, the probability the first is defective is 16/20, and the probability the second is then also defective is 15/19, so the probability both are defective is

(16/20)(15/19) = 12/19

and the probability at least one is defective is 1 - 12/19 = 7/19.

The probability that at least one of these is defective: 1 - none defective

None defective: Select both from non - defectives

Total : 20
Defectives: 4
Non-defective: 20 - 4 = 16

P(non-defective) = \(\frac{^{16}^{C_2} }{ ^{20}^{C_2}} = \frac{12}{19}\)

The probability that at least one of these is defective: \(1 - \frac{12}{19} = \frac{7}{19}\)

Answer B

Your answer is almost right. When you pick one defective bulb and one non-defective bulb, you need to count the number of ways to pick the non-defective bulb as well. So you can pick 1 bulb of each type in (4C1)(16C1) ways, and the answer is

[(4C1)(16C1) + 4C2 ] / 20C2

which is equal to 7/19.



Good explanation ..thnks...

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