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Counting & probability: defective bulbs

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by euro » Fri Oct 01, 2010 10:50 am
[Q] There are a few bulbs in a box. If 30% of them are defective, what is the probability that exactly 4 out of 7 bulbs being picked one after another from the box and immediately put back are non-defective?

(A) [30 x 7^3 x 3^3 ] / 10^7

(B) [30 x 7^3 x 3^4 ] / 10^7

(C) [30 x 7^4 x 3^3 ] / 10^7

(D) [35 x 7^3 x 3^4 ] / 10^7

(E) [35 x 7^4 x 3^4 ] / 10^7

The OA is [spoiler](E)[/spoiler]
Please explain.
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by sanabk » Fri Oct 01, 2010 3:46 pm
4 out of 7 bulbs non-defective (ND)

D=30%=3/10
ND=70%=7/10

7c4(ND x ND x ND x ND x D x D x D x D)=35 x (7/10)^4 x (3/10)^3=[35 x 7^4 x 3^3}]/10^7

E
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by mj78ind » Fri Oct 01, 2010 7:45 pm
euro wrote:[Q] There are a few bulbs in a box. If 30% of them are defective, what is the probability that exactly 4 out of 7 bulbs being picked one after another from the box and immediately put back are non-defective?

(A) [30 x 7^3 x 3^3 ] / 10^7

(B) [30 x 7^3 x 3^4 ] / 10^7

(C) [30 x 7^4 x 3^3 ] / 10^7

(D) [35 x 7^3 x 3^4 ] / 10^7

(E) [35 x 7^4 x 3^4 ] / 10^7

The OA is [spoiler](E)[/spoiler]
Please explain.
Think E has a typo, the prob is given by Binomial - 7C4(7/10)^4*(3/10)^3

Hence E should read 3^3
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