Let p is the probability of an item to be defective nbsp; Given p = 0.1, q = 0.9, n = 10 amp; Let X = Number of defective items So by using Binomial Distr ...

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Binomial Distribution

A lot had 10

Question

A lot had 10% defective items. Ten items are chosen randomly from this lot. The probability that exactly 2 of the chosen items are defective is

0.0036

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0.1937

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0.2234

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0.3874

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Solution

The correct option is B 0.1937
Let p is the probability of an item to be defective
Given p = 0.1, q = 0.9, n = 10 & Let X = {Number of defective items}

So by using Binomial Distribution

P(X = 2 success) $=^{n} C_{2} p^{2} q^{n - 2}$
$=^{10} C_{2} (0.1)^{2} (0.9)^{8}$
= 0.1937

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A lot had 10% defective items. Ten items are chosen randomly from this lot. The probability that exactly 2 of the chosen items are defective is

The correct option is B 0.1937Let p is the probability of an item to be defective Given p = 0.1, q = 0.9, n = 10 & Let X = {Number of defective items} So by using Binomial Distribution P(X = 2 success) = nC2 p2 qn−2 = 10C2(0.1)2 (0.9)8 = 0.1937

The correct option is B 0.1937
Let p is the probability of an item to be defective
Given p = 0.1, q = 0.9, n = 10 & Let X = {Number of defective items}

So by using Binomial Distribution

P(X = 2 success) $=^{n} C_{2} p^{2} q^{n - 2}$
$=^{10} C_{2} (0.1)^{2} (0.9)^{8}$
= 0.1937