The items produced by a company contain 10% defective items. Show that the probability of getting 2 defective items in a sample of 8 items is $fraction numerator 28 space cross times space 9 to the power of 6 over denominator 10 to the power of 8 end fraction.$

Solution

Here n = 8
$straight p space equals space 10 over 100 space equals 1 over 10 comma space space straight q space equals space 1 minus straight p space equals space 1 minus 1 over 10 space equals 9 over 10$
Required probability = $straight P left parenthesis 2 right parenthesis space equals space straight C presuperscript 8 subscript 2 space open parentheses 9 over 10 close parentheses to the power of 6 space open parentheses 1 over 10 close parentheses squared space equals space fraction numerator 8 space cross times 7 over denominator 1 cross times 2 end fraction cross times 9 to the power of 6 over 10 to the power of 6 cross times 1 over 10 squared space equals space fraction numerator 28 space cross times space 9 to the power of 6 over denominator 10 to the power of 12 end fraction$

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Probability

The items produced by a company contain 10% defective items. Show that the probability of getting 2 defective items in a sample of 8 items is $fraction numerator 28 space cross times space 9 to the power of 6 over denominator 10 to the power of 8 end fraction.$

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