The items produced by a company contain 5% defective items. Show that the probability of getting 2 defective items in a sample of 10 items is $fraction numerator 45 space cross times space 19 to the power of 8 over denominator 20 to the power of 10 end fraction.$

Solution

Here n = 10
$straight p space equals space 5 over 100 space equals 1 over 20 comma space space straight q space equals space 1 space minus space straight p space equals 1 space minus 1 over 20 space equals space 19 over 20$
Required probability = $straight P left parenthesis 2 right parenthesis space equals space straight C presuperscript 10 subscript 2 space open parentheses 19 over 20 close parentheses to the power of 8 space open parentheses 1 over 20 close parentheses squared$
$equals space fraction numerator 10 space cross times space 9 over denominator 1 space cross times 2 end fraction cross times 19 to the power of 8 over 20 to the power of 8 cross times 1 over 20 squared space equals fraction numerator 45 space cross times space 19 to the power of 18 over denominator 20 to the power of 10 end fraction$

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Probability

The items produced by a company contain 5% defective items. Show that the probability of getting 2 defective items in a sample of 10 items is $fraction numerator 45 space cross times space 19 to the power of 8 over denominator 20 to the power of 10 end fraction.$

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