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Question: The bulbs manufactured by a company are known to be 5% defective. If company sells in boxes of 100 a...

The bulbs manufactured by a company are known to be 5% defective. If company sells in boxes of 100 and guarantees

that not more than 10 bulbs will be defective. Assuming

poisson distribution, the probability that a box will fail to

meat the requirement, is –

A

1 + ∑x=09e−55xx!\sum _ { x = 0 } ^ { 9 } \frac { e ^ { - 5 } 5 ^ { x } } { x ! }

B

1 – ∑x=010e−55xx!\sum _ { x = 0 } ^ { 10 } \frac { e ^ { - 5 } 5 ^ { x } } { x ! }

C

1 – ∑x=110e−55xx!\sum _ { x = 1 } ^ { 10 } \frac { e ^ { - 5 } 5 ^ { x } } { x ! }

D

1 – ∑x=110e−44xx!\sum _ { x = 1 } ^ { 10 } \frac { e ^ { - 4 } 4 ^ { x } } { x ! }

Answer

1 – ∑x=010e−55xx!\sum _ { x = 0 } ^ { 10 } \frac { e ^ { - 5 } 5 ^ { x } } { x ! }

Explanation

Solution

Probability of defective = p = 5100\frac { 5 } { 100 } = 120\frac { 1 } { 20 }