Nine items are taken at random from a box of 21 items. The box is rejected if more than 3 item is found to be faulty. If there are 5 items, find the probability that the box is accepted.

From PU/ Probability and Queuing Theory

Asked by prakash lawagun on 12 Aug, 2020

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arjun adhikari on 12 Aug, 2020 Like 1 Dislike

Out of 21 items, 16 are good and 5 are faulty. 9 items are chosen at random from the box.

So, 9 items can be choosen from 21 items in 21C9 different ways.

In this condition, box wont be accepted if we find 4 faulty items or 5 faulty items out of 9 items choosen.

Out of 9 items:

4 faulty items can be choosen in 5C4 + 16C5 ways.

5 faulty items can be choosen in 5C5 + 10C4 ways.

The box will be rejected if we choose 4 or 5 faulty items out of 9 items. So the probability of box being rejected

P(r) = (5C4 + 16C5)/21C9 Or (5C5 + 16C4)/21C9

= ((5C4 + 16C5)+(5C5 + 16C4))/21C9

=> P(r) = 0.02107

So, the probability of being accepted, 

P(a) = 1 - P(r)

= 1- 0.02107

= 0.978926

Hence, the probabily of being accepted is: 0.978926.

 

 

 

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Like 0 Dislike line 6 5C5 + 16C4 hola nui? Alson Garbuja on 8 Dec, 2021

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