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The probability of failure of the telephone exchange for each call is 0.004.Received 500 calls. Determine the probability that there were 9 failures.

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Given that we know the probability of a failure in a telephone call, we want to find the probability that in 500 calls, 9 of them will fail. The probability of having exactly 9 failures is 0.00018

Working with probabilities.

We know that:

  • A call has a 0.004 probability of failing
  • A call has a 1 - 0.004 = 0.996 probability of not failing.

So if in 500 calls 9 of them fail, then we have the product of 0.004 9 times and the product of 0.996 491 times.

This is written as:

[tex]p = 0.004^9*0.996^{491}[/tex]

But this would be only for one set of 9 calls that can fail, actually, any group of 9 calls out of the set of 500 can fail, so we need to take in account the "permutations".

The number of different groups of 9 calls that we can take from the group of 500 is given by:

[tex]N = \frac{500!}{(500- 9)!*9!} = \frac{500*499*498*497*496*495*494*493*492}{9*8*7*6*5*4*3*2*1} = 5*10^{18}[/tex]

We need to multiply p by this number, so we get:

[tex]P = 0.004^9*0.996^{491}*5*10^{18} = 0.00018[/tex]

So the probability of having exactly 9 failures is 0.00018

If you want to learn more about probability, you can read:

https://brainly.com/question/251701

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