Question

Sec 2.1 Pitman "Probability"

A man fires 8 shots at a target. Assume that the shots are independent, and each shot hits the bull’s eye with a probability of 0.7

Given that he hits the bull’s eye at least twice, what is the chance that he hits the bull’s eye exactly 4 times?

P(hits the bull’s eye exactly 4 times | hits the bull’s eye at least twice)

The answer has P(hits the bull’s eye exactly 4 times) / P(hits the bull’s eye at least twice)

How is this determined???

Answer 1

Conditional Probability of event A given that event B has already occured

P(hits the bull’s eye exactly 4 times) / P(hits the bull’s eye at least twice)

P(r success in n events) = (Probability Of Success) ^r * (Probability Of Failure) ^{n - r}

Here, C stands for combination

Now,

P(hits the bull’s eye exactly 4 times out of 8)

Here success is hitting bull's eye

And

P(hits the bull’s eye at least twice)

= P(hitting bull`s eye 2 times) + P(hitting bull`s eye 3 times) + P(hitting bull`s eye 4 times) + P(hitting bull`s eye 5 times) + P(hitting bull`s eye 6 times) + P(hitting bull`s eye 7 times) + P(hitting bull`s eye 8 times)

OR

= 1- P(hitting bull`s eye 0 times) + P(hitting bull`s eye 1 time)

So,

P(hits the bull’s eye at least twice)

= 1- P(hitting bull`s eye 0 times) + P(hitting bull`s eye 1 time)

  

So,

P(hits the bull’s eye exactly 4 times) / P(hits the bull’s eye at least twice)

If you have any query related to question or the solution provided feel free to ask.