adityapareshshah wrote:
quantumliner wrote:
Given:
Probability of hitting the target is 0.7
Probability of Not hitting the target is 0.3
Now,probability that he will hit the target only three times in his first five attempts, this means, out of 5 attempts he will hit the target 3 times and miss the target 2 times ==> which can be represented as = 0.7 * 0.7 * 0.7 * 0.3 * 0.3
Now the three successful and unsuccessful attempts can happen in any sequence and which needs to be accounted for above.
So one of the sequence may be - Success,Failure,Success,Success,Failure
Another sequence may be - Failure, Success,Failure,Success,Success
So to account the different sequences we can considers that there are five places where three Success needs to be placed and this can be represented by 5C3 = 5!/3!*2! = 10
Therefore,
the approximate probability that he will hit the target only three times in his first five attempts
= 0.7 * 0.7 * 0.7 * 0.3 * 0.3 * 10
= 0.3087
Answer is B. 0.3
Hi
quantumliner,
I have a little silly doubt. Need to get concepts right.
We are considering the sequence for occurrence of Success and hence multiplying the probability by 10, but we aren't considering the failure sequence here.
What am I missing?
adityapareshshahWhen I said we need to consider the sequence of success, it also includes the sequence of failure.
For example, lets take the sequence - "Success,Failure,Success,Success,Failure". This sequence is a unique sequence for both success and failure.
In another sequence "Failure, Success,Failure,Success,Success", This again is a unique sequence for both success and failure.
In simple words, we are asked to arrange 3 success and 2 failure in a row of 5 seats. Hopefully this helps.