Question 485391: Respected Sir / Mam ,
Please help me to solve this question. I will be very grateful for your help .
My question is
If the overall percentage of success in the exam is 60, what is the probability that out of a group of 4 students, at least one has passed?
As the answer given is : 0.9744
Please provide me the steps for this question.
Thank you
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Like many things in this life there is a hard way and an easy way to answer this question.
If 4 students take the test, then there are 5 situations that could possibly arise:
None pass, 1 passes, 2 pass, 3 pass, or all 4 pass.
Note for future reference that since these five outcomes are inclusive of all possible outcomes, the sum of the probabilities of each must be 1, because it is a certainty that one of the outcomes will occur.
Now, your question asks for the probability that at least one student passes. Looking at it in a straight-forward way, that would be the sum of the probabilities that exactly 1 passes plus exactly 2 pass plus...and so on.
Since "Pass/Fail" is an either/or outcome, and the probability of success for any given instance of one student taking one test is given as 60%, we know that the probability should be calculated using the binomial distribution.
The probability of  successes in  trials where  is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)\left(p\right)^k\left(1\,-\,p\right)^{n\,-\,k})
Where ) is the number of combinations of things taken at a time and is calculated by !})
But you would need to perform this calculation four times and then sum the results. First you would need to calculate:
\ =\ \left(4\cr 1\right\)\left(0.6\right)^1\left(0.4\right)^{3})
To get the probability for exactly 1, and then you would have to do exactly 2, and so on. In summary:
\ =\ \sum_{i=1}^4\left(4\cr i\right\)\left(0.6\right)^i\left(0.4\right)^{n\,-\,i})
But fortunately, there is a much simpler way. From the point of view of your question, there are only two possible outcomes. Either 1 or more pass, or nobody passes. So if you take the probability that nobody passes and subtract that from 1, you get the probability that at least 1 passes.
\ =\ 1\ -\ P_4(0,0.6)\ =\ 1\ -\ \left(4\cr 0\right\)\left(0.6\right)^0\left(0.4\right)^{4})
Since we know that pick 0 is 1 for all positive integers and that  for all real numbers  ,
The above reduces to:
^4)
The arithmetic is yours to do.
John
My calculator said it, I believe it, that settles it
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