Question 258730: A candidate got interview calls from6 Institutes of Management. Assume that his chance of getting admission in each of the Institute is 0.25. Assume each institute conducts separate interview process and results are independent of each other. What is the probability that the candidate will get admission in at least 3 institutes?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! p(admission) = .25
interviews from 6 institutions
probability that the candidate will get admission in at 3 institutions?
this would be equal to:
probability of getting admission in exactly 3 plus:
probability of getting admission in exactly 4 plus:
probability of getting admission in exactly 5 plus:
probability of getting admission in exactly 6.
this is also equal to:
1 minus:
probability of getting admission in exactly 0 plus:
probability of getting admission in eacttly 1 plus:
probability of getting admission in exactly 2.
we'll do both to show you how it works, but in practice, you would normally choose the option that takes less work to get the result from.
probability of getting exactly 0 admissions = 1 * .75^6
probability of getting exactly 1 admission = 6 * .25 * .75^5
probability of getting exactly 2 admissions = 15 * .25^2 * .75^4
probability of getting exactly 3 admissions = 20 * .25^3 * .75^3
probability of getting exactly 4 admissions = 15 * .25^4 * .75^2
probability of getting exactly 5 admissions = 6 * .25^5 * .75
probability of getting exactly 6 admissions = 1 * .25^6
the multipliers are derived as follows:
you have a total of 6 to draw from.
you want to know how many ways you can draw different sets from this number.
the number of ways you can draw 0 is 1 based on the combination formula of c(6,0) = 6! / (6!*0!)
the number of ways you can draw sets of 1 is 6 based on the combination formula of c(6,1) = 6! / (5!*1!)
the number of ways you can draw sets of 2 is 15 based on the combination formula of c(6,2) = 6! / (4!*2!)
the number of ways you can draw sets of 3 is 20 based on the combination formula of c(6,3) = 6! / (3!*3!)
the number of ways you can draw sets of 4 is 15 based on the combination formula of c(6,4) = 6! / (2!*4!)
the number of ways you can draw sets of 5 is 6 based on the combination formula of c(6,5) = 6! / (1!*5!)
the number of ways you can draw sets of 6 is 1 based on the combination formula of c(6,6) = 6! / (0!*6!)
note that 0! is always equal to 1.
the total probability should always be equal to 1.
the numbers from all the probabilities come out as follows:
p(0) = 1 * .75^6 = .177978516
p(1) = 6 * .25 * .75^5 = .355957031
p(2) = 15 * .25^2 * .75^4 = .296630859
p(3) = 20 * .25^3 * .75^3 = .131835938
p(4) = 15 * .25^4 * .75^2 = .032958984
p(5) = 6 * .25^5 * .75 = .004394531
p(6) = 1 * .25^6 = .000244141
probability of getting all possible combinations is equal to 1 as it should be.
probability of getting 0 or 1 or 2 = p(0) + p(1) + p(2) = .830566406
probability of getting 3 or 4 or 5 or 6 = p(3) + p(4) + p(5) + p(6) = .169433594
your answer is .169433594.
this was achieved by taking p(0) + p(1) + p(2) and subtracting the result from 1.
1 minus (p(0) + p(1) + p(2) = 1 - (.17... + .35... + .29...) = 1 - .830566406 = .169433594.
it was also achieved by taking p(3) + p(4) + p(5) + p(6) = .13... + .03... + .004... + .0002... = .169433594.
either way you get the same answer.
as an example of the ways that a particlar probability can happen, consider the following example:
what is the probability of getting exactly 2 admissions from 6 interviews if the probability of getting an admission is .25.
the probability of getting exactly 2 admissions is equal to .25^2 * .75^4 = .019775391.
this can happen in c(6,2) ways = 6! / (4!*2!) = 15 ways.
those possible ways are:
sn = school number (ex: s1 = school number 1)
1 = admitted
0 = not admitted
way s1 s2 s3 s4 s5 s6
1 1 1 0 0 0 0
2 1 0 1 0 0 0
3 1 0 0 1 0 0
4 1 0 0 0 1 0
5 1 0 0 0 0 1
6 0 1 1 0 0 0
7 0 1 0 1 0 0
8 0 1 0 0 1 0
9 0 1 0 0 0 1
10 0 0 1 1 0 0
11 0 0 1 0 1 0
12 0 0 1 0 0 1
13 0 0 0 1 1 0
14 0 0 0 1 0 1
15 0 0 0 0 1 1
the probability for each one of these happening is .25^2 * .75^4.
they are all mutually exclusive (either one way or the other, never both at the same time).
the total probability is therefore equal to the sum of each individual one happening which is p(way 1) + p(way 2) ..... + p(way 25) which is eaual to:
15 * .24^2 * .75^4.
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