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In a school, 60% of pupils have access to the internet at home
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24 Feb 2012, 11:17
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In a school, 60% of pupils have access to the internet at home. A group of 8 students is chosen at random. Find the probability that exactly 5 have access to the internet.
If a pupil is selected at random and asked if he/she has an internet connection at home, the answer would be yes or no and therefore it is a binomial experiment. The probability of the student answering yes is 60% = 0.6. Let X be the number of students answering yes when 8 students are selected at random and asked the same question. The probability that X = 5 is given by the binomial probability formula as follows:
P(X = 5) = 8C5 (0.6)5 (1-0.6)3 = 0.278691
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Re: In a school, 60% of pupils have access to the internet at home
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24 Feb 2012, 19:17
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morya003 wrote:
Hi Could someone please help to explain the below questions and answers. Thank you in advance.
In a school, 60% of pupils have access to the internet at home. A group of 8 students is chosen at random. Find the probability that exactly 5 have access to the internet.
If a pupil is selected at random and asked if he/she has an internet connection at home, the answer would be yes or no and therefore it is a binomial experiment. The probability of the student answering yes is 60% = 0.6. Let X be the number of students answering yes when 8 students are selected at random and asked the same question. The probability that X = 5 is given by the binomial probability formula as follows:
P(X = 5) = 8C5 (0.6)5 (1-0.6)3 = 0.278691
That's a great question.
First of all, this topic (calculation of a binomial probability) is more advanced than what you would probably see on the GMAT. This is the type of question they would ask, for example, on an AP Statistics test. Nevertheless, there are some important principles here.
Let's say, for the sake of argument, with each student, the probability of them having internet at home is 0.60. Technically, that's not true, and the answer given is a bit off, because we are selecting without replacement, which means this isn't a perfect binomial situation, but let's ignore that, because that would make the problem 10x harder than it already is.
Idea #1: as a rough-and-ready rule, the word "and" in probability means "multiply"
Idea #2: as a rough-and-ready rule, the word "or" in probability means "add"
So, probability that student #1 has internet is 0.60. That means probability of not having internet is 1 - 0.60 = 0.40.
So, how can it be that exactly five of the six students interviewed have internet. Here is one randomly chosen scenario:
Student #1 has internet Student #2 has internet Student #3 has internet Student #4 does not has internet Student #5 has internet Student #6 does not has internet Student #7 does not has internet Student #8 has internet
All those are connect with the word "and", so we multiply all those probabilities --> 0.60*0.60*0.60*0.40*0.60*0.40*0.40*0.60 = [(0.40)^3]*[(0.60)^5]
Now, notice, I picked only one possible scenarios. That raises the question: how many different scenarios are there where exactly three of the eight members don't have internet and the others do? That's identical to the question: how many different scenarios are there where exactly five of the eight members have internet and the three student don't?
Idea #3: If we have a set of n, and r of them are being chosen or designated in some way, then the number of combinations is given by:
# of combinations of "n choose r" = nCr = (n!)/[(r!)((n-r)!)]
where the "!" symbol is the factorial, and 5! (read "five factorial") is equal to the product of all the positive integers from itself down to one. 5! = 5*4*3*2*1 = 120
Here, we want 8C3, or 8C5 (both of which are equal to one another). 8C5 = (8!)/[(5!)(3!)] = (8*7*6*5*4*3*2*1)/[(5*4*3*2*1)(3*2*1)] = (8*7*6)/(3*2*1) = 8*7 = 56
There are 56 ways to give exactly five people a designation in a set of eight. (Or, equivalently, 56 ways to give exactly three people a designation in a set of eight.)
So, for each combination, the probability is [(0.40)^3]*[(0.60)^5], and there are 8C5 = 56 ways for this to happen, so the total probability is
P = (8C5)*[(0.40)^3]*[(0.60)^5] = (56)*[(0.40)^3]*[(0.60)^5] = 0.29869184
No one would expect you to compute this without a calculator, and of course, you can't have a calculator on the GMAT, so this is not a GMAT appropriate kind of question. Nevertheless, the ideas about probability and the ideas about combinations are highly relevant to GMAT math.
Here's a question much more representative of what you might see on the real GMAT.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
gmatclubot
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