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A certain family consists of 8 adult members, including Susi

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Zarrolou wrote:
$$P(Selected)=1-P(NoSelected)=1-\frac{7}{8}*\frac{6}{7}*\frac{5}{6}=\frac{3}{8}.$$

B

Zarrolou,

Not sure whether 1 - P(Noselected) is applicable here since Susie can't be running all the three stores.. I started this path and shortly realized that Susie selected to run all the three stores should be excluded.. And surprisingly this also gives the right answer..

Not sure whether this problem is GMAT level.

P.S Sorry, Just realized that the probability of Susie selected to manage stores A, B and C is zero.
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Alternative Method

Total No of Ways in which 3 ppl can be selected from 8 = 3C8 = 56

Assume Susie is not selected at all for running any of the stores... Which means the total number of options = Total no of ways of choosing 3 people from 7 = 3C7 = 35

So probability susie is not selected = 35/56 = 5/8

Prob susie is selected = 1-5/8 = 3/8
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selection for stores : for A or for B or for C hence the sum is needed for all individual probablities

sushie selected for A : 1/8
selected for B : not for A * selected for store B = 7/8*1/7 := 1/8
selected for C : not for A * not for B * selected for C = 7/8*6/7*5/6 := 1/8

hence 1/8 + 1/8+ 1/8 = 3/8 (ans)
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Transcendentalist wrote:
Alternative Method

Total No of Ways in which 3 ppl can be selected from 8 = 3C8 = 56

Assume Susie is not selected at all for running any of the stores... Which means the total number of options = Total no of ways of choosing 3 people from 7 = 3C7 = 35

So probability susie is not selected = 35/56 = 5/8

Prob susie is selected = 1-5/8 = 3/8

In this case, there are 2 alternatives as discussed above. It is always good to use the second approach as first case becomes too tedious when the number of cases increase. For example, if we have 5 shops instead of 3, we will have 5 cases. In case we have 100 shops, it becomes impossible to employ first method. So it is always prudent to find the case when Susie is not included and subtract it from 1.

I hope it helps!!!
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Re: A certain family consists of 8 adult members, including Susi [#permalink]
Gian wrote:
A certain family consists of 8 adult members, including Susie.Each day, three stores A,B and C are operated by the family as follows:one of the 8 family members is selected randomly to operate store A,one of the remaining 7 family members is selected to operate store B, and finally one of the remaining 6 members is selected to operate store C. What is the probability that Susie will be selected to operate one of the three stores?

a) 1/3
b) 3/8
c) 1/24
d) 1/336
e) 1/512

We can look at this problem in terms of only 2 possible events: Either Susie will be selected to operate one of the stores, or she will not be selected to operate any stores. This means that:

P(Susie selected to operate one store) + P(Susie not selected to operate any stores) = 1

P(Susie selected to operate one store) = 1 - P(Susie not selected to operate any stores)

Thus, if we can determine the probability that Susie is not selected to operate any stores, then we’ll quickly be able to calculate the probability that she’ll be selected to operate one store.

We see that there are 8 family members, including Susie. Thus:

P(Susie does not get selected to operate store A) = 7/8

P(Susie does not get selected to operate store B) = 6/7

P(Susie does not get selected to operate store C) = 5/6

Total probability that Susie does not get selected to operate any store:

(7/8) x (6/7) x (5/6) = 6/8 x 5/6 = 5/8

Thus, the probability that Susie is selected to operate one of the stores is 1 - 5/8 = 3/8.

Alternative solution:

The probability that Susie will be selected to operate one of the three stores is the probability that she will operate store A or store B or store C. That is:

P(Susie will operate any one of the three stores) = P(she will operate store A) + P(she will operate store B) + P(she will operate store C)

Let’s find the probability that she will operate any one of these 3 stores (keep in mind that if she works at one store, she can’t work at the other two):

P(she will operate store A) = P(A, not B, not C) = 1/8 x 7/7 x 6/6 = 1/8

P(she will operate store B) = P(not A, B, not C) = 7/8 x 1/7 x 6/6 = 1/8

P(she will operate store C) = P(not A, not B, C) = 7/8 x 6/7 x 1/6 = 1/8

Therefore:

P(Susie will operate any one of the three stores) = 1/8 +1/8 + 1/8 = 3/8

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Re: A certain family consists of 8 adult members, including Susi [#permalink]
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Re: A certain family consists of 8 adult members, including Susi [#permalink]
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