There are 12 people in a room: 4 couples and 4 singles. picking 4 people at random, which is the probability of picking only one couple?

Probability =

\frac{Desired outcomes}{Total Outcomes}Total Outcomes = selection of 4 persons among 12 persons = 12C4 = 495

Desired Outcomes = Selection of 1 Couple AND selection of 2 persons from remaining 10 persons (Condition - The two selected should not be a couple)

1 couple can be selected from 4 couples in 4C1 ways.

Now we are left with 3 couples and 4 singles (Total 10 Persons) from which we have to select 2 persons. (Condition - The two selected should not be a couple)

This can be calculated as = Total number of ways selecting 2 persons from 10 persons - Number of ways selecting a couple from available couples

= 10C2 - 3C2 = 42

Lets see how this arrived

Let these 10 persons be m1, f1, m2, f2, m3, f3, s1, s2, s3, s4

If you chose m1, you can choose any one of the remaining 9 persons except for f1--------> So the available options are 8

If you chose f1, you can choose any one of the remaining 9 persons except for m1--------> So the available options are 8

If you chose m2, you can choose any one of the remaining 9 persons except for f2, m1, f1 --------> So the available options are 6

If you chose f2, you can choose any one of the remaining 9 persons except for m2, m1, f1 --------> So the available options are 6

If you chose m3, you can choose any one of the remaining 9 persons except for f3, m1, m2, f1, f2 --------> So the available options are 4

If you chose f3, you can choose any one of the remaining 9 persons except for m3, m1, m2, f1, f2 --------> So the available options are 4

If you chose s1, you can choose any one of the remaining 9 persons except for m1, m2, m3, f1, f2, f3, f4 --------> So the available options are 3

If you chose s2, you can choose any one of the remaining 9 persons except for m1, m2, m3, f1, f2, f3, f4, s1--------> So the available option is 2

If you chose s3, you can choose any one of the remaining 9 persons except for m1, m2, m3, f1, f2, f3, f4, s1, s2--------> So the available options are 1

If you chose s3, you can choose any one of the remaining 9 persons except for m1, m2, m3, f1, f2, f3, f4, s1, s2, s3--------> So the available options are 0

Total 8+8+6+6+4+4+3+3+3 = 42 options

Total Desired outcomes [Fundamental Principle of Multiplication] = 4C1 X 42 = 168

Probablity =

\frac{168}{495}What is the OA ?

Narenn

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