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Probability

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Probability

by thx5828 » Sat Jun 26, 2010 4:26 pm
John picks 3 different numbers between 10 and 20 inclusive. Assuming John picks randomly, what is the probability that John will pick the numbers 12 and 15 as two of his three selections?

Approach:

Probability = Total Desired Outcomes / Total Number of Outcomes

Total Number of Outcomes = 11c3 = 165

Total Desired =

12 and 15 must be in each set. So possibilities are:
12, 15, (any of the other 9) = 9 possibilities
12, (other 9), 15 = 9 possibilities
15, 12, (other 9) = 9 more
15, (other 9), 12 = 9 more
(other 9), 12, 15 = 9 more
(other 9), 15, 12 = 9 more
For a total of 9 x 6 = 54 desired; 54 / 165 = 18 / 55, not an option. Help!

A) 2/42
B) 3/55
C) 1/15
D) 4/36
E) 2/11

OA: B
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by ljr » Sat Jun 26, 2010 5:39 pm
You are right about the 165 possible outcomes.

Treat it as a combination because order doesn't matter here. If you treat it as a combination you should get 9 possible desired outcomes. (= 9!/8!)

9/165 = 3/55 (answer B).
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