tauchmeister wrote:
Having trouble on this question to,
Each of the integers from 0 to 9, inclusive, is
written on a separate slip of blank paper and the
ten slips are dropped into a hat. If 3 of the slips
are the drawn, without replacement, what is the
probability that all 3 have a even number written
on it?
a) 1/12
b) 1/10
c) 1/8
d) 1/2
e) 5/9
Initially there are 10 numbers from 0 to 9.
5 of them are even and 5 are odd. Note that 0 is an even number.
Probability of picking an even number on first pick = 5/10
Now you have 9 numbers left since you do not replace the number you picked in your first pick. Out of the 9, you have 4 even numbers (since you already picked an even number) and 5 odd numbers.
Probability of picking an even number on second pick = 4/9
Now you have 8 numbers left and 3 of them are even since 2 even are already picked.
Probability of picking an even number on third pick = 3/8
Probability of picking even numbers of all three picks = (5/10)*(4/9)*(3/8) = 1/12
Answer (A)
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Karishma
Veritas Prep GMAT Instructor
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