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15 Apr 2014, 16:45
Kudos
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A
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Difficulty:
25%
(medium)
Question Stats:
80% (01:37) correct
20%
(02:15)
wrong
based on 194
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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
A. 1/12
B. 1/10
C. 1/8
D. 1/2
E. 5/9
m3equals333
GMAT 3: 710 Q48 V40
Posts: 141
15 Apr 2014, 18:38
Kudos
Bookmarks
Key is that there is no replacement, so each successive choice will become more skewed towards picking a neg (i.e. the pool of positives decreases, while the pool of negatives stay the same)
P(+ on 1st pick) = 5/10
P(+ on 2nd pick) = 4/9
P(+ on 3rd pick) = 3/8
5/10*4/9*3/8=1/12
P(+ on 1st pick) = 5/10
P(+ on 2nd pick) = 4/9
P(+ on 3rd pick) = 3/8
5/10*4/9*3/8=1/12
15 Apr 2014, 20:47
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tauchmeister
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)
