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24 May 2023, 01:30
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B
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Difficulty:
25%
(medium)
Question Stats:
81% (01:47) correct
19%
(01:41)
wrong
based on 53
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Not Attempted Yet
Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store, 2 sets are selected randomly. What is the probability that at least 1 set is broken?
A. 11/28
B. 13/28
C. 15/28
D. 1/5
E. 1/8
A. 11/28
B. 13/28
C. 15/28
D. 1/5
E. 1/8
NiranjanaV
GMAT 1: 730 Q48 V42
Posts: 53
24 May 2023, 02:18
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Probability (at least 1 broken TV set of 2) = 1 - P (0 broken TV sets of 2)
P(x) = 1 - (6/8 * 5/7)
P(x) = 1 - (15/28) = 13/28
Answer - [B]
P(x) = 1 - (6/8 * 5/7)
P(x) = 1 - (15/28) = 13/28
Answer - [B]
Updated on: 28 Jun 2023, 06:21
Originally posted by 8.toastcrunch on 24 May 2023, 23:21.
Last edited by 8.toastcrunch on 28 Jun 2023, 06:21, edited 2 times in total.
Last edited by 8.toastcrunch on 28 Jun 2023, 06:21, edited 2 times in total.
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1 is the total probability.
Chose fixed T.V. and probability of 6 fixed T.V. = 6C2 divided by 8C2 = 15/28
Now 1-(15/28) = 13/28
ANSWER - [B]
Chose fixed T.V. and probability of 6 fixed T.V. = 6C2 divided by 8C2 = 15/28
Now 1-(15/28) = 13/28
ANSWER - [B]
