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Shadwal
Joined: 22 Aug 2019
Last visit: 15 Jan 2022
Posts: 7
Given Kudos: 16
Location: India
Schools: ISB '23 (S)
GMAT 1: 770 Q51 V42
GPA: 3.25
14 Aug 2021, 08:07
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
88% (00:44) correct
13%
(02:09)
wrong
based on 8
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What is the probability that a student randomly selected from a class of 60 students will be a male who has brown hair?
(1) One-half of the students have brown hair.
(2) One-third of the students are males."
Why is the answer by GMAC 'E' and not 'C'? The two events (gender and hair colour) are completely independent. So why can't we simply multiply 1/2 x 1/3 and get the probability of 1/6?
Let's say the question asked,
"What is the probability of selecting a capital boy from a set of 4 students?
1. 50% of students are boys
2. 50% of students are capital
Now, if we combine the two statements, we know that 50% of students are boys and 50% of students are capital.
Possible cases:
Case A: bbGG - 1 possibility
Case B: BBgg - 1 possibility
Case C: BbGg - 4 possibilities (since the students are different entities)
Overall 1+1+4 = 6 possible ways in which out of 4 students, 50% are boys and 50% are capital.
Probability of selecting a capital boy
= P(Case A)xP(capital B given case A) + P(Case B)xP(capital B given case B) + P(Case C)x(capital B given case C)
=1/6x0 + 1/6x1/2 + 4/6x1/4
=0 + 1/12 + 1/6
=1/4
i.e. P(boys)xP(capital)
The assumption is that each combination (BBgg, BbGg, bBGg, etc.) is equally likely to happen - the same assumption we use when we combine independent events (in the original case, the gender and the hair colour of a student).
If the question asked how many boys have brown hair, we cannot say because there are multiple possibilities. But since the question asks what is the probability that a randomly selected student will be a boy with brown hair, we can surely tell.
Please tell me where I went wrong.
(1) One-half of the students have brown hair.
(2) One-third of the students are males."
Why is the answer by GMAC 'E' and not 'C'? The two events (gender and hair colour) are completely independent. So why can't we simply multiply 1/2 x 1/3 and get the probability of 1/6?
Let's say the question asked,
"What is the probability of selecting a capital boy from a set of 4 students?
1. 50% of students are boys
2. 50% of students are capital
Now, if we combine the two statements, we know that 50% of students are boys and 50% of students are capital.
Possible cases:
Case A: bbGG - 1 possibility
Case B: BBgg - 1 possibility
Case C: BbGg - 4 possibilities (since the students are different entities)
Overall 1+1+4 = 6 possible ways in which out of 4 students, 50% are boys and 50% are capital.
Probability of selecting a capital boy
= P(Case A)xP(capital B given case A) + P(Case B)xP(capital B given case B) + P(Case C)x(capital B given case C)
=1/6x0 + 1/6x1/2 + 4/6x1/4
=0 + 1/12 + 1/6
=1/4
i.e. P(boys)xP(capital)
The assumption is that each combination (BBgg, BbGg, bBGg, etc.) is equally likely to happen - the same assumption we use when we combine independent events (in the original case, the gender and the hair colour of a student).
If the question asked how many boys have brown hair, we cannot say because there are multiple possibilities. But since the question asks what is the probability that a randomly selected student will be a boy with brown hair, we can surely tell.
Please tell me where I went wrong.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
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14 Aug 2021, 10:27
Kudos
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Shadwal
Discussed here: what-is-the-probability-that-a-student-randomly-selected-from-a-class-13843.html?fl=similar Hope it helps.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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