If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women is p > 1/2?
1. More than 1/2 of the 10 employees are women
2. The probability that both representatives selected will be men is less than 1/10
OA = E
Probability DS
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Given: Possible number of way to select 2 out of 10 = 10C2 = 45tonebeeze wrote:If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women is p > 1/2?
1. More than 1/2 of the 10 employees are women
2. The probability that both representatives selected will be men is less than 1/10
OA = E
Statement 1: More than 1/2 of the 10 employees are women.
Number of can be 6, 7, 8, 9 or 10.
- For number of women = 6, Possible number of selecting two women out of 6 = 6C2 = 15. Thus probability = 15/45 = 1/3 < 1/2
For number of women = 9, Possible number of selecting two women out of 9 = 9C2 = 36. Thus probability = 36/45 = 4/5 > 1/2
Statement 2: The probability that both representatives selected will be men is less than 0.1. This means (Probability that both representative will be women + Probability that one of the representative will be man and other woman) > (1 - 0.1) = 0.9
But we don't have any idea about "Probability that one of the representative will be man and other woman".
So, (2) is NOT SUFFICIENT.
Combining 1 & 2 Together: No relevant new information.
Combining (1) and (2) is also NOT SUFFICIENT.
The correct answer is E.
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If you have 7 women, the probability of picking two women will be (7/10)(6/90) = 42/90, which is less than 0.5, and if you have 8 women, the probability of picking two women will be (8/10)(7/9) = 56/90, which is more than 0.5. So the question is really just asking 'is the number of women greater than 7?', or equivalently, 'is the number of men less than 3?'.
So Statement 1 is not sufficient.
2. The probability that both representatives selected will be men is less than 1/20
then Statement 2 would actually be sufficient and the answer would be B. The number really matters here.
Using Statement 2 as given, if we have 3 men, the probability of picking two men will be (3/10)(2/9) = 6/90. This is less than 1/10, so it's possible we have 3 men and 7 women, and the answer to the question is 'no', and it's also possible we have fewer than 3 men in which case the answer is 'yes', so Statement 2 is not sufficient either, and the answer is E.
So Statement 1 is not sufficient.
You cannot dismiss Statement 2 that quickly: Statement 2 tells you, roughly, 'there aren't very many men'. That is exactly the kind of information we're looking for; if the number of men is less than 3, the information is sufficient. If you change Statement 2 here to read:Anurag@Gurome wrote: Statement 2: The probability that both representatives selected will be men is less than 0.1. This means (Probability that both representative will be women + Probability that one of the representative will be man and other woman) > (1 - 0.1) = 0.9
But we don't have any idea about "Probability that one of the representative will be man and other woman".
2. The probability that both representatives selected will be men is less than 1/20
then Statement 2 would actually be sufficient and the answer would be B. The number really matters here.
Using Statement 2 as given, if we have 3 men, the probability of picking two men will be (3/10)(2/9) = 6/90. This is less than 1/10, so it's possible we have 3 men and 7 women, and the answer to the question is 'no', and it's also possible we have fewer than 3 men in which case the answer is 'yes', so Statement 2 is not sufficient either, and the answer is E.
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