Out of a group of four men and six women, three people will be randomly chosen to participate in a marketing survey. What is the probability that at least two of the people chosen is a man?
thanks
Probability Problem
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ans is 5/12
probability to choose 0 man ot 1 man
which is 3/6 (all woman) + 1/4*2/6 (one man and 2 woman) = 7/12
1 - 7/12 = 5/12 prob of choosing atleast 2 man
probability to choose 0 man ot 1 man
which is 3/6 (all woman) + 1/4*2/6 (one man and 2 woman) = 7/12
1 - 7/12 = 5/12 prob of choosing atleast 2 man
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ans is 5/12
probability to choose 0 man ot 1 man
which is 3/6 (all woman) + 1/4*2/6 (one man and 2 woman) = 7/12
1 - 7/12 = 5/12 prob of choosing atleast 2 man
probability to choose 0 man ot 1 man
which is 3/6 (all woman) + 1/4*2/6 (one man and 2 woman) = 7/12
1 - 7/12 = 5/12 prob of choosing atleast 2 man
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Kidboc1 - What is the answer? Please provide answer options.
Using a similar method to Varun, I got ans=1/3.
The probabily of it being all woman is: [(6/10)(5/9)(4/8)] = 1/6
The probabilty of only having 1 man out of 3 (this is probably the trickiest part) is:
If man is chosen first:[(4/10)+(6/9)+(5 / 8)] = 1/6
If man is chosen second: [(6/10)+(4/9)+(5 / 8)] = 1/6
If man is chosen third: [(6/10)+(5/9)+(4 / 8)] = 1/6
Probability of NOT having at least 2 man = 1/6+[(1/6)+(1/6)+(1/6)] = (1/6)+(1/2) = 4/6 = 2/3
Probability of having at least 2 men = 1-(2/3) = 1/3
Using a similar method to Varun, I got ans=1/3.
The probabily of it being all woman is: [(6/10)(5/9)(4/8)] = 1/6
The probabilty of only having 1 man out of 3 (this is probably the trickiest part) is:
If man is chosen first:[(4/10)+(6/9)+(5 / 8)] = 1/6
If man is chosen second: [(6/10)+(4/9)+(5 / 8)] = 1/6
If man is chosen third: [(6/10)+(5/9)+(4 / 8)] = 1/6
Probability of NOT having at least 2 man = 1/6+[(1/6)+(1/6)+(1/6)] = (1/6)+(1/2) = 4/6 = 2/3
Probability of having at least 2 men = 1-(2/3) = 1/3
Last edited by ssy on Mon Sep 22, 2008 1:24 am, edited 1 time in total.
I think answer is 1/3
Part 1:
Total way to select 3 people from 10 = 10c3 = 120
Part 2:
We have 2 men and 1 woman
4c2.6c1 = 36 possibilities.
We can have 3 men team only
4c3.6c0 = 4 possibilities.
So total way = 36 +4 = 40
Prob . 40 / 120 = 1/3
Hope i am clear and its very easy method .
Thanks
Vishal Shah
Part 1:
Total way to select 3 people from 10 = 10c3 = 120
Part 2:
We have 2 men and 1 woman
4c2.6c1 = 36 possibilities.
We can have 3 men team only
4c3.6c0 = 4 possibilities.
So total way = 36 +4 = 40
Prob . 40 / 120 = 1/3
Hope i am clear and its very easy method .
Thanks
Vishal Shah
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Hi guys, thanks for all your input, I did this a different way. I calculated all the possibilities for men being chosen.
MMM
4/10 * 3/9 * 2/8
MWM
4/10 * 6/9 * 3/8
MMW
4/10 * 3/9 * 6/8
WMM
6/10 * 4/9 * 3/8
I then added the answers and got 1/3
MMM
4/10 * 3/9 * 2/8
MWM
4/10 * 6/9 * 3/8
MMW
4/10 * 3/9 * 6/8
WMM
6/10 * 4/9 * 3/8
I then added the answers and got 1/3