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

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

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

I am not quite sure why my number 8 have turned into smiley faces in my post above!

Please replace all smiley faces with the number 8.

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

Thank you all for your responses. 1/3 is the correct answer.

Thank you all for your responses. 1/3 is the correct answer.

agree 1/3 is correct ans my method is flaw

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

Using Combination.

Result is ( 4c3+4c2 . 6c1 ) / 10c3 => 1/3