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From the question bank again: WHat is the probability of chosing one man, two women, and one child from 10 men, 8 women, and 10 children. Here's my stab at it. Hope I'm correct (10c1 8c2 10c1)/28c4

From the question bank again: WHat is the probability of chosing one man, two women, and one child from 10 men, 8 women, and 10 children. Here's my stab at it. Hope I'm correct (10c1 8c2 10c1)/28c4

Yes, without restrictions, total # of outcomes = 28C4.

With restrictions, 1 man to be chosen from 10 men = 10C1

2 women from 8 = 8C2

1 child from 10 = 10C1

The probability = (10C1 * 8C2 * 10C1)/28C4

The problem assumes that you are selecting four and only four from 28.

The problem could have stated that "When selecting 4 people from a group of 10 men, 8 women and 10 children, what is the probability that ..."
this is probably implied from the statement.

Yes, without restrictions, total # of outcomes = 28C4.

With restrictions, 1 man to be chosen from 10 men = 10C1

2 women from 8 = 8C2

1 child from 10 = 10C1

The probability = (10C1 * 8C2 * 10C1)/28C4

The problem assumes that you are selecting four and only four from 28.

The problem could have stated that "When selecting 4 people from a group of 10 men, 8 women and 10 children, what is the probability that ..." this is probably implied from the statement.

Yes, without restrictions, total # of outcomes = 28C4.

With restrictions, 1 man to be chosen from 10 men = 10C1

2 women from 8 = 8C2

1 child from 10 = 10C1

The probability = (10C1 * 8C2 * 10C1)/28C4

The problem assumes that you are selecting four and only four from 28.

The problem could have stated that "When selecting 4 people from a group of 10 men, 8 women and 10 children, what is the probability that ..." this is probably implied from the statement.

This answer assumes that you are selecting a Man, two Women and a Child in that order. i.e. MWWC. In other words, you are introducing a restriction that is not in the original problem.

Your answer does not account for additional selections which are possible arrangements of "MWWC" itself. Note that "MWWC" can be arranged in 4!/2! ways = 4*3 = 12 ways.

Note that the difference between the original answer and your answer is just this - 4*3 = 12.

Just by rearranging the terms, your answer can also be expressed as (10C1 * 8C2 * 10C1 / 28C4) * (4*3).

When the order is not important, you just select 1 man from 10, 2 women from 8 and 1 children from 10 and multiply.

Even though it really depends on the problem, a general rule of thumb, the approach you have taken is more appropriate for problems involving a deck of cards or problems involving picking blue/red/black marbles from a bag/box or problems involving picking a blue/red/black sox from a bag/box etc.

Even though it really depends on the problem, a general rule of thumb, the approach you have taken is more appropriate for problems involving a deck of cards or problems involving picking blue/red/black marbles from a bag/box or problems involving picking a blue/red/black sox from a bag/box etc.

Yes, without restrictions, total # of outcomes = 28C4.

With restrictions, 1 man to be chosen from 10 men = 10C1

2 women from 8 = 8C2

1 child from 10 = 10C1

The probability = (10C1 * 8C2 * 10C1)/28C4

The problem assumes that you are selecting four and only four from 28.

The problem could have stated that "When selecting 4 people from a group of 10 men, 8 women and 10 children, what is the probability that ..." this is probably implied from the statement.

-mathguru

Can you please tell me why my method is wrong....

(10/28) * (8/27 * 7/26) * (10/25)

I think you are right.....because the question is NOT for the number of combinations that we can get 1 man, 2 women and 1 children from 28 people.....THe question is only the probability of choosing 1 man, 2 women y 1 child from 10 men, 8 women and 10 children....

On the other hand Mathguru is really good....but I think the question is not more than that it is....