MHB What are the Probabilities of Committee Compositions and Coin Selections?

[rainbow1]

from a group of 6 men and 10 women, a committee of 4 is selected at random. write your answer as a fully reduced fraction.

a. find P(at least 1 woman). b. find P(0 men)
c.Find P(all men or all women).a coin purse contains 10 pennies, 4 dimes, and 4 quarters. four coins are selected at random with replacement. Write your answer as a fully reduced fraction. a. find P ( dime then penny then penny then quarter).
b. find P(quarter then penny then nickel then nickel)c. find P(quarter then dime then dime then dime)

 

[joypav]

I'll give you some help and then see if you can finish b and c..

a. find P(at least 1 woman)
P(at least 1 woman) = 1 - P(no women) = $1 - \frac{\text{(choose 4 of 6 men)(choose 0 of 10 women)}}{\text{(choose 4 of 16 total)}} = 1 - \frac{{6 \choose 4}{10 \choose 0}}{{16 \choose 4}}$

It will be difficult to calculate the probability of choosing at least 1 woman. Because that is the probability of choosing 1, 2, 3, or 4 women for a 4 person committee. So instead, we will do 1 minus the probability of choosing no women at all.

That means we are picking all men! So we want to choose all 4 committee members to be men out of the 6 men available. And we want to choose 0 women committee members out of the 10 available.

Then we divide by the total number of possibilities... choosing 4 from the 16 (10 women + 6 men) people we have.

I'll let you do the calculations.

b. find P(0 men)
P(0 men) = $\frac{\text{(choose 0 of 6 men)(choose 4 of 10 women)}}{\text{(choose 4 of 16 total)}}$

Try completing b and c.

 

[HOI]

For the first one, as joypav said, it is 1 minus the probability of all men (you could do it by calculating the probabilities of "exactly one woman", "exactly two women", "exactly three women", "exactly four women" and then adding but that is much more difficult).

Rather than use a formula I would argue:
There are originally 6 men and 10 women, a total of 16 people. The probability the first person selected is a man is 6/16= 3/8. There are then 5 men and 10 women remaining. The probability the next person selected is a man is 5/15= 1/3. There are then 4 men and 10 women remaining. The probabilty the third person is selected is a man is 4/14= 2/7. Finally there are 3 men and 10 women. The probability the fourth person selected is a man is 3/13.

The probability all four selected are men is (3/8)(1/3)(2/7)(3/13)= 6/(8(7)(3))= 1/(4(7))= 1/28. The probability "at least one woman" is selected is 1- 1/28= 27/28.

The probability there are 0 men selected for the committee is exactly the probability that there a 4 women. Calculating as above, the probabilty the first person selected is a woman is 10/16= 5/8, the probability the second person selected is a woman is 9/15= 3/5, the probabiity the third person is selected is 8/14= 4/7, and the probability the fourth person selected is a woman is 7/13. The proability there is no man on the committee is (5/8)(3/5)(4/7)(7/13)= 3(4)/(8(13))= 3/(2(13))= 3/26.

The proababilty the committee consists of "all men" or "all women" is the sum of the two probabilities separately:
1/28+ 3/26.