Question 326205: Can anyone tell me if I worked this problem right? A committee of two is selected at random from a set consisting of three Democrats, four Republicans and one Independent.
(a)What is the probability that the committee consists of no Republicans?
I count the combination's of not getting an R; so I have d1,d2,d3,I to select from. I can have d1,d2 - d1,d3 - d1,I - d2,d3 – d2,I – d3,I 6 combo’s
Six chances to pull 2 (or one of those combinations) I take the median:
8*7*6/ 1*2*3 = 56 so I’ve got 6/56 or 3/28
(b) What is the probability that the committee consists of all Republicans?
Combo’s r1, r2, r3, r4: r1,r2 – r1,r3 – r1,r4 – r2,r3 – r2,r4 – r3,r4 6 combo’s again so the chances are the same.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
(a) There is a pool of 8 people to choose from, and 4 of them are not Republicans. So the probability that the first person chosen will not be a Republican is  .
Once the first person is chosen and it is, in fact, not a Republican, then you now have a pool of 7 people from which to choose, 3 of which are not Republicans. So the probability that the second person chosen will not be a Republican given that the first was not a Republican is  .
The total probability is the product: \left(\frac{3}{7}\right)\ =\ \frac{12}{56}\ =\ \frac{3}{14})
(b) There is a pool of 8 people to choose from, and 4 of them are Republicans. So the probability that the first person chosen will be a Republican is .
Once the first person is chosen and it is, in fact, a Republican, then you now have a pool of 7 people from which to choose, 3 of which are Republicans. So the probability that the second person chosen will be a Republican given that the first was a Republican is .
The total probability is the product:
John
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