Kendra F. answered • 04/03/17

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**a.)**

Probability of drawing a white ball = 5 white balls / 14 total balls

1st draw:

= (5/14)

None are replaced, (put back in the bag) so now there are 4 white balls in a bag of 13 total balls.

2nd draw:

= (4/13)

Again, no balls are replaced so now there are 3 white balls in a bag of 12 total.

3rd draw:

= (3/12)

Multiply the probabilities together to find the chance of drawing white 3 times in a row.

= (5/14)*(4/13)*(3/12)

= 0.027

**b.)**

It's the same method here. One white, two red drawn. No ball replacement.

1st draw: (white)

= (5/14)

2nd draw: (red)

= (9/13)

3rd draw: (red)

= (8/12)

Probability of drawing 1 white, two red

= (5/14)*(9/13)*(8/12)

= 0.165

Mick P.

Kendra's (b) and (c) answers are incorrect.

(a) C(5,3)/C(14,3) = 10/364 = .027

(b) (C(5,1)*C(9,2))/C(14,3) = (5*36)/364 = .495

(c) 1-(C(9,3)/C(14,3)) = 1 - (84/364) = .769

In Kendra's (b) answer, she only calculates for the occurrence of WRR and forgets the occurrences of RWR and RRW.

For (c), remember "at least one W" means we want all scenarios except all reds; so 1 - probability of all reds.

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04/25/18

KKK M.

Mick has right. But read first Kendra's explanation of a. (and maybe of b. too, remembering Mick's comment), as her explanations are great. Mick's (c): There are C(9,3) = 9!/3!(9-3)! = 84 ways of choosing 3 red balls of the nine, i.e., 84 different combinations of 3 red balls, but C(14,3)=364 different combinations of 3 balls. Each of them is equally likely, so the chance of obtaining one of the former ones is 84/364. Therefore, the chance of not obtaining 3 red balls is 1 - 84/364 = .769.
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02/24/21

KKK M.

Note: very often it is easier to compute P(X does not happen) than P(X does happen). Then use 1-P (as in (c)).
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02/24/21

Kendra F.

c.)"at least one white ball" implies the following scenarios:and = multiplyor = addc.)Probability of getting "at least one" white is the probability of each scenario added together.04/03/17