Patrick C.

asked • 04/01/17

A bag contains 5 white balls and 9 red balls. Three balls are drawn, without replacement, from the bag.

(a) What is the probability that all three balls are white?


(b) What is the probability that exactly one ball is white?



(c) What is the probability that at least one ball is white?

1 Expert Answer

By:

Kendra F. answered • 04/03/17

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Kendra F.

c.) "at least one white ball" implies the following scenarios:
 
1 white, 2 red = 0.165
or..
2 white, 1 red = ?
or..
3 white, 0 red = 0.027
 
and = multiply
(as in (b.) drawing a white and red and red)
or = add
Solve for 2 white, 1 red
1st draw: (white)
= (5/14)
2nd draw: (white)
= (4/13)
3rd draw: (red)
= (9/12)
 
probability of drawing 2 white, one red:
= (5/14)*(4/13)*(9/12)
= 0.082
 
c.) Probability of getting "at least one" white is the probability of each scenario added together.
 
= 0.165 + 0.082 + 0.027
=  0.274
 
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04/03/17

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

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