bag of baloon: probability!!!

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bag of baloon: probability!!!

by acorra » Sat Apr 25, 2009 5:52 am
In a bag of baloons, the ratio of red baloons to other colours is 1:5, if 3 baloons are selected from the bag at random, what is the probability that:
a) at least one baloon is red
b) exactly one baloon is red

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Re: bag of baloon: probability!!!

by Ian Stewart » Sat Apr 25, 2009 6:59 am
acorra wrote:In a bag of baloons, the ratio of red baloons to other colours is 1:5, if 3 baloons are selected from the bag at random, what is the probability that:
a) at least one baloon is red
b) exactly one baloon is red

OA will follow soon
You cannot answer this kind of question if you only know the ratio of red balloons to other balloons. Take question a), for example. Notice that we either get zero red balloons, or we get at least one red balloon, so

Prob(0 red balloons) + Prob(1 or more red balloons) = 1

Prob(1 or more red balloons) = 1 - Prob(0 red balloons)

So, if we can find the probability that all three balloons are not red, we can find the probability that at least one balloon is red by subtracting from one.

Now, say we have exactly 1 red balloon, and exactly 5 non-red balloons. The probability that the first is not red is 5/6, that the second is again not red is 4/5, and that the third is again not red is 3/4. So the probability that all three are non-red is

(5/6)(4/5)(3/4) = 1/2

So the probability of 'at least one red' is 1 - 1/2 = 1/2

On the other hand, if we have 2 red balloons, and 10 non-red balloons, the probability that all are non-red is

(10/12)(9/11)(8/10) = 6/11

and the probability that at least one is red is 5/11.

So the ratio alone is not sufficient to get a numerical answer here - the answer is different depending on the number of balloons in the bag. Of course we could solve the problem using an unknown. If we have x red balloons, we have 5x non-red balloons. The probability of selecting three non-red balloons is therefore:

(5x/6x)[(5x - 1)/(6x - 1)][(5x - 2)/(6x - 2)]

and subtracting that from 1 (and canceling the common factor of x in the numerator and denominator), you could get an algebraic expression for the probability in question, in terms of x, the number of red balloons. But that's about the best you can do here.

Throughout, I've assumed that the selections are made 'without replacement', something which is strongly suggested by the wording of the question. If the selections are made 'with replacement' (i.e. after selecting each balloon, we put it back in the bag before making our next selection), then the answer would be 1 - (5^3/6^3), since the probability would be 5/6 on each selection that a balloon is non-red. As the question is worded, however, that interpretation (that the selections are made with replacement) doesn't make much sense.
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