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There is a bag with 70 balls. 8 balls are blue. If you pick up 5 balls (no replacement), what is the probability that at least one is blue?

My first approach was to obtain the probability (A) that no blue ball is picked up: - P(A)= 62/70 · 61/69 · 60/68 · 59/67 · 58/66

And then, the probability that at least one blue ball is picked up is P(B) = 1 - P(A). I just wanted to be sure if I'm correct. Thanks.

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3 Answers 3

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Let $A$ be the event that none of the balls are blue.

Then $$P(A) = \frac{62}{70} \cdot \frac{61}{69} \cdot \frac{60}{68} \cdot \frac{59}{67} \cdot \frac{58}{66}$$

The probability you desire is $ 1 - P(A)$

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Suppose you only get 1, you'll have a 100 * 8/70 chance of it being blue, right? In the next round there are no 70, but 69, therefore you will have 800/69 odds that it will be blue.

If you continue with this logic, you would have an 800/65 odds in the fifth round.

There is no reason to think that by drawing the 5 balls at the same time you are going to have a higher probability.

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Calculate the probability that none of the balls is blue. Then subtract that probability from $1$ to get the probability that at least one ball is blue.

As you are not replacing the balls the total number of balls keep decearsing.

Desired probablity is $$P=1- \frac{62}{70} \cdot \frac{61}{69} \cdot \frac{60}{68} \cdot \frac{59}{67} \cdot \frac{58}{66}$$

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