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Suppose I have a bag of 8 balls, half of which are yellow, the other half of which are blue.

If I pick 4 balls at random, what's the probability that all four are yellow?

I have two thinking processes...but I don't know which one is wrong.

1. The probability is (4/8) \* (3/7) \* (2/6) \* (1/5) = 0.0143; or
2. Number of ways to pick 4 balls is 8C4 = 70. And the number of ways to pick 4 YELLOW balls are 4\*3\*2\*1 = 4!. So the probability is 4! / 8C4 = 0.34

The two answers differ by a factor of 4!. So i figure it must be something to do with whether the order of balls matters, but I am having brain fog now and can't figure out which way of thinking is the correct one!!
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The first method is correct.  The second method makes the mistake of assuming that the order of the balls does not matter in the first computation, but does matter in the second.  You could get a valid answer with either assumption if they were consistent.

If order matters, we get that the number of ways to pick 4 balls out of 8 where order matters is 8P4 = 8!/4! = 8 * 7 * 6 * 5 = 1680 (which you'll note is precisely the unsimplified denominator from your first method), and the number of ways to pick 4 yellow balls is 4 * 3 * 2 * 1 = 4!, so the probability is 24/1680 = 0.0143

If order does not matter, then the number of ways to pick 4 balls out of 8 is 8C4 = 70, and the number of ways to pick 4 yellow balls is 1, because you have to have all four of them.  So the probability is 1/70 = 0.0143.

The inconsistency between these two methods is what caused your error.
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The correct answer is 1. Your reasoning is correct by repeated applications of Bayes' theorem that relates *joint* and *conditional probabilities*.

    P(B1=Y, B2=Y) = P(B2=Y| B1=Y) P(B1=Y)

As you correctly point out in point out in 1, P(B1 = Y) = 4/8, and P(B2 = Y | B1 = Y) = 3/7, i.e. that the probability of ball 2 being yellow *conditioned upon* or *knowing that* the first ball is yellow is 3/7.

We can chain this rule together for multiple events:

    P(B1=Y, B2=Y, B3=Y) = P(B3=Y| B1=Y, B2=Y) P(B1=Y, B2=Y)
                        = P(B3=Y| B1=Y, B2=Y) P(B2=Y |B1 = Y) P(B1 = Y)

And so on, yielding your answer (4/8)(3/7)(2/6)(1/5).

Your second answer is wrong because you count incorrectly. There are 4!4! ways to draw 4 yellow balls then 4 remaining balls. There are 8! ways to draws 8 balls. 4!4!/8! = 4\*3\*2\*~~4~~\*~~3~~\*~~2~~/(8\*7\*6\*5\*~~4~~\*~~3~~\*~~2~~) = (4/8)(3/7)(2/6)(1/5).
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There is only 1 way to pick 4 yellow balls in situation 2.
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