Help on Math Question!?

Question

An urn contains 6 red and 4 white balls. Three are drawn without replacement. If the third ball is red, determine the probability that the second was white.

And can I plz get an explanation? Thanks!

Answer 1

It doesn't matter what the the third one was, the probability of having a white ball is still 4 sixths (2/5ths)

Answer 2

The probability of the third ball being red has nothing to do with the ball already in your hand.

If you draw a red ball first at P = 6/10, then you draw a white ball at 4/9, the total probability is 6/10*4/9 = 24/90 = 12/45 = 4/15
If you draw a white ball first at P = 4/10, then you draw a white ball at 3/9 the total probability of this outcome is 12/90 = 6/45 = 2/15.

Add the two different possible ways to draw the white ball together = 6/15 = 2/5.

By the way the probabilities of the other two outcomes are:
Red then red = 6/10*5/9. P =1/3 = 5/15
White then red = 4/10*6/9 = 24/90 = 4/15.

Answer 3

If the first was white (4/9), then the odds are 3/8.
If the first was red (5/9), then the odds are 4/8 = 1/2.

Total = 4/9 * 3/8 + 5/9 * 1/2 = 1/6 + 5/18 = 8/18 = 4/9

Answer 4

Since you looked at the third ball, that makes one less ball that could have been the second ball. Of those one less balls, 4 are white and 5 are red for a total of 9. So p(second ball white given third ball red) = 4/9

Answer 5

There's not enough information to solve this since we don't know what the 1st draw was.

Probability of drawing one of a specific color is
[Number of balls of that color]/[Total number of balls]

The chance of two independent events happening is the multiplication of the two probabilities.

The chance to draw a white ball on the first pull is 40% (4/10)
The chance to draw a white ball on the second pull, if the first one drawn was white is 33.3% (3/9)

The chance of drawing white on the first two draws is much less (40%*33.3% = 13.3%!)

Hope that helps you out a bit.