3. A bag contains 8red balls and 6 white balls. Two balls
are randomly drawn one after another without replace
mondo. Find the probability, of getting balls of diffecent colors

Work Shown:

We have 8 red and 6 white giving 8+6 = 14 total

The probability of selecting red is 8/14.

After that ball is not replaced, we have 14-1 = 13 balls left. The number of white balls will not go down since only the red ball count goes down.

The probability of selecting white on the second selection is 6/13

So we can say

P(red first, then white) = P(red first)*(white second, given red first)

P(red first, then white) = (8/14)*(6/13)

P(red first, then white) = (8*6)/(14*13)

P(red first, then white) = 48/182

P(red first, then white) = 24/91

Let's label this as A.

A = 24/91

That way we can come back to it later.

Now let's say we pick white first, then red

So we'd have the following.

P(white first, then red) = P(white first)*(red second, given white first)

P(white first, then red) = (6/14)*(8/13)

P(white first, then red) = (6*8)/(14*13)

P(white first, then red) = 48/182

P(white first, then red) = 24/91

Let B = 24/91

Now we add up the values of A and B

A+B = 24/91 + 24/91

A+B = (24+24)/91

A+B = 48/91

make sure you have version 2.50

try looking in the very back of the book : )

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step-by-step explanation:

i used the pitagoras theoreme