An urn contains n white and m black balls, where n and m are positive numbers.
(a) if two balls are randomly withdrawn, what is the probability that they are the same color?
(b) if a ball is randomly withdrawn and then replaced before the second one is drawn, what is the probability that the withdrawn balls are the same color?
(c) show that the probability in part (b) is always larger than the one in part (a).

a) (n²-n+m²-m)÷((n+m)×(n+m-1))

b)(n²+m²)÷(n+m)

c) part b is larger ,check below.

Step-by-step explanation:

a)If there are n white and m black balls ,total will be n+m balls. Let's find the probability of choosing two balls white.

First ball as white n÷(n+m)  second ball to be white again is n-1÷(n+m-1) the reason why we take away 1 is because we already chose one white ball in the beginning .So the probability will be product of these to get the both balls white.

Let's find the the probability of choosing both black.Same strategy ,first ball black  is m÷(n+m),second ball black is m-1÷(n+m-1).So the probability will be product of these to get the both balls black. We should add the final products and simplify

b)Because this time they are replacing the ball ,we will not take away 1.So to get both white probability is n÷(n+m) times n÷(n+m) .The probability of both black is m÷(n+m) times m÷(n+m).Add the products .

c) b will be always larger,We should compare the final products.

In b  (n²+m²) is divided by (n+m)

In a (n²-n+m²-m)÷((n+m)×(n+m-1))  less amount divided by a bigger value ,so it will result always with a smaller quotient

a

step-by-step explanation:

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step-by-step explanation: