A box contains 6 blue balls, 4 black balls and 5 red balls at the same size . A ball is selected at random from the box and then replaced . A second ball is then selected. Find the probability of obtaining ; (i) two red balls ; (ii) two blue balls or two black balls ; (iii) one black ball and one red balls.

(i)   1/9.

(ii)  52/225

(iii)  41/225.

Step-by-step explanation:

There are 15 balls in the bag.

So:

(i) Prob(2 red balls) = 5/15 * 5/15

= 1/3 * 1/3

= 1/9.

(ii)  (i) Prob(2 blues) = 6/15 * 6/15

= 36/225.

Prob( 2 blacks) = 4/15 * 4/15

=  16/225

So the probability of 2 blues or 2 blacks

4/25 + 16/225

= 52/225.

(iii) Prob ( red ball then black) =  5/15 * 5/15 = 25/225.

Prob(black ball then red ball) = 4/15 *  4/15 = 16/225

So the probability of 1 black and 1 red = 41/225.

i think it might be a but you might want a second opinion

sinn is the total amount of students now and m were the students before, you subtract them both (n-m) and divide that by m. then, since to get a percentage from a fraction you have to multiply by 100, your answer should look like this: n-m/m*100

step-by-step explanation: