 Good morning friends. I am Purva and today we will work out the following question. Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that first both balls are red, second first ball is black and second is red, third one of them is black and other is red. Let us begin with the solution now. Now we are given that two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. So let S be the sample space of the experiment. Then S is a set consisting of 10 black balls. Here B represents black balls and 8 red balls. Here R represents red balls. Now this is equal to 10 plus 8 which is equal to 18 outcomes. Now let A be the event of drawing a red ball. Then we have probability of A is equal to 8 upon 18 because there are 8 red balls and total number of possible outcomes are 18. So we get probability of A is equal to 8 upon 18 which is equal to 4 upon 9. Let B be the event of drawing a black ball. Then we have probability of B is equal to 10 upon 18 because there are 10 black balls and this is equal to 5 upon 9. Now in the first part we have to find the probability that both balls are red. So probability of drawing both red balls is probability of A into probability of A because both the balls are drawn with replacement. This is equal to 4 upon 9 into 4 upon 9 because we have probability of A is equal to 4 upon 9 and we get this is equal to 16 upon 81. In the second part we have to find the probability that first ball is black and second is red. So the probability of drawing first black ball and then red ball is given by probability of B into probability of A that is probability of drawing a black ball into probability of drawing a red ball and this is equal to probability of B is equal to 5 upon 9 into probability of A is equal to 4 upon 9 and we get this is equal to 20 upon 81. Now finally we have to find the probability that one of them is black and other is red. Now probability of drawing one black and one red ball is given by probability that first ball is black and second is red or probability that first ball is red and second is black. So let the first ball drawn be black and second be red. Then probability of drawing first black ball and then red ball is given by probability of B into probability of A we have just seen in the second part and this is equal to 5 upon 9 into 4 upon 9 which is equal to 20 upon 81. We mark this as 1. Now let the first ball drawn is red and second is black. Then the probability of drawing first red ball and then black ball is given by probability of A into probability of B because probability of drawing a red ball is given by probability of A and probability of drawing a black ball is given by probability of B and this is equal to 4 upon 9 into 5 upon 9 which is equal to 20 upon 81. We mark this as 2. Thus probability that one of them is black and other is red is given by 1 plus 2 that is probability that first ball is black and second is red plus probability that first ball is red and second is black and this is equal to 20 upon 81 plus 20 upon 81 which is equal to 40 upon 81. Thus we have got the answer for the first part as 16 upon 81, answer for the second part as 20 upon 81 and answer for the third part as 40 upon 81. Hope you have understood the solution. Bye and take care.