 Hello! I am welcome to the session. I am Ripika here. Let's discuss the question which says An arm contains 25 bells at which 10 bells bear a mark X and the remaining 15 bear a mark Y. A ball is drawn at random from the arm. Its mark is noted down and it is replaced. If six bells are drawn in this way find the probability that all will bear X mark rock more than two will bear Y mark at least one ball will bear Y mark and the number of bells with X mark and Y mark will be equal. So let's start the solution. Now according to the given question an arm contains 25 balls of which 10 balls bear a mark X and the remaining 15 bear a mark Y. Again a ball is drawn at random from the arm. Its mark is noted down and it is replaced. So let A be the event of getting a ball marked X be the event of getting a ball marked Y. So we have probability of A is equal to 10 over 25 and this is equal to 2 over 5 that is P is equal to 2 over 5. Now probability of B is given by 1 minus 2 over 5 and this is equal to 3 over 5. So we have Q is equal to 3 over 5. Now we have probability of X successes is equal to NCX into Q raise to power N minus X into P raise to power X where X is equal to 0 to N and Q is equal to 1 minus P. Now here we have N is equal to 6 P is equal to 2 over 5 Q is equal to 3 over 5 therefore probability of X successes is equal to 6CX into 3 over 5 raise to power 6 minus X into 2 over 5 raise to power X Now in part 1 we have to find the probability that all the 6 balls drawn will be X mark. So in part 1 probability of 6 successes is equal to 6C6 into 3 over 5 raise to power 6 minus 6 into 2 over 5 raise to power 6 and this is equal to 1 into 1 into 2 over 5 raise to power 6 which is equal to 2 over 5 raise to power 6 hence the probability that all the balls drawn will be X is 2 over 5 raise to power 6 So this is the answer for part 1 Now in part 2 we have to find the probability that not more than 2 will be a Y mark Now 6 balls are drawn So if no ball is drawn that is 0 ball is drawn with Y mark then 6 balls are drawn with X mark Similarly if 1 ball is drawn with Y mark then 5 balls are drawn with X mark. Again if 2 balls are drawn with Y mark then 4 balls are drawn with X mark So the probability that not more than 2 will be a Y mark is the same as the probability that at least 4 will be a X mark is equal to probability of X greater than equal to 4 This is again equal to probability of X equal to 4 plus probability of X equal to 5 plus probability of X equal to 6 This is equal to 6 C 4 into 3 over 5 raise to power 6 minus 4 into 2 over 5 raise to power 4 plus 6 C 5 into 3 over 5 raise to power 6 minus 5 into 2 over 5 raise to power 5 plus 6 C 6 into 3 over 5 raise to power 6 minus 6 into 2 over 5 raise to power 6 Now this is again equal to 6 C 4 which is 6 into 5 over 2 that is 15 into 3 over 5 square into 2 over 5 raise to power 4 plus 6 C 5 which is 6 only into 3 over 5 into 2 over 5 into 2 over 5 raise to power 5 plus 6 C 6 which is 1 into 3 over 5 raise to power 0 which is also 1 into 2 over 5 raise to power 6 Now let us take 2 over 5 raise to power 4 common from these terms so we have this is equal to 2 over 5 raise to power 4 into 15 into 9 over 25 plus 6 into 3 over 5 into 2 over 5 plus 2 over 5 into 2 over 5 Now this is again equal to 2 over 5 raise to power 4 into 27 over 5 plus 36 over 25 plus 4 over 25 and this is again equal to 2 over 5 raise to power 4 into 135 plus 36 plus 4 over 25 and this is equal to 2 over 5 raise to power 4 into 175 over 25 and this is equal to 7 into 2 over 5 raise to power 4 hence the answer for part 2 is 7 into 2 over 5 raise to power 4 Now in part 3 we have to find the probability that at least one ball will be a Weimar Now the probability that at least one ball will be a Weimar This is equal to 1 minus probability that no ball will be a Weimar So no ball will be a Weimar means all balls will be a X-MAR So this is equal to 1 minus probability that all balls will be a Now we have the probability that all balls will be a X-MAR is 2 over 5 raise to power 6 So this is equal to 1 minus 2 over 5 raise to power 6 So the answer for this part is 1 minus 2 over 5 raise to power 6 In part 4 we have to find the probability that the number of balls with X-MAR and Y-MAR will be equal Now 6 balls are wrong So the probability that the number of balls with X-MAR and Y-MAR will be equal is given by probability of X equal to 3 This is equal to 6C3 raise to power 5 raise to power 3 into 2 over 5 raise to power 3 Now this is equal to 6C3 which is 6 into 5 into 4 over 3 into 2 into 1 into 3 over 5 raise to power 3 which is 27 over 125 into 2 over 5 cube which is 8 over 125 This is equal to 4 into 27 into 8 over 25 into 125 This is equal to 864 over 3125 So the answer for this part is 864 over 3125 So this completes our session I hope the solution is clear to you Bye and have a nice day