 Hello and welcome to the session. I am Deepika here. Let's discuss the question which says an experiment succeeds twice as often as it fails. Find the probability that in the next six trials there will be at least four successes. Now we know that probability of x successes is equal to n c x into q raise to power n minus x into p raise to power x where x is from 0 to n and q is equal to 1 minus p. So this is a key idea behind our question. We will take the help of this key idea to solve the above question. So let's start the solution. Now according to the question an experiment succeeds twice as often as it fails. So according to the given question we have p is equal to 2 q where p is the probability of success and q is the probability of failure. So this implies p is equal to 2 into 1 minus p because q is equal to 1 minus p and this implies p is equal to 2 minus 2 p and this implies 3 p is equal to 2 or p is equal to 2 over 3. Now p is equal to 2 over 3 therefore q which is equal to 1 minus p is equal to 1 minus 2 over 3 and this is equal to 1 over 3. Now we have to find the probability that in the next six trials there will be at least four successes. Now we have probability of x successes is equal to n c x into q raise to power n minus x into p raise to power x where x is from 0 to n and q is equal to 1 minus p. So here we have n is equal to 6 p is equal to 2 over 3 and q is equal to 1 over 3 therefore probability of x successes is equal to 6 c x into 1 over 3 raise to power 6 minus x into 2 over 3 raise to power x. Now probability of at least four successes is given by probability of x greater than equal to 4. Now this is equal to probability of four successes plus probability of five successes plus probability of six successes. So this is equal to 6 c 4 into 1 over 3 raise to power 6 minus 4 into 2 over 3 raise to power 4 plus 6 c 5 into 1 over 3 raise to power 6 minus 5 into 2 over 3 raise to power 5 plus 6 c 6 into 1 over 3 raise to power 6 minus 6 into 2 over 3 raise to power 6. Now this is equal to 6 c 4 which is 6 into 5 over 2 into 1 over 3 raise to power 6 minus 4 which is 1 over 3 square into 2 over 3 raise to power 4 plus 6 c 5 which is 6 into 1 over 3 raise to power 6 minus 5 which is 1 over 3 into 2 over 3 raise to power 5 plus 6 c 6 which is 1 into 1 over 3 raise to power 0 which is again 1 into 2 over 3 raise to power 6. Now this is again equal to 5 over 3 into 2 over 3 raise to power 4 plus 2 into 2 over 3 raise to power 5 plus 2 over 3 raise to power 6. Now let us take 2 over 3 raise to power 4 common among these terms. So we have this is equal to 2 over 3 raise to power 4 into 5 over 3 plus 2 into 2 over 3 plus 2 over 3 into 2 over 3. Now this is equal to 2 over 3 raise to power 4 into 15 plus 12 plus 4 over 9 and this is again equal to 31 over 9 into 2 over 3 raise to power 4 has the probability of at least 4 successes is 31 over 9 into 2 over 3 raise to power 4. So this is the answer for that question. This completes our session. I hope the solution is clear to you. Bye and have a nice day.