 Hello, and welcome to the session I am Deepika here. Let's discuss the question which says, In an examination, 20 questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers true. If it falls tails, he answers false. Find the probability that he answers at least 12 questions correctly. So let's start the solution. Now according to the question, in an examination 20 questions of true-false type are asked. A student tosses a fair coin to determine his answer to each question. Now we know that the repeated tosses of a coin are Bernoulli trials because here there are 20 questions and tossing coin for each of them is independent of each other and also number of trials is finite that is 20. So the trials are Bernoulli trials. Again according to the question, if the coin falls heads, he answers true. If it falls tails, he answers false. And we have to find the probability that he answers at least 10 questions correctly. So let X denote the number of tosses. So X has a binomial distribution with n is equal to 20 and p is equal to 1 over 2 because the probability of getting a correct answer is 1 over 2. Now we have probability of X successes is equal to ncx q raised to power n minus x into p raised to power x where x is from 0 to n and q is equal to 1 minus p. Now here we have n is equal to 20. Probability of the correct answer is 1 over 2. So p is equal to 1 over 2 and q which is equal to 1 minus p is equal to 1 minus 1 over 2 which is again equal to 1 over 2. So we have probability of X successes is equal to 20cx into 1 over 2 raised to power 20 minus x into 1 over 2 raised to power x and this is again equal to 20cx into 1 over 2 raised to power 20. Now we have to find the probability of at least 12 successes this is given by probability of X greater than equal to 12. So this is equal to probability of 12 successes plus probability of 13 successes plus so on till probability of 20 successes so this is equal to 20c12 into 1 over 2 raised to power 20 plus 20c13 into 1 over 2 raised to power 20 plus so on till 20c20 into 1 over 2 raised to power 20. So this is again equal to let us take 1 over 2 raised to power 20 common from these terms. So we have probability of at least 12 successes is equal to 1 over 2 raised to power 20 into 20c12 plus 20c13 plus so on till 20c20. So this is a probability that the student answers at least 12 questions correctly. So the answer for the above question is 1 over 2 raised to power 20 into 20c12 plus 20c13 plus so on till 20c20. So this completes our session. I hope the solution is clear to you. Bye and have a nice day.