 Hello and welcome to the session. In this session we discuss the following question that says on a multiple choice examination with three possible answers out of which only one is correct for each of the five questions what is the probability that a candidate would get four or more correct answers just by guessing. Consider the probability function of the binomial distribution where we have probability of x is equal to x is equal to mcx into p raised to the power x into q raised to the power n minus x where we have this x is equal to 0 1 and so on up to n and q is equal to 1 minus p or you can say p plus q is equal to 1. So this n is the number of trials in an experiment capital X of t random variable which denotes the number of successes in the trials. Then this p is the probability of getting a success in a single trial and q is the probability of getting a failure in a single trial. This is the key idea that we use in this question. Let's now move on to the solution. So in an examination there are five multiple choice questions and there are three possible answers for each question out of which only one answer would be correct and we are supposed to find the probability that the candidate would get four or more correct answers just by guessing. First of all we suppose let the number of correct answers given by the candidate just by guessing then this x would be the random variable which would denote the number of successes in the trials. Now as the number of questions are five so n would be equal to five that is number of trials would be five p would be the probability of getting a correct answer by just guessing on a multiple choice question with three possible answers for each question. So in this case p would be equal to 1 upon 3 and since we know that q plus p is equal to 1 so this means q would be equal to 1 minus 1 upon 3 that is equal to 2 upon 3. So probability of x equal to x which is the probability of getting x correct answers by guessing in five multiple choice questions would be given by x into p to the power x into q to the power n minus x where we have the value of n as 5 p as 1 upon 3 and q as 2 upon 3 so this would be equal to 5 cx into p to the power of x into q to the power of 5 minus x where x is equal to 0 1 2 3 4 and 5. As in the question we are supposed to find the probability that a candidate would get four or more correct answers just by guessing that is we are supposed to find the probability of x greater than equal to 4 which is the probability of getting or more or you can say four or five correct answers just by guessing. So probability of x greater than equal to 4 would be equal to probability of x equal to 4 plus the probability of x equal to 5 and these two probabilities can be found out by putting x equal to 4 and 5 respectively in this formula. So this is equal to 5 c4 into 1 upon 3 to the power of 4 into 2 upon 3 to the power of 5 minus 4 which is 1 plus 5 c5 into 1 upon 3 to the power of 5 into 2 upon 3 to the power of 5 minus 5 which is 0. Now 5 c4 is equal to 5 so 5 into 1 upon 3 whole to the power of 4 into 2 upon 3 plus 5 c5 is 1 so 1 into 1 upon 3 whole to the power of 5 and 2 upon 3 to the power of 0 is 1. So taken 1 upon 3 whole to the power of 4 common we get 1 upon 3 whole to the power of 4 into 5 into 2 upon 3 is 10 upon 3 plus 1 upon 3. So this is equal to 1 upon 3 whole to the power of 4 into 11 upon 3 which is equal to 1 upon 81 into 11 upon 3 or you can say 11 upon 243 this is the probability of x greater than equal to 4. So thus we can say the required probability is equal to 11 upon 243. So 11 upon 243 is our final answer. This completes the session hope you have understood the solution of this question.