 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. Question says, in a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is 5 upon 6. What is the probability that he will knock down fewer than 2 hurdles? Let's now start with the solution. Now we know repeated trials for crossing of hurdles are Bernoulli trials. We know Bernoulli trials are finite in number and they are independent. Also they have exactly 2 outcomes. Now let us assume that X denotes the number of hurdles knocked down by the player in 10 trials. We know he has to cross 10 hurdles. So total number of trials is equal to 10 only. Now we are given probability that player will clear each hurdle is equal to 5 upon 6. So the probability that player does not clear the hurdle is equal to 1 minus 5 upon 6 which is further equal to 1 upon 6. Now clearly we can see X has Bernoulli distribution with n is equal to 10 and p is equal to 1 upon 6. Therefore probability of X successes is equal to ncx multiplied by q raised to the power n minus x multiplied by p raised to the power x where x is equal to 0 to n. Now here we have n is equal to 10, p is equal to 1 upon 6 and q is equal to 1 minus p that is 1 minus 1 upon 6 or we can simply write q is equal to 5 upon 6. Now substituting corresponding values of n, p and q in this expression we get probability of X successes is equal to 10cx multiplied by q is equal to 5 upon 6 So here we can write 5 upon 6 raised to the power 10 minus x multiplied by p that is 1 upon 6 raised to the power x. Now we know probability of knocking down fewer than 2 hurdles is equal to probability of knocking down 0 hurdle plus probability of knocking down 1 hurdle or we can simply write probability of X less than 2 is equal to probability of X is equal to 0 plus probability of X is equal to 1. Now to find this probability or we can say to find probability for X is equal to 0 we will substitute 0 for X in this expression and we get 10c0 multiplied by 5 upon 6 raised to the power 10 minus 0 multiplied by 1 upon 6 raised to the power 0. We will write this plus sign as it is now we will substitute 1 for X in this expression to find this probability and we get 10c1 multiplied by 5 upon 6 raised to the power 10 minus 1 multiplied by 1 upon 6 raised to the power 1. Now simplifying further we get 5 upon 6 raised to the power 10 plus 10 multiplied by 5 upon 6 raised to the power 9 multiplied by 1 upon 6 Now 5 upon 6 is common in both of these terms so we can write 5 upon 6 whole raised to the power 9 multiplied by 5 upon 6 plus 10 upon 6 is equal to probability of X less than 2. Now this is further equal to 5 upon 6 raised to the power 9 multiplied by 15 upon 6 now we will cancel common factor 3 from numerator and denominator both and we get 5 upon 2 multiplied by 5 upon 6 whole raised to the power 9 is equal to probability of X less than 2 or we can write probability of X less than 2 is equal to 5 raised to the power 10 upon 2 multiplied by 6 raised to the power 9 so probability that the player will knock down fewer than 2 hurdles is equal to 5 raised to the power 10 upon 2 multiplied by 6 raised to the power 9 This is our required answer this completes the session hope you understood the solution take care and have a nice day.