00:01
Hello everyone.
00:03
So for this question, the random variable x denotes that a student has graduated, has graduated, and it is given that x follows binomial distribution with parameters n np, such that the value of n is equal to 10, and p, the probability of success, that is the student has graduated, is a student, 90 .9 percentage, that is 0 .909.
00:53
Right? so here we calculate the value of q also, which is 1 minus b, that is 1 minus 0 .909, which will be equals to 0 .091.
01:10
Now, we will note down the probability mass function that is probability of x is equals to k is n c k p raised to the power k q raised to the power n minus k here the values of n p and q are given so we can rewrite this as 10 k multiplied by p that is 0 .909 raised to the power k multiplied by q that is 0 .091 raised to the power 10 minus k.
01:48
Coming to part first, if 10 of the students from these special programs are randomly selected, find the probability that at least nine of them graduated, that is probability of x greater than equals to nine.
02:07
So this will be equals to probability of x equals to 9 plus probability of x equals to 10.
02:21
Now in the probability mass function ever we'll be substituting the value of k as 9 and 10.
02:26
That is we get 10c9, multiplied by 0 .909, raised to the power 9, multiplied by 0 .091, raised to the power 10 minus, 9.
02:46
Plus 10c10 multiplied by 0 .909 raise to the power 10 multiplied by 0 .091 raise to the power 10 minus 10.
03:00
So this becomes on solving the first part that is 10c9 multiplied by 0 .909 raised to the power 9 multiplied by 0 .09 raised to the power 9 multiplied by 0 .09 multiplied by 0 .091.
03:23
Raised to the power 10 minus 1 that is 1.
03:29
So this will be equals to 0 .386 plus 10c 10 is 1 multiplied by 0 .909 raised to the power 10 multiplied by 0 .091 raised to the power 0 .0 that is 1.
03:51
So this becomes 0 .385.
03:56
Now adding these 2 that is 0 .386 and 0 .386 and 0 .386...