SOLUTION: Of all the new vehicles of a certain model that are sold, 20% require repairs to be done under warranty during the first year of service. A particular dealership sells 14 such ve

This is a binomial distribution type problem, where the probability under the question is the sum


     P = P(fewer than 5 of 14) = P(0) + P(1) + P(2) + P(3) + P(4) =       (1)


The number of trials is              14;
The indexes of success trials        k = 0,1,2,3,4
The probability of success trial     p = 0.2;
                                     q = 1 - p
C(n,k) = n! / (k! * (n-k)!)          are binomial coefficients.


The sum  (1)  is a cumulative sum.


Instead of calculating every term of (1) individually and then summing them up, you may use Excel function 

BINOM.DIST(4, 14, 0.2, TRUE)  to calculate the cumulative sum value in one click.


In this way,  you get


    P =  = 0.87016.      ANSWER

This is a binomial distribution type problem, where the probability under the question is the sum


     P(more than 2 of 14) = P(3) + P(4) + P(5) + . . . + P(14) =       (1)


The number of trials is              14;
The indexes of success trials        k = 3,4,5, . . . ,14
The probability of success trial     p = 0.2;
                                     q = 1 - p
C(n,k) = n! / (k! * (n-k)!)          are binomial coefficients.



Instead of calculating every term of (1) individually and then summing them up, you may use Excel standard function BINOM.DIST

which allows calculating the entire aggregate of such addends in one click.


But function BINOM.DIST works for so called cumulative sums ONLY, that are the sums of addends for k from 0 to some integer number.


Therefore, I should convert the sum (1) to the cumulative form.


The conversion is this formula  

     = 1 -        (2)


In the right side of (2),  the sum is in the cumulative form, so I can apply the standard Excel function BINOM.DIST to calculate it


     = BINOM.DIST(2, 14, 0.2, TRUE) = 0.448051.


In this way, the value of  (2)  is equal to  1 - 0.448051 = 0.551949.    


ANSWER.  The probability under the question (b) is  P = 0.551949.