AIf 12 of all disk drives produced on an assembly line are d

A)If 12% of all disk drives produced on an assembly line are defective, what is the probability that there will be exactly two defects in a random sample of 24 of these?

B)What is the probability that there will be one defect in a random sample of 24?

C)What is the probability that there will be less than seven defects in a random sample of 24?

**For A I did

=FACT(24)/FACT(22)*(0.12)^2*(0.88)^22

For B I did

=FACT(24)/FACT(23)*(0.12)^1*(0.88)^23

and I\'m confused on C. Can someone help me out on excel with C, and double check my A&B?

Solution

Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

a)
P( X = 2 ) = ( 24 2 ) * ( 0.12^2) * ( 1 - 0.12 )^22
= 0.2387
b)
P( X = 1 ) = ( 24 1 ) * ( 0.12^1) * ( 1 - 0.12 )^23
= 0.1522
c)
P( X < 7) = P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 24 6 ) * 0.12^6 * ( 1- 0.12 ) ^18 + ( 24 5 ) * 0.12^5 * ( 1- 0.12 ) ^19 + ( 24 4 ) * 0.12^4 * ( 1- 0.12 ) ^20 + ( 24 3 ) * 0.12^3 * ( 1- 0.12 ) ^21 + ( 24 2 ) * 0.12^2 * ( 1- 0.12 ) ^22 + ( 24 1 ) * 0.12^1 * ( 1- 0.12 ) ^23 + ( 24 0 ) * 0.12^0 * ( 1- 0.12 ) ^24   
= 0.9806