Question

Thirty percent of all customers who enter a store will make a purchase. suppose that six customers enter the store and that these customers make independent purchase decisions. (1) use the binomial formula to calculate the probability that exactly five customers make a purchase. (round your answer to 4 decimal places.) probability (2) use the binomial formula to calculate the probability that at least three customers make a purchase. (round your answer to 4 decimal places.) probability (3) use the binomial formula to calculate the probability that two or fewer customers make a purchase. (round your answer to 4 decimal places.) probability (4) use the binomial formula to calculate the probability that at least one customer makes a purchase. (round your answer to 4 decimal places.) probability

Answer 1

Using binomial distrubution,
p=0.30
$P(X=x) = C(n,x)p^x (1-p)^{n-x}$
where
$C(n,x)=\frac{n!}{x!(n-x)!}$

(1) exactly x=5 customers out of n=6 make purchase
$P(X=x)=C(n,x)p^x (1-p)^{n-x}$
$P(X=5)=C(6,5)(0.3)^5 (0.7)^{6-5}$
$=6*0.00243*0.7$
$=0.10206$

(2) at least three customers make a purchase
$P(X\ge 3)=\sum_{i=3}^{6}C(6,i)(0.3)^i (1-0.3)^{6-i}$
=0.185220+0.059535+0.010206+0.000729=0.255690
(3) Two or less customers make purchase
Similarly, using binomial distribution
$P(X\le2)=\sum_{i=3}^{6}C(6,i)(0.3)^i(1-0.3)^{6-i}$
=0.117649+0.302526+0.324135
=0.744310

In fact, we do not need to use the binomial distribution to find the answer.
From results in (2), we have
P(X<=2)=1-P(X>=3)=1-0.255690=0.744310

(4) At least one customer makes a purchase
Use the same formulas above for P(X>=1)
Here we will just use
P(X>=1) 
= 1- P(X<1)
= 1-P(X=0)
$1-P(X=0)=1-C(6,0)(0.3)^0 (1-0.3)^{6-0}$ 
=1-0.117649
=0.882351