To solve this problem, we make use of the binomial
probability equation:
P = [n! / (n – r)! r!] p^r * q^(n – r)
where,
n is the total number of adult samples = 77
r is the selected number of adults who believe in
reincarnation
p is the of believing in reincarnation = 40% = 0.40
q is 1 – p = 0.60
a. What is the probability that exactly 66 of the
selected adults believe in reincarnation?
So we use r = 66
P = [77! / (77 – 66)! 66!] 0.40^66 * 0.60^(77 – 66)
P = 1.32 x 10^-16
b. What is the probability that all of the selected
adults believe in reincarnation?
So we use r = 77
P = [77! / (77 – 77)! 77!] 0.40^77 * 0.60^(77 – 77)
P =2.28 x 10^-31
c. What is the
probability that at least 66 of the selected adults believe in reincarnation?
So we use r = 66 to 77
P (r=66) = 1.32 x 10^-16
P (r=67) = [77! / (77 – 67)! 67!] 0.40^67 * 0.60^(77 – 67)
= 1.44 x 10^-17
P (r=68) = [77! / (77 – 68)! 68!] 0.40^68 * 0.60^(77 – 68)
= 1.42 x 10^-18
P (r=69) = [77! / (77 – 69)! 69!] 0.40^69 * 0.60^(77 – 69)
= 1.23 x 10^-19
P (r=70) = [77! / (77 – 70)! 70!] 0.40^70 * 0.60^(77 – 70)
= 9.38 x 10^-21
P (r=71) = [77! / (77 – 71)! 71!] 0.40^71 * 0.60^(77 – 71)
= 6.17 x 10^-22
P (r=72) = [77! / (77 – 72)! 72!] 0.40^72 * 0.60^(77 – 72)
= 3.43 x 10^-23
P (r=73) = [77! / (77 – 73)! 73!] 0.40^73 * 0.60^(77 – 73)
= 1.56 x 10^-24
P (r=74) = [77! / (77 – 74)! 74!] 0.40^74 * 0.60^(77 – 74)
= 5.64 x 10^-26
P (r=75) = [77! / (77 – 75)! 75!] 0.40^75 * 0.60^(77 – 75)
= 1.50 x 10^-27
P (r=76) = [77! / (77 – 76)! 76!] 0.40^76 * 0.60^(77 – 76)
= 2.64 x 10^-29
P (r=77) = 2.28 x 10^-31
The total is the sum of all:
P(total) = 1.48 x 10^-16
d. If 77 adults are randomly selected, is 66 a
significantly high number who believe in reincarnation?
C.Yes, because the probability
that 66 or more of the selected adults believe in reincarnation is less than
0.05
