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
because 36 goes into 60 1 time with 24 left over
