Data shows that 93% of the people in a certain population are right-handed. A group of 9 people from this population are selected at random. What is the probability that exactly 7 of the people in the group are right-handed?
2
prob of right-handed = .93
prob of NOT right-handed = .07
prob (7 of 9 are right-handed)
= C(9,7)(.93)^7(.07)^2
= appr. 0.10614
To find the probability of exactly 7 people out of a group of 9 being right-handed, we can use the concept of binomial probability.
The binomial probability formula is:
P(x) = (nCx) * p^x * q^(n-x)
Where:
P(x) is the probability of exactly x successes in n trials,
nCx is the number of combinations of n items taken x at a time,
p is the probability of success in a single trial,
q is the probability of failure in a single trial (q = 1 - p),
x is the number of successes we want to find,
and n is the total number of trials.
In this case:
n = 9 (total number of people in the group)
x = 7 (number of right-handed people we want to find)
p = 0.93 (probability of a person being right-handed)
q = 1 - p = 1 - 0.93 = 0.07 (probability of a person being left-handed)
Using the binomial probability formula, we can calculate the probability:
P(7) = (9C7) * (0.93^7) * (0.07^(9-7))
To calculate (9C7), we use the formula for combinations:
(9C7) = 9! / (7! * (9-7)!)
where n! is the factorial of n.
Calculating (9C7):
(9C7) = 9! / (7! * 2!)
= (9 * 8 * 7!) / (7! * 2 * 1)
= (9 * 8) / (2 * 1)
= 36
Plugging this into the binomial probability formula:
P(7) = 36 * (0.93^7) * (0.07^2)
Calculating (0.93^7) and (0.07^2):
(0.93^7) ≈ 0.5477463623
(0.07^2) = 0.0049
Plugging these values back into the formula:
P(7) = 36 * 0.5477463623 * 0.0049
≈ 0.09641
Therefore, the probability that exactly 7 people out of a group of 9 are right-handed is approximately 0.09641 or about 9.641%.