To solve this problem, we can use the binomial formula:
P(x) = [n! / (x!(n-x)!)] * P^x * Q^(n-x)
where:
- P(x) is the probability of getting exactly x successes
- n is the total number of trials (in this case, the number of family members, 5)
- x is the number of successes we want to achieve (in this case, 2)
- P is the probability of success (in this case, the probability of catching the flu, 0.15)
- Q is the probability of failure (in this case, 1 - P, which is 0.85)
Now, let's substitute the given values into the formula:
P(2) = [5! / (2!(5-2)!)] * 0.15^2 * 0.85^(5-2)
Simplifying the factorial terms:
P(2) = [5! / (2!3!)] * 0.15^2 * 0.85^3
Simplifying the other terms:
P(2) = [5*4*3! / 2!3!] * 0.0225 * 0.614125
Simplifying further:
P(2) = [5*4] * 0.0225 * 0.614125
P(2) = 20 * 0.0225 * 0.614125
P(2) = 0.277/2
P(2) = 0.138
Therefore, the probability that 2 out of 5 members of the Dawson family get the flu this year is approximately 13.8%.
So, the correct answer is c. 13.8%.
The probability of catching the flu this year is 0.15. Use the binomial model to find the probability that 2 out 5 members of the Dawson family get the flu this year.
P(x) = [n!/ x!(n-x)!] P^xQ^n-x
a. 2.43%
b. 10.7%
c. 13.8%
d. 23.7%
1 answer