The Jones family was one of the first to come to the U.S. They had 5 children. Assuming that the probability of a child being a girl is .5, find the probability that the Jones family had.
Please proofread, correct, and repost.
at least 3 girls: 0.5
at most 4 girls: 0.96875
To find the probability that the Jones family had a specific number of girls, we can use the concept of binomial probability. In this case, the number of trials is 5 (since the Jones family had 5 children), and the probability of success (having a girl) is 0.5.
To calculate the probability that the Jones family had 0, 1, 2, 3, 4, or 5 girls, we can use the binomial probability formula:
P(x) = (nCx) * p^x * q^(n-x)
Where:
- P(x) is the probability of having x girls
- n is the total number of trials (children), which is 5 in this case
- x is the number of successful trials (girls)
- p is the probability of success (having a girl), which is 0.5
- q is the probability of failure (having a boy), which is 1 - p = 1 - 0.5 = 0.5
Now, let's calculate the probabilities:
P(0 girls) = (5C0) * (0.5^0) * (0.5^(5-0)) = (1) * (1) * (1/32) = 1/32 = 0.03125
P(1 girl) = (5C1) * (0.5^1) * (0.5^(5-1)) = (5) * (0.5) * (0.5^4) = 5/32 = 0.15625
P(2 girls) = (5C2) * (0.5^2) * (0.5^(5-2)) = (10) * (0.5^2) * (0.5^3) = 10/32 = 0.3125
P(3 girls) = (5C3) * (0.5^3) * (0.5^(5-3)) = (10) * (0.5^3) * (0.5^2) = 10/32 = 0.3125
P(4 girls) = (5C4) * (0.5^4) * (0.5^(5-4)) = (5) * (0.5^4) * (0.5^1) = 5/32 = 0.15625
P(5 girls) = (5C5) * (0.5^5) * (0.5^(5-5)) = (1) * (0.5^5) * (0.5^0) = 1/32 = 0.03125
Therefore, the probabilities that the Jones family had 0, 1, 2, 3, 4, or 5 girls are: 0.03125, 0.15625, 0.3125, 0.3125, 0.15625, and 0.03125, respectively.