The Wilson family was one of the first to come to the U.S. They had 9 children. Assuming that the probability of a child being a girl is .5, find the probability that the Wilson family had:
at least 2 girls?
at most 3 girls?
To find the probability of at least 2 girls in a family of 9 children, we can calculate the probability of having exactly 0 or 1 girl and subtract it from 1.
The probability of having exactly 0 girls can be calculated using the binomial probability formula:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) is the probability of having k successes (in our case, girls)
- C(n, k) is the number of combinations of choosing k items from a set of n items
- p is the probability of success (having a girl)
- (1 - p) is the probability of failure (having a boy)
- n is the total number of trials (in our case, the number of children)
Using this formula, the probability of having exactly 0 girls in a family of 9 children is:
P(X = 0) = C(9, 0) * 0.5^0 * (1 - 0.5)^(9 - 0) = 1 * 1 * 0.5^9 = 0.001953125
Similarly, the probability of having exactly 1 girl in a family of 9 children is:
P(X = 1) = C(9, 1) * 0.5^1 * (1 - 0.5)^(9 - 1) = 9 * 0.5 * 0.5^8 = 0.017578125
Therefore, the probability of having at least 2 girls in the Wilson family is:
P(at least 2 girls) = 1 - P(X = 0) - P(X = 1) = 1 - 0.001953125 - 0.017578125 = 0.98046875
To find the probability of at most 3 girls, we can calculate the probability of having 0, 1, 2, or 3 girls in the family.
Using the same formula, the probability of having exactly 2 girls is:
P(X = 2) = C(9, 2) * 0.5^2 * (1 - 0.5)^(9 - 2) = 36 * 0.25 * 0.5^7 = 0.1640625
The probability of having exactly 3 girls is:
P(X = 3) = C(9, 3) * 0.5^3 * (1 - 0.5)^(9 - 3) = 84 * 0.125 * 0.5^6 = 0.2734375
Therefore, the probability of having at most 3 girls in the Wilson family is:
P(at most 3 girls) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.001953125 + 0.017578125 + 0.1640625 + 0.2734375 = 0.45603125
To find the probability that the Wilson family had at least 2 girls, we need to calculate the probability of all possible outcomes where they have 2, 3, 4, 5, 6, 7, 8, or 9 girls and add them together.
First, let's find the probability of having exactly 2 girls. Since the probability of having a girl is 0.5, the probability of having a boy is also 0.5. So, the probability of having exactly 2 girls and 7 boys is:
P(2 girls) = (0.5)² * (0.5)⁷
Similarly, the probability of having exactly 3 girls is:
P(3 girls) = (0.5)³ * (0.5)⁶
The probability of having 4 girls is:
P(4 girls) = (0.5)⁴ * (0.5)⁵
And so on, until we reach the probability of having 9 girls:
P(9 girls) = (0.5)⁹ * (0.5)⁰ (since the probability of having 0 boys is 1)
To find the probability of at least 2 girls, we need to sum up all these probabilities:
P(at least 2 girls) = P(2 girls) + P(3 girls) + P(4 girls) + ... + P(9 girls)
Similarly, we can find the probability of having at most 3 girls by summing up the probabilities of having 0, 1, 2, or 3 girls:
P(at most 3 girls) = P(0 girls) + P(1 girl) + P(2 girls) + P(3 girls)
Now, let's calculate these probabilities:
P(2 girls) = (0.5)² * (0.5)⁷ = 0.5⁹
P(3 girls) = (0.5)³ * (0.5)⁶ = 0.5⁹
P(4 girls) = (0.5)⁴ * (0.5)⁵ = 0.5⁹
P(5 girls) = (0.5)⁵ * (0.5)⁴ = 0.5⁹
P(6 girls) = (0.5)⁶ * (0.5)³ = 0.5⁹
P(7 girls) = (0.5)⁷ * (0.5)² = 0.5⁹
P(8 girls) = (0.5)⁸ * (0.5)¹ = 0.5⁹
P(9 girls) = (0.5)⁹ * (0.5)⁰ = 0.5⁹
P(at least 2 girls) = P(2 girls) + P(3 girls) + P(4 girls) + ... + P(9 girls) = 8 * 0.5⁹
P(at most 3 girls) = P(0 girls) + P(1 girl) + P(2 girls) + P(3 girls) = 4 * 0.5⁹
After evaluating the calculations, we find that:
P(at least 2 girls) ≈ 0.363
P(at most 3 girls) ≈ 0.527
Therefore, the probability that the Wilson family had at least 2 girls is approximately 0.363, and the probability that they had at most 3 girls is approximately 0.527.
Prob(at least 2 girls)
= 1 - prob(no girls) - prob(exactly 1 girl)
= 1 - C(9,0) (1/2)^0 (1/2)^9 - C(9,1) (1/2) (1/2)^8
= 1 - (1/2)^9 - 9(1/2)^9
= 1 - 1/512 - 9/512
= 251/512 or appr .98
prob(at most 3 girls)
prob(exactly 1 girl) + prob(exactly 2 girls) + prob(exactly 3 girls)
= .....