The Jones 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 Jones family had at least 5 girls.
Use binomial distribution because:
1. Bernoulli trials (either girl or boy)
2. number of trials (n=9) known.
3. probability (p=0.5) of outcome known and remains constant throughout trials
4. trials are independent of each other.
n=9
p=0.5
P(X=k)=C(n,k)*p^k*(1-p)^(n-k)
where
C(n,k) is the number of combinations of k objects taken from n.
At least 5 girls means X=5,6,7,8,9
So
P(X>=5)
=P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)
=0.24609+0.16406+0.07031+0.01758+0.00195
=0.5000
To find the probability that the Jones family had at least 5 girls, we need to calculate the probability of having 5, 6, 7, 8, or 9 girls and sum them up.
The probability of having exactly k girls out of 9 children can be calculated using the binomial probability formula:
P(k girls) = (9 choose k) * (0.5)^k * (0.5)^(9-k)
where (n choose k) represents the binomial coefficient, equal to n! / (k! * (n-k)!)
Using this formula, we can calculate the probabilities for each case:
P(5 girls) = (9 choose 5) * (0.5)^5 * (0.5)^(9-5)
P(6 girls) = (9 choose 6) * (0.5)^6 * (0.5)^(9-6)
P(7 girls) = (9 choose 7) * (0.5)^7 * (0.5)^(9-7)
P(8 girls) = (9 choose 8) * (0.5)^8 * (0.5)^(9-8)
P(9 girls) = (9 choose 9) * (0.5)^9 * (0.5)^(9-9)
Finally, we add up these probabilities to get the desired result:
P(at least 5 girls) = P(5 girls) + P(6 girls) + P(7 girls) + P(8 girls) + P(9 girls)
To find the probability that the Jones family had at least 5 girls, we can use the binomial distribution formula. However, since the number of children is small (9), we can calculate it directly.
First, let's determine the number of possible outcomes. Since each child can be either a girl or a boy, there are 2^9 = 512 possible outcomes, representing all the different combinations of genders for the 9 children.
Next, let's determine the number of favorable outcomes, i.e., the number of combinations in which the Jones family had at least 5 girls. We can consider four cases:
Case 1: 5 girls and 4 boys
The number of ways to choose 5 girls from 9 is given by the combination formula: C(9, 5) = 9! / (5! * (9-5)!) = 126.
Similarly, the number of ways to choose 4 boys from 9 is C(9, 4) = 9! / (4! * (9-4)!) = 126.
Therefore, the total number of outcomes with 5 girls and 4 boys is 126 * 126 = 15,876.
Case 2: 6 girls and 3 boys
Using the combination formula, C(9, 6) = 9! / (6! * (9-6)!) = 84, representing the number of ways to choose 6 girls from 9.
Similarly, C(9, 3) = 9! / (3! * (9-3)!) = 84, representing the number of ways to choose 3 boys from 9.
The total number of outcomes with 6 girls and 3 boys is 84 * 84 = 7,056.
Case 3: 7 girls and 2 boys
Using the combination formula, C(9, 7) = 9! / (7! * (9-7)!) = 36, representing the number of ways to choose 7 girls from 9.
Similarly, C(9, 2) = 9! / (2! * (9-2)!) = 36, representing the number of ways to choose 2 boys from 9.
The total number of outcomes with 7 girls and 2 boys is 36 * 36 = 1,296.
Case 4: 8 girls and 1 boy
Using the combination formula, C(9, 8) = 9! / (8! * (9-8)!) = 9, representing the number of ways to choose 8 girls from 9.
Similarly, C(9, 1) = 9! / (1! * (9-1)!) = 9, representing the number of ways to choose 1 boy from 9.
The total number of outcomes with 8 girls and 1 boy is 9 * 9 = 81.
Finally, we need to account for the outcome where all 9 children are girls, which is just one possibility.
So, the total number of favorable outcomes (outcomes with at least 5 girls) is 15,876 + 7,056 + 1,296 + 81 + 1 = 24,310.
Now we can find the probability by dividing the number of favorable outcomes by the number of possible outcomes.
Probability of having at least 5 girls = 24,310 / 512 = 0.0475 (rounded to 4 decimal places)
Therefore, the probability that the Jones family had at least 5 girls is approximately 0.0475 or 4.75%.