It has been found out that the probability that a child is a male in family is 0.4. In a family of 6 children, what is the probability that; i. exactly 5 are boys ? ii. at least 4 are boys ? iii. all boys

Formula (if you don't use the binomial probability table):

P(x) = (nCx)(p^x)[q^(n-x)]

Values:
n = 6
p = 0.4
q = 1 - p = 1 - 0.4 = 0.6

i. Find P(5)

ii. Find P(4), P(5), and P(6). Add together for your total probability.

iii. I'll let you try this one on your own.

I hope this will help get you started.

.036864

.0781

To find the probabilities in this scenario, we can use the binomial probability formula. The formula is:

P(x) = (nCx) * p^x * (1-p)^(n-x)

Where:
P(x) is the probability of getting exactly x successes
n is the total number of trials
p is the probability of success in a single trial

Now let's solve the problems step by step:

i. To find the probability that exactly 5 children are boys in a family of 6 children, we will use the binomial probability formula with x = 5.

P(5 boys) = (6C5) * 0.4^5 * (1-0.4)^(6-5)

First, we calculate (6C5) which represents the number of ways to choose 5 boys out of 6 children:

(6C5) = 6!/((6-5)! * 5!) = 6

Next, we substitute the values into the formula:

P(5 boys) = 6 * 0.4^5 * 0.6^1

P(5 boys) = 0.07776

Therefore, the probability that exactly 5 children are boys in a family of 6 children is 0.07776, or 7.78%.

ii. To find the probability that at least 4 children are boys in a family of 6 children, we need to calculate the probability of getting 4 boys, plus the probability of getting 5 boys, plus the probability of getting all 6 boys.

P(at least 4 boys) = P(4 boys) + P(5 boys) + P(6 boys)

We already calculated P(5 boys) as 0.07776.

P(4 boys) can be calculated using the same formula as before, with x = 4:

P(4 boys) = (6C4) * 0.4^4 * (1-0.4)^(6-4)

(6C4) = 6!/((6-4)! * 4!) = 15

P(4 boys) = 15 * 0.4^4 * 0.6^2 = 0.2304

P(6 boys) can be calculated similarly, with x = 6:

P(6 boys) = (6C6) * 0.4^6 * (1-0.4)^(6-6)

(6C6) = 6!/((6-6)! * 6!) = 1

P(6 boys) = 1 * 0.4^6 * 0.6^0 = 0.046656

Now we can calculate P(at least 4 boys):

P(at least 4 boys) = 0.2304 + 0.07776 + 0.046656

P(at least 4 boys) = 0.354816

Therefore, the probability that at least 4 children are boys in a family of 6 children is 0.354816, or 35.48%.

iii. To find the probability of having all boys in a family of 6 children, we can directly calculate it using the probability formula with x = 6:

P(all boys) = (6C6) * 0.4^6 * (1-0.4)^(6-6)

(6C6) = 6!/((6-6)! * 6!) = 1 (there is only one way to have all boys)

P(all boys) = 1 * 0.4^6 * 0.6^0 = 0.046656

Therefore, the probability that all children in a family of 6 children are boys is 0.046656, or 4.67%.