Mr. and Mrs. Doran have a genetic history such that the probability that a child being born to them with a certain trait is 1/4. If they have seven children, what is the probability that exactly six of their seven children will have that trait? Round your answer to the nearest thousandth.

Question

Mr. and Mrs. Doran have a genetic history such that the probability that a child being born to them with a certain trait is 1/4. If they have seven children, what is the probability that exactly six of their seven children will have that trait? Round your answer to the nearest thousandth.

Answer 1

The probability of an event, p, occurring exactly r times:

n Cr .pr . qn-r
Binomial Probability-1
n = number of trials
r = number of specific events you wish to obtain
p = probability that the event will occur
q = probability that the event will not occur
(q = 1 – p, the complement of the event)

Answer 2

p = 1/4

q = 1-1/4 = 3/4
n = 7
r = 6
C(n,r) = n!/ [ r!(n-r)! ] = 7!/ [6!*1*] = 7
7 * (1/4)^6 ( 3/4)^1
= 7 * 0.00024 * .75
= 0.00128

Answer 3

I copied

screwy notation
better
nCr * p^r * q^ (n-r)

Answer 4

To find the probability, we need to calculate the probability of having exactly six children with the certain trait out of the seven children.

The probability of having a child with the certain trait is 1/4. Therefore, the probability of having a child without the certain trait is 1 - 1/4 = 3/4.

We can use the binomial probability formula to calculate the probability:

P(X = k) = (nCk) * p^k * (1 - p)^(n - k)

Where:
P(X = k) is the probability of having exactly k successes,
n is the total number of trials (number of children in this case),
k is the total number of successful trials,
p is the probability of success in any given trial,
and (nCk) is the binomial coefficient, calculated as n! / (k! * (n - k)!)

In this case, we want to find the probability of having exactly 6 children with the certain trait out of 7 children.

Using the formula, substituting the values, we get:

P(X = 6) = (7C6) * (1/4)^6 * (3/4)^(7-6)

Calculate as follows:

P(X = 6) = (7!) / (6!(7-6)!) * (1/4)^6 * (3/4)^1

Simplifying:

P(X = 6) = 7 * (1/4)^6 * (3/4)

P(X = 6) = 7 * (1/4)^6 * 3/4

P(X = 6) = 7 * 1/4096 * 3/4

P(X = 6) = 0.0168 (rounded to four decimal places)

Therefore, the probability that exactly six of their seven children will have that certain trait is approximately 0.0168 or 1.68% when rounded to the nearest thousandth.