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.
Answers
Anonymous
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)
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)
Anonymous
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
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
Anonymous
I copied
screwy notation
better
nCr * p^r * q^ (n-r)
screwy notation
better
nCr * p^r * q^ (n-r)