f both parents are carriers (heterozygous) for a gene responsible for a genetically inherited disorder the probability of baby having 0.25 to have 6 kids, calculate the probability that at least 2 lids will have the disorder.

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if both parents are carriers (heterozygous) for a gene responsible for a genetically inherited disorder the probability of baby having 0.25 to have 6 kids, calculate the probability that at least 2 lids will have the disorder. 

the correct answer is between 0.41 and 0.49 and we are supposed to use the binomial distribution equation, but I am having trouble plugging the numbers in correctly can someone help?

both parents are carriers - prob of ther baby having
the disorder is 0.25
they decide to have le kids
Calculate the probability that at least 2 kids
will have the disorder
use. binomial distrib.eg.
P² (1-p) ^-x
f(x) = n!
x: (n-x)!
f(2)= 6!
0.25! (6-0.25)
720
p (2)
225(1-2)6-0,25
Correct answer is btw) 0.41-0.49
Transcribed Image Text:both parents are carriers - prob of ther baby having the disorder is 0.25 they decide to have le kids Calculate the probability that at least 2 kids will have the disorder use. binomial distrib.eg. P² (1-p) ^-x f(x) = n! x: (n-x)! f(2)= 6! 0.25! (6-0.25) 720 p (2) 225(1-2)6-0,25 Correct answer is btw) 0.41-0.49
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