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A biologist examines turtles for a genetic defect, which is normally found in 12.5% of turtles. She collects and examines 12 turtles. What's the probability that she finds the trait in
(a) none of the 12 turtles?
(b) exactly 2 of the 12 turtles?
(c) at least 3 of the 12 turtles?
Solution-A;
p=12.5%=0.125
q=1-p=1-0.125=0.875
n=12
problem on binomial distribution
P(none of the 12 turtles)=P(X=0)
we have P(X=x)=ncx*p^x*q^n-x
P(X=0)=12c0*0.125^0*0.875^12-0
=0.20142
probability that she finds the trait in none of the 12 turtles is 0.20142
Solution-b:
P(X=2)
P(X=2)=12c2*0.125^2*0.875^12-2
=0.2713
probability that she finds the trait in exactly 2 of the 12 turtles
0.2713
Solution-C:
P(X>=3)
=1-P(X<2)
=1-(P(X=0)+P(X=1)+P(X=2)
=1-{(12c0*0.125^0*0.875^12-0)+(12c1*0.125^1*0.875^12-1)+(12c2*0.125^2*0.875^12-2)}
=1-{0.20142+ 0.34529+0.2713}
=1-0.8180
=0.182
probability that she finds the trait in at least 3 of the 12 turtles=
0.182
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