Question

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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?

Answer #1

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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