Question

A blue jay has a nest with 7 eggs. There is a 50% chance that they will hatch to be male.

What is the probability that the nest will hatch 3 males?

What is the probability that the nest will hatch more than 3 males?

What is the probability that the nest will hatch at most 3 males?

Answer 1

Since Blue jay has a nest with 7 eggs and there is a 50% chance that they will hatch to be male.

Number of trials = n = 7

And probability of egg will hatch male = p = 0.5

(a) the probability that the nest will hatch 3 males

Number of success = X = 3

Applying the binomial probability formula

$p(X) = \binom{n}{X}*p^{X}*(1-p)^{n-X}$

Where

$\binom{n}{X} = \frac{n!}{X!(n-X)!}$

Therefore

$p(X) =\frac{n!}{X!(n-X)!}*p^{X}*(1-p)^{n-X}$

$p(3) =\frac{7!}{3!(7-3)!}*0.5^{3}*(1-0.5)^{7-3}$​​​​​​

So the probability that the nest will hatch 3 males is 0.2734

(b)  the probability that the nest will hatch more than 3 males

So

$p(4) =\frac{7!}{4!(7-4)!}*0.5^{4}*(1-0.5)^{7-4} = 0.2734$

$p(5) =\frac{7!}{5!(7-5)!}*0.5^{5}*(1-0.5)^{7-5} = 0.1641$

$p(6) =\frac{7!}{6!(7-6)!}*0.5^{6}*(1-0.5)^{7-6} = 0.0547$

$p(7) =\frac{7!}{7!(7-7)!}*0.5^{7}*(1-0.5)^{7-7} = 0.0078$

​​​​​​

So the probability that the nest will hatch more than 3 males is 0.5

(c)the probability that the nest will hatch at most 3 males

$p(X\leq 3) = p(3)+p(2)+p(1)+p(0)$

0r,

$p(X\leq 3) = 1-p(X> 3)$

Since we have calculated p(X>3) in last part

p(X < 3) = 1-0.5 = 0.5

So the probability that the nest will hatch at most 3 males is 0.5.