The American Red Cross says that about 25% of the population has Type B blood. A blood drive is being held at your institute. Suppose that you have selected 20 blood donors in a random way. The probability that exactly 2 of the blood donors have Type B blood is
Given: p = 0.25; x = 2; n = 20
b(n,p,x)=(nx)∗px∗(1−p)(n−x)b(n,p,x) = \binom{n}{x}*p^x* {\lparen1-p\rparen}^{\lparen n-x\rparen}b(n,p,x)=(xn)∗px∗(1−p)(n−x)
Here (nx)≡Cnx=n!x!(n−x)!\binom{n}{x}\equiv C^x_n = \frac{n!}{x!(n-x)!}(xn)≡Cnx=x!(n−x)!n!
Then b(20,0.25,2)=20!2!(20−2)!∗0.252∗(1−0.25)20−2=b(20,0.25,2) = \frac{20!}{2!(20-2)!}*0.25^2*(1-0.25)^{20-2} =b(20,0.25,2)=2!(20−2)!20!∗0.252∗(1−0.25)20−2=
=19∗202∗0.0625∗0.7518≈190∗0.625∗0.0056377≈0.066947= \frac{19*20}{2}*0.0625*0.75^{18} \approx 190*0.625* 0.0056377 \approx 0.066947=219∗20∗0.0625∗0.7518≈190∗0.625∗0.0056377≈0.066947
Answer b≈0.066947b \approx 0.066947b≈0.066947
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