Answer to Question #210626 in Statistics and Probability for Mark

Question

Let X be a binomial random variable with p = 0.1 and n = 10. Calculate the following probabilities.

(a) P(X ≤ 2) 

(b) P(X > 8)

(c) P(X = 4) 

(d) P(5 ≤ X ≤ 7)

Answer 1

(a)

$n=10, p=0.1, q=1-p=1-0.1=0.9$

$$P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)$$

$$=\dbinom{10}{0}(0.1)^0(0.9)^{10-0}+\dbinom{10}{1}(0.1)^1(0.9)^{10-1}$$

$$+\dbinom{10}{2}(0.1)^2(0.9)^{10-2}=0.3486784401$$

$$+0.387420489+0.1937102445$$

$$=0.9298091736$$

(b)

$n=10, p=0.1, q=1-p=1-0.1=0.9$

$$P(X>8)=P(X=9)+P(X=10)$$

$$=\dbinom{10}{9}(0.1)^9(0.9)^{10-9}+\dbinom{10}{10}(0.1)^{10}(0.9)^{10-10}$$

$$=0.000000009+0.0000000001$$

$$=0.0000000091$$

(c)

$n=10, p=0.1, q=1-p=1-0.1=0.9$

$$P(X=4)=\dbinom{10}{4}(0.1)^4(0.9)^{10-4}=0.011160261$$

(d)

$n=10, p=0.1, q=1-p=1-0.1=0.9$

$$P(5\leq X\leq 7)=P(X=5)+P(X=6)+P(X=7)$$

$$=\dbinom{10}{5}(0.1)^5(0.9)^{10-5}+\dbinom{10}{6}(0.1)^6(0.9)^{10-6}$$

$$+\dbinom{10}{7}(0.1)^7(0.9)^{10-7}=0.0014880348$$

$$+0.000137781+0.000008748$$

$$=0.0016345638$$