Answer to Question #255958 in Statistics and Probability for Sumit

Question

The latest nationwide political poll indicates that for Americans who are randomly selected, the probability that they are conservative is 0.55, the probability that they are liberal is 0.30, and the probability that they are middle – of –the road is 0.15. Assuming that theses probabilities are accurate, answer the following question pertaining to a randomly chosen group of 10 Americans.

Answer 1

a) What is the probability that four are liberal?

$n=10, p=0.3, q=1-p=1-0.3=0.7$

$$P(X=4)=\dbinom{10}{4}(0.3)^4(0.7)^{10-4}=$$

$$=0.200120949$$

b) What is the probability that none are conservative?

$n=10, p=0.55, q=1-p=1-0.55=0.45$

$$P(Y=0)=\dbinom{10}{0}(0.55)^0(0.45)^{10-0}=$$

$$\approx0.000340506289$$

c) What is the probability that two are middle – of –the road?

$n=10, p=0.15, q=1-p=1-0.15=0.85$

$$P(W=2)=\dbinom{10}{2}(0.15)^2(0.85)^{10-2}=$$

$$\approx0.2758966566$$

d) What is the probability that at least eight are liberal?

$n=10, p=0.3, q=1-p=1-0.3=0.7$

$$P(X\geq8)=P(X=8)+P(X=9)+P(X=10)$$

$$=\dbinom{10}{8}(0.3)^8(0.7)^{10-8}+\dbinom{10}{9}(0.3)^9(0.7)^{10-9}$$

$$+\dbinom{10}{10}(0.3)^10(0.7)^{10-10}=$$

$$=0.0015903864$$