(a) What is the probability that at most 4 of the calls involve a fax message? (b) What is the probability that exactly 4 of the calls involve a fax message? (c) What is the probability that at least 4 of the calls involve a fax message? (d) What is the probability that more than 4 of the calls involve a fax message?
Answer :
Given :
The probability of success $(p) = 0.20$
Sample size $(n) = 20$
Solution :
A) The probability that at most 4 of the calls involve a fax message :
$$\text{P}(x \leq 4) = \sum_{0}^{4} \binom{n}{x} p^x (1-p)^{n-x}$$ $$= \sum_{0}^{4} \binom{20}{x} 0.2^x (1-0.2)^{20-x}$$ $$= 0.630$$
B) The probability that exactly 4 of the calls involve a fax message :
$$\text{P}(x = 4) = \binom{n}{x} p^x (1-p)^{n-x}\bigg|_{x=4}$$ $$= \binom{20}{4} 0.2^4 (1-0.2)^{20-4}$$ $$= 0.218$$
C) The probability that at least 4 of the calls involve a fax message :
$$\text{P}(x \geq 4) = 1 – \text{P}(x \leq 3)$$ $$= 1 – \sum_{x=0}^{3} \binom{n}{x} p^x (1-p)^{n-x}$$ $$= 1 – \sum_{x=0}^{3} \binom{20}{x} 0.2^x (1-0.2)^{20-x}$$ $$= 1 – 0.4114$$ $$= 0.589$$
D) The probability that more than 4 of the calls involve a fax message :
$$\text{P}(x > 4) = 1 – \text{P}(x \leq 4)$$ $$= 1 – \sum_{x=0}^{4} \binom{n}{x} p^x (1-p)^{n-x}$$ $$= 1 – \sum_{x=0}^{4} \binom{20}{x} 0.2^x (1-0.2)^{20-x}$$ $$= 1 – 0.6296$$ $$= 0.370$$
