# Solved:The 29% of the loans given by a bank are high-cost loans. If 5 cases of

#### ByDr. Raju Chaudhari

Sep 26, 2020

The 29% of the loans given by a bank are high-cost loans. If 5 cases of loans were selected, what is the probability that

i) none are high-cost loans?
ii) all 5 are high-cost loans?
iii) at least 2 are high-cost loans?

#### Solution

Let $X$ denote the number of high-cost loans out of 5 cases.

The probability of high-cost loans is $p = 0.29$.

The random variable $X$ follows Binomial distribution with parameters $n = 5$ and $p=0.29$.

Thus $X\sim B(5, 0.29)$.

The probability mass function of $X$ is
$$P(X=x) = \binom{5}{x} (0.29)^x (1-0.29)^{5-x}, \; x=0,1,\cdots, 5.$$

i. The probability that none are high-cost loans is

 \begin{aligned} P(X=0) &=\binom{5}{0} (0.29)^{0} (1-0.29)^{5-0}\\ &= 0.1804 \end{aligned}

ii. The probability that all 5 are high-cost loans is

 \begin{aligned} P(X= 5) &= \binom{5}{5} (0.29)^{5} (1-0.29)^{5-5}\\ &=0.0021 \end{aligned}

iii. The probability of at least 2 are high-cost loans is

 \begin{aligned} P(X\geq 2) &= 1- P(X\leq 1)\\ &=1- \big(P(X=0)+P(X=1)\big)\\ &=1- \big(\binom{5}{0} (0.29)^{0} (1-0.29)^{5-0}+\binom{5}{1} (0.29)^{1} (1-0.29)^{5-1}\big)\\ &=1-(0.1804+0.3685)\\ &=0.4511 \end{aligned}