Question

It is known that 60% of mice inoculated with a
serum are protected from a certain disease. If 5 mice
are inoculated, find the probability that
(a) none contracts the disease;
(b) fewer than 2 contract the disease;
(c) more than 3 contract the disease.

Answer 1

Answer on Question #72631, Math / Statistics and Probability

It is known that 60% of mice inoculated with a serum are protected from a certain disease. If 5 mice are inoculated, find the probability that

(a) none contracts the disease;

(b) fewer than 2 contract the disease;

(c) more than 3 contract the disease.

Solution

Let $X$ be the random variable representing the number of mice that contract the disease out of 5 mice inoculated with a serum.

If a mouse contracts the disease, we consider that as a success.

If a mouse does not contract the disease, will be a failure.

Then $q = 0.6$, $p = 1 - q = 0.4$. Trials are independent.

Hence, $X \sim \text{Bin}(n, p)$, where $n = 5$ and $p = 0.4$.

$$P(X = x) = b((x; 5, 0.4) = \binom{5}{x} (0.4)^x (0.6)^{5 - x}, \text{ where } x = 0, 1, 2, 3, 4, 5$$

(a) none contracts the disease

$$P(X = 0) = \binom{5}{0} (0.4)^0 (0.6)^{5 - 0} = (0.6)^5 = 0.07776$$

(b) fewer than 2 contract the disease

$$\begin{array}{l} P(X < 2) = P(X \leq 1) = P(X = 0) + P(X = 1) = \\ = 0.07776 + \binom{5}{1} (0.4)^1 (0.6)^{5 - 1} = 0.07776 + 5(0.4)(0.6)^4 = 0.33696 \\ \end{array}$$

(c) more than 3 contract the disease

$$\begin{array}{l} P(X > 3) = 1 - P(X \leq 3) = \\ = 1 - \left(P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)\right) = \\ = 1 - (0.33696 + \binom{5}{2} (0.4)^2 (0.6)^{5 - 2} + \binom{5}{3} (0.4)^3 (0.6)^{5 - 3}) = \\ = 1 - (0.33696 + \frac{5!}{2! (5 - 2)!} (0.4)^2 (0.6)^3 + \frac{5!}{3! (5 - 3)!} (0.4)^3 (0.6)^{5 - 3}) = \\ = 1 - 0.91296 = 0.08704 \\ \end{array}$$

Or

$$\begin{array}{l} P(X > 3) = P(X \geq 4) = P(X = 4) + P(X = 5) = \\ = \binom{5}{4} (0.4)^4 (0.6)^{5 - 4} + \binom{5}{5} (0.4)^5 (0.6)^{5 - 5} = \\ = 5(0.4)^4 (0.6) + (0.4)^5 = 0.08704 \\ \end{array}$$

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