It is known that 60% of mice inoculated with serum are protected from a certain disease. If 5 mice are inoculated, find the probability that
i. none contract the disease
ii. more than 3 contract the disease.
Given that, the probability that the mice are protected from a certain disease is 60%.
Also, the sample size of mice = 5
Let p be the probability of the mice not contracting with a certain disease.
Then, q is the probability of the mice contracting with a certain disease.
⇒ p = 60%
And we know that,
p + q = 1
⇒ q = 1 – p
Let X be a random variable representing a number of mice contracting with the disease.
Then, the probability of r mice contracting with the disease out of n mice inoculated is given by the following binomial distribution.
P (X = r) = nCrqrpn-r
Putting the values,
n = 5,
(i). We need to find the probability that none of the mice contract with the disease.
For this, put r = 0.
We get the probability as,
Probability = P (X = 0)
From equation (A),
⇒ P (X = 0) = 0.07776
∴, the probability that none of the mice contracts with the disease is 0.07776.
(ii). We need to find the probability that more than 3 mice contract the disease.
So, probability = P (X = 4) + P (X = 5)
⇒ Probability = 0.2125
∴, the probability that more than 3 mice contract the disease is 0.2125.