Answer :

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,



&


We get


…(A)


(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.


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