Question #278746

An influenza vaccine is produced by two different companies. It is known that a vaccine produced by company 1 is effective with probability 0.89, while a vaccine produced by company 2 is effective with probability 0.93. We also know that company 1 supplies 40% of the vaccines, while company 2 supplies 60% of the vaccines ordered by the government.

(i) What is the probability that a vaccine is effective, given that it was produced by company 2?

(ii) What is the probability that a randomly chosen vaccine from the government’s order is not effective?

(iii) What is the probability that given a vaccine is not effective that it was produced by company 1?


1
Expert's answer
2021-12-13T14:35:44-0500

Define the following events,

Let C1C_1 and C2C_2 be the events that the influenza vaccine is manufactured by company 1 and company 2 respectively. Also, let EE be the event that the influenza vaccine produced is effective and EE' is the event that the influenza vaccine produced is not effective.

The following probabilities are given,

p(EC1)=0.89, p(EC2)=0.93, p(C1)=0.40, p(C2)=0.60p(E|C_1)=0.89,\space p(E|C_2)=0.93,\space p(C_1)=0.40,\space p(C_2)=0.60


i)i)

The probability that a vaccine is effective, given that it was produced by company 2 is given as,

p(EC2)=0.93p(E|C_2)=0.93.


ii)ii)

We determine the probability that a randomly selected vaccine is effective. To do so, we shall apply the law of total probability as follows,

p(E)=p(EC1)p(C1)+p(EC2)p(C2)=0.890.40+0.930.60=0.356+0.558=0.914p(E)=p(E|C_1)*p(C_1)+p(E|C_2)*p(C_2)=0.89*0.40+0.93*0.60=0.356+0.558=0.914

The probability that a randomly selected vaccine is not effective is given as,

p(E)=1p(E)=10.914=0.086p(E')=1-p(E)=1-0.914=0.086

Therefore, the probability that a randomly selected vaccine is not effective is 0.086.


iii)iii)

Here, we determine the conditional probability, p(C1E)p(C_1|E') defined as,

p(C1E)=p(C1E)p(E)p(C_1|E')={p(C_1\cap E')\over p(E')}

P(C1E)=0.4×0.11=0.044 and p(E)=0.086P(C_1\cap E')=0.4\times0.11=0.044\space and \space p(E')=0.086

Thus, p(C1E)=0.0440.086=0.5116(4dp)p(C_1|E')={0.044\over 0.086}=0.5116(4dp)

Therefore, the probability that a randomly selected vaccine is produced by company 1 given it is not effective is 0.5116.


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