Question #277620

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?


Expert's answer

Let C1C_1 and C2C_2 be the events that the influenza vaccine is manufactured by company 1 and 2 respectively. Also, let EE be the event that the influenza vaccine produced is effective. 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 as stated above.


ii)ii)

We need to 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)

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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