Respuesta :
Answer:
The conditional probability of that shipment’s having come from C given that the hospital finds at least one of the five vials ineffective is 0.1778.
Step-by-step explanation:
To find the conditional probability of that shipment’s having come from C you have to use the Bayes Theorem:
[tex]P[A_n|B]=\frac{P(B|A_n)P(A_n)}{Sum[P(A_i|B)P(A_i)]}[/tex]
In this case, you can describe the probabilities as follows:
P(A), P(B), P(C) the probability that the shipment is from A, B or C respectively
P(D) the probability that at least one of the vials is ineffective.
From the problem, you know that
P(A)=0.4
P(B)=0.5
P(C)=0.1
This kind of problem follows a binomial distribution given by the formula:
[tex]P(X=k)=\left(\begin{array}{c}n\\k\\\end{array}\right) p^k(1-p)^{n-k}[/tex]
Now, to find the binomial probability of finding an ineffective vial in each vaccine from each company you apply this formula. The easiest way is to find the probability that none of the vials is defective and use the complement property. In this case n= 5, k = 0 and p is the probability of each Company having an ineffective vaccine.
For Company A, p=0.03
[tex]P(X=0)=\left(\begin{array}{c}5\\0\\\end{array}\right) 0.03^0(1-0.03)^{5-0}\\P(X=0)=0.8587[/tex]
Using the complement rule, the probability of finding at leats one vial ineffective is 1-0.8587=0.1413
For Company B, p=0.02
[tex]P(X=0)=\left(\begin{array}{c}5\\0\\\end{array}\right) 0.02^0(1-0.02)^{5-0}\\P(X=0)=0.9039[/tex]
Using the complement rule, the probability of finding at leats one vial ineffective is 1-0.9039=0.0961
For Company C, p=0.05
[tex]P(X=0)=\left(\begin{array}{c}5\\0\\\end{array}\right) 0.05^0(1-0.05)^{5-0}\\P(X=0)=0.7738[/tex]
Using the complement rule, the probability of finding at least one vial ineffective is 1-0.7738=0.2262
Now using the Bayes Theorem to find the conditional probability that the vaccine come from Company C given that one of the 5 vials reviewed by the company is ineffective:
[tex]P(C|D)=\frac{P(C)P(D|C)}{P(A)P(D|A)P(B)P(D|B)P(C)P(D|C)}\\P(C|D)=\frac{0.1(0.2262)}{0.4(0.1413)+0.5(0.0961)+0.1(0.2262)}=0.1778[/tex]