Question

1. a hospital receives \\(1/5\\) of its flu vaccine shipments from company x and the remainder of its shipments from other companies. each shipment contains a very large number of vaccine vials.

for company xs shipments, \\(10\\%\\) of the vials are ineffective. for every other company, \\(2\\%\\) of the vials are ineffective. the hospital tests 30 randomly selected vials from a shipment and finds that one vial is ineffective.

calculate the probability that this shipment came from company x.

(a) 0.10
(b) 0.14
(c) 0.37
(d) 0.63
(e) 0.86

Explanation:

⚡ Using what you learned: Conditional Probability · 🆕 New: Bayes' Theorem
Updating beliefs when new evidence is observed

Step 1: Define the Events and Prior Probabilities

Let \( X \) be the event that the shipment comes from Company X, and \( X^c \) be the event that it comes from another company.

The prior probabilities are:

$$ P(X) = \frac{1}{5} = 0.20 $$

$$ P(X^c) = 1 - 0.20 = 0.80 $$

Step 2: Define the Likelihoods using Binomial Distribution

Let \( E \) be the event that exactly 1 vial out of 30 randomly selected vials is ineffective. Since the shipments contain a very large number of vials, we model the number of ineffective vials using a binomial distribution with \( n = 30 \).

For Company X, the probability of a vial being ineffective is \( p_X = 0.10 \):

$$ P(E \mid X) = \binom{30}{1} (0.10)^1 (0.90)^{29} = 30 \times 0.10 \times (0.90)^{29} $$

For other companies, the probability of a vial being ineffective is \( p_O = 0.02 \):

$$ P(E \mid X^c) = \binom{30}{1} (0.02)^1 (0.98)^{29} = 30 \times 0.02 \times (0.98)^{29} $$

Step 3: Calculate the Likelihood Values

Calculate the numerical values for both likelihoods:

$$ P(E \mid X) = 3 \times (0.90)^{29} \approx 3 \times 0.04710 = 0.14130 $$

$$ P(E \mid X^c) = 0.6 \times (0.98)^{29} \approx 0.6 \times 0.55487 = 0.33292 $$

Step 4: Apply Bayes' Theorem

We want to find the posterior probability \( P(X \mid E) \):

$$ P(X \mid E) = \frac{P(E \mid X) P(X)}{P(E \mid X) P(X) + P(E \mid X^c) P(X^c)} $$

Substitute the values into the formula:

$$ P(X \mid E) = \frac{0.14130 \times 0.20}{(0.14130 \times 0.20) + (0.33292 \times 0.80)} $$

$$ P(X \mid E) = \frac{0.02826}{0.02826 + 0.26634} $$

$$ P(X \mid E) = \frac{0.02826}{0.29460} \approx 0.0959 $$

Rounding to two decimal places gives \( 0.10 \).

Answer:

(A) 0.10