Question

Three companies C1, C2, C3 produce car tyres. A car manufacturing company buys 40% of its requirement from C1, 35% from C2 and 25% from C3. The company knows that 2% of the tyres supplied by C1, 3% by C2 and 4% by C3 are defective. If a tyre chosen at random from the consignment received is found defective then the probability that it was supplied by C2 is

• A. $7/19$
• B. $12/19$
• C. $10/57$
• D. $26/57$

Hint:

Bayes' theorem problems typically involve finding a "reverse" conditional probability. The structure is always: identify the initial probabilities (priors), the conditional probabilities (likelihoods), calculate the total probability of the event, and then apply the theorem.

Answer 1

The Correct Option is A

Let $C_1, C_2, C_3$ be the events that a tyre is from company C1, C2, and C3, respectively.
Let $D$ be the event that a chosen tyre is defective.
We are given the following probabilities:
$P(C_1) = 0.40$
$P(C_2) = 0.35$
$P(C_3) = 0.25$
We are also given the conditional probabilities of a tyre being defective, given the company:
$P(D|C_1) = 0.02$
$P(D|C_2) = 0.03$
$P(D|C_3) = 0.04$
We need to find the probability that a defective tyre was supplied by C2, which is $P(C_2|D)$.
We use Bayes' theorem: $P(C_2|D) = \frac{P(D|C_2)P(C_2)}{P(D)}$.
First, we calculate the total probability of a tyre being defective, $P(D)$, using the law of total probability:
$P(D) = P(D|C_1)P(C_1) + P(D|C_2)P(C_2) + P(D|C_3)P(C_3)$.
$P(D) = (0.02)(0.40) + (0.03)(0.35) + (0.04)(0.25)$.
$P(D) = 0.0080 + 0.0105 + 0.0100 = 0.0285$.
Now we can calculate $P(C_2|D)$:
$P(C_2|D) = \frac{(0.03)(0.35)}{0.0285} = \frac{0.0105}{0.0285}$.
To simplify the fraction, multiply the numerator and denominator by 10000:
$P(C_2|D) = \frac{105}{285}$.
Divide both by 15: $P(C_2|D) = \frac{105 \div 15}{285 \div 15} = \frac{7}{19}$.