Question

When an experiment is conducted, one and only one of three mutually exclusive events S1, S2, and S3 can occur, with
P(S1) = 0.3,

P(S2) = 0.5,
and
P(S3) = 0.2.
The probabilities that a fourth event A occurs, given that event S1, S2, or S3 has occurred, are
P(A|S1) = 0.3 P(A|S2) = 0.1 P(A|S3) = 0.2.
If event A is observed, find
P(S3|A).
(Round your answer to four decimal places.)
P(S3|A) =

Answer 1

Answer:

$P(S_3|A) = \frac{2}{9}$

Explanation:

Given

$P(S1) = 0.3$           $P(S_2) = 0.5$      $P(S_3) = 0.2$

$P(A|S_1) = 0.3$       $P(A|S_2) = 0.1$     $P(A|S_3) = 0.2$

Required

Determine $P(S_3|A)$

Using Bayes' theorem:

$P(S_3|A) = \frac{P(S_3) * P(A|S_3)}{P(S_1) * P(A|S_1) + P(S_2) * P(A|S_2) + P(S_3) * P(A|S_3)}$

So, we have:

$P(S_3|A) = \frac{0.2 * 0.2}{0.3 * 0.3+ 0.5* 0.1 + 0.2 * 0.2}$

$P(S_3|A) = \frac{0.04}{0.18}$

Multiply by 100/100

$P(S_3|A) = \frac{4}{18}$

$P(S_3|A) = \frac{2}{9}$