Mathematics - Probability Question with Solution | TestHub

Let $E_1$, $E_2$, and $E_3$ be the possible events for the composition of balls in the bag (total 4 balls), given that at least two balls drawn are white. The possibilities are:

* $E_1$: 2 white balls and 2 non-white (black) balls.

* $E_2$: 3 white balls and 1 non-white ball.

* $E_3$: 4 white balls and 0 non-white balls.

Assuming each composition is equally likely, the prior probabilities are

$$P(E_1)=P(E_2)=P(E_3)=\frac{1}{3}.$$

Let $A$ be the event that two balls drawn are white.

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Step 2: Calculate conditional probabilities

We calculate the probability of drawing two white balls given each event using combinations, where $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is the number of ways to choose $k$ items from $n$:

* $P(A|E_1)=\frac{\left(\genfrac{}{}{0ex}{}{2}{2}\right)}{\left(\genfrac{}{}{0ex}{}{4}{2}\right)}=\frac{1}{6}$.

* $P(A|E_2)=\frac{\left(\genfrac{}{}{0ex}{}{3}{2}\right)}{\left(\genfrac{}{}{0ex}{}{4}{2}\right)}=\frac{3}{6}=\frac{1}{2}$.

* $P(A|E_3)=\frac{\left(\genfrac{}{}{0ex}{}{4}{2}\right)}{\left(\genfrac{}{}{0ex}{}{4}{2}\right)}=1$.

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Step 3: Apply Bayes' theorem and simplify

Using Bayes' theorem to find the posterior probability $P(E_3|A)$:

$$P(E_3|A)=\frac{P(A|E_3)\cdot P(E_3)}{P(A|E_1)\cdot P(E_1)+P(A|E_2)\cdot P(E_2)+P(A|E_3)\cdot P(E_3)}$$

Substitute the calculated values into the formula:

$$P(E_3|A)=\frac{1\cdot \frac{1}{3}}{\frac{1}{6}\cdot \frac{1}{3}+\frac{1}{2}\cdot \frac{1}{3}+1\cdot \frac{1}{3}}=\frac{\frac{1}{3}}{\frac{1}{18}+\frac{3}{18}+\frac{6}{18}}=\frac{\frac{1}{3}}{\frac{10}{18}}=\frac{\frac{1}{3}}{\frac{5}{9}}=\frac{1}{3}\cdot \frac{9}{5}=\frac{3}{5}$$