Question

Consider four coins labelled as 1, 2, 3 and 4. Suppose that the probability of obtaining a 'head' in a single toss of the \(i\)th coin is \( \frac{1}{4} \), \( i = 1, 2, 3, 4 \). A coin is chosen uniformly at random and flipped. Given that the flip resulted in a 'head', the conditional probability that the coin was labelled either 1 or 2 equals

• A. \( \frac{1}{10} \)
• B. \( \frac{2}{10} \)
• C. \( \frac{3}{10} \)
• D. \( \frac{4}{10} \)

Hint:

When calculating conditional probabilities, use the formula \( P(A | B) = \frac{P(A \cap B)}{P(B)} \) and be sure to account for all possible outcomes and events involved.

Answer 1

The Correct Option is C

Let the coin chosen be \(C_i\) with
\[P(C_i)=\frac14,\qquad P(H\mid C_i)=\frac{i}{4},\quad i=1,2,3,4.\]

Step 1: Compute \(P(H)\)

\[P(H)=\sum_{i=1}^4 P(C_i),P(H\mid C_i) =\frac14\left(\frac14+\frac24+\frac34+\frac44\right) =\frac14\cdot\frac{10}{4} =\frac{10}{16}.\]

Step 2: Compute \(P(C_1\text{ or }C_2\mid H)\)

\[P(C_1\text{ or }C_2\mid H) =\frac{P(H\mid C_1)P(C_1)+P(H\mid C_2)P(C_2)}{P(H)}.\]

\[=\frac{\frac14\cdot\frac14+\frac24\cdot\frac14}{\frac{10}{16}} =\frac{\frac{3}{16}}{\frac{10}{16}} =\frac{3}{10}.\]

\[\boxed{\frac{3}{10}}\]

Hence, the correct answer is option (C).