Question

A boy tosses fair coin 3 times. If he gets 2X for X heads, then his expected gain equal to

• A. 1
• B. 3/2
• C. 3
• D. 4

Answer 1

Answer: (C)

3

Answer 2

To determine the expected gain, we first need to identify the possible number of heads and their respective probabilities when a fair coin is tossed 3 times.

Let X be the number of heads obtained in 3 tosses. Since the coin is fair, the probability of getting a head (H) is $P(H) = 1/2$, and the probability of getting a tail (T) is $P(T) = 1/2$. The number of tosses is $n=3$. The number of heads X can be 0, 1, 2, or 3.

The probability of getting k heads in n tosses of a fair coin is given by the binomial probability formula: $P(X=k) = C(n, k) * (1/2)^k * (1/2)^(n-k) = C(n, k) * (1/2)^n$. In this case, $P(X=k) = C(3, k) * (1/2)^3 = C(3, k) / 8$.

Let's calculate the probability for each possible number of heads:

1. X = 0 (0 heads): $P(X=0) = C(3, 0) * (1/8) = 1 * (1/8) = 1/8$
2. X = 1 (1 head): $P(X=1) = C(3, 1) * (1/8) = 3 * (1/8) = 3/8$
3. X = 2 (2 heads): $P(X=2) = C(3, 2) * (1/8) = 3 * (1/8) = 3/8$
4. X = 3 (3 heads): $P(X=3) = C(3, 3) * (1/8) = 1 * (1/8) = 1/8$

The problem states that the boy gets 2X for X heads. This means the gain for each outcome is:

• If X = 0, Gain = $2 * 0 = 0$
• If X = 1, Gain = $2 * 1 = 2$
• If X = 2, Gain = $2 * 2 = 4$
• If X = 3, Gain = $2 * 3 = 6$

The expected gain is calculated as the sum of (gain * probability) for all possible outcomes: $E[\text{Gain}] = \sum_{k=0}^{3} (\text{Gain for X=k}) * P(X=k)$ $E[\text{Gain}] = (0 * P(X=0)) + (2 * P(X=1)) + (4 * P(X=2)) + (6 * P(X=3))$ $E[\text{Gain}] = (0 * 1/8) + (2 * 3/8) + (4 * 3/8) + (6 * 1/8)$ $E[\text{Gain}] = 0 + 6/8 + 12/8 + 6/8$ $E[\text{Gain}] = (0 + 6 + 12 + 6) / 8$ $E[\text{Gain}] = 24 / 8$ $E[\text{Gain}] = 3$

Alternatively, using the linearity of expectation: The expected number of heads for n tosses of a fair coin is $E[X] = n * p$, where p is the probability of getting a head. Here, $n=3$ and $p=1/2$. So, $E[X] = 3 * (1/2) = 3/2$. The gain is given by $G(X) = 2X$. By linearity of expectation, $E[G(X)] = E[2X] = 2 * E[X]$. $E[\text{Gain}] = 2 * (3/2) = 3$.

Both methods yield the same result.