Question

Find the mean number of heads in three tosses of fair coin:

• A. 1.5
• B. 2.5
• C. 4.5
• D. 3.5

Answer 1

The Correct Option is A

In three tosses of a fair coin, we have a total of \(2^3 = 8\) possible outcomes. 
These outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT 
Counting the number of heads in each outcome, we have: 
HHH: 3 heads 
HHT: 2 heads 
HTH: 2 heads 
HTT: 1 head 
THH: 2 heads 
THT: 1 head 
TTH: 1 head 
TTT: 0 heads 
To find the mean, we sum up the number of heads in each outcome and divide by the total number of outcomes: 
\(\frac{{3 + 2 + 2 + 1 + 2 + 1 + 1 + 0}}{8} = \frac{12}{8} = 1.5\)
 Therefore, the mean number of heads in three tosses of a fair coin is 1.5. The correct option is (1) 1.5.

Answer 2

When a fair coin is tossed 3 times, the number of heads \( X \) can be 0, 1, 2, or 3.

The probabilities are given by the binomial distribution: \[ P(X = r) = \binom{3}{r} \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{3-r} = \binom{3}{r} \left( \frac{1}{2} \right)^3 = \binom{3}{r} \cdot \frac{1}{8} \] So, the probabilities are:
\[ P(0) = \binom{3}{0} \cdot \frac{1}{8} = 1 \cdot \frac{1}{8} = \frac{1}{8} \] \[ P(1) = \binom{3}{1} \cdot \frac{1}{8} = 3 \cdot \frac{1}{8} = \frac{3}{8} \] \[ P(2) = \binom{3}{2} \cdot \frac{1}{8} = 3 \cdot \frac{1}{8} = \frac{3}{8} \] \[ P(3) = \binom{3}{3} \cdot \frac{1}{8} = 1 \cdot \frac{1}{8} = \frac{1}{8} \] Now compute the expected value (mean): \[ E(X) = 0 \cdot \frac{1}{8} + 1 \cdot \frac{3}{8} + 2 \cdot \frac{3}{8} + 3 \cdot \frac{1}{8} \] \[ = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = 1.5 \] Correct Answer: 1.5

Answer 3

Let \(X\) be the random variable representing the number of heads obtained in three tosses of a fair coin.

The possible values for \(X\) are 0, 1, 2, 3. 

Since the coin is fair, the probability of getting a head (H) in a single toss is \(P(H) = p = 0.5\), and the probability of getting a tail (T) is \(P(T) = q = 1 - p = 0.5\).

This is a binomial distribution with the number of trials \(n=3\) and probability of success (getting a head) \(p=0.5\).

The probability distribution of \(X\) is calculated as follows using the binomial probability formula \(P(X=k) = \binom{n}{k} p^k q^{n-k}\):

• P(X=0) (0 Heads: TTT): \[P(X=0) = \binom{3}{0} (0.5)^0 (0.5)^{3-0} = 1 \times 1 \times (0.5)^3 = 0.125 = \frac{1}{8}\]
• P(X=1) (1 Head: HTT, THT, TTH): \[P(X=1) = \binom{3}{1} (0.5)^1 (0.5)^{3-1} = 3 \times (0.5) \times (0.5)^2 = 3 \times (0.5)^3 = 0.375 = \frac{3}{8}\]
• P(X=2) (2 Heads: HHT, HTH, THH): \[P(X=2) = \binom{3}{2} (0.5)^2 (0.5)^{3-2} = 3 \times (0.5)^2 \times (0.5) = 3 \times (0.5)^3 = 0.375 = \frac{3}{8}\]
• P(X=3) (3 Heads: HHH): \[P(X=3) = \binom{3}{3} (0.5)^3 (0.5)^{3-3} = 1 \times (0.5)^3 \times 1 = (0.5)^3 = 0.125 = \frac{1}{8}\]

The mean (or expected value) of a discrete random variable \(X\) is given by the formula:

\[ \mathbf{E(X) = \sum_{i} x_i P(X=x_i)} \]

So, the mean number of heads is:

\[ E(X) = (0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2)) + (3 \times P(X=3)) \]

\[ E(X) = \left(0 \times \frac{1}{8}\right) + \left(1 \times \frac{3}{8}\right) + \left(2 \times \frac{3}{8}\right) + \left(3 \times \frac{1}{8}\right) \]

\[ E(X) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} \]

\[ E(X) = \frac{3 + 6 + 3}{8} = \frac{12}{8} = \frac{3}{2} = \mathbf{1.5} \]

Alternatively, for a binomial distribution \(B(n, p)\), the mean is given directly by the formula \(E(X) = np\).

\[ E(X) = 3 \times 0.5 = 1.5 \]

Comparing this with the given options, the correct option is:

1.5