Question

The mean of the number of heads in a simultaneous toss of three coins is :

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

Answer 1

The Correct Option is C

To determine the mean number of heads when three coins are tossed simultaneously, we consider the possible outcomes and their respective probabilities.

The possible outcomes of tossing three coins are as follows, where H represents heads and T represents tails:

• HHH
• HHT
• HTH
• THH
• HTT
• THT
• TTH
• TTT

The probability of each outcome is \(\frac{1}{8}\) since there are \(2^3 = 8\) possible outcomes.

Let's determine the number of heads in each outcome:

• HHH: 3 heads
• HHT: 2 heads
• HTH: 2 heads
• THH: 2 heads
• HTT: 1 head
• THT: 1 head
• TTH: 1 head
• TTT: 0 heads

To find the mean, compute the expected value of the number of heads. The formula for the mean (\( \mu \)) is given by:

\[ \mu = \sum (x \times P(x)) \]

Number of Heads (x)	Probability P(x)
3	\(\frac{1}{8}\)
2	\(\frac{3}{8}\)
1	\(\frac{3}{8}\)
0	\(\frac{1}{8}\)

Calculate the mean:

\[ \mu = (3 \times \frac{1}{8}) + (2 \times \frac{3}{8}) + (1 \times \frac{3}{8}) + (0 \times \frac{1}{8}) \]

\[ \mu = \frac{3}{8} + \frac{6}{8} + \frac{3}{8} + 0 \]

\[ \mu = \frac{12}{8} = \frac{3}{2} \]

Therefore, the mean number of heads is \(\frac{3}{2}\).