Questions: Fill in the probability distribution table for the number of heads that occur when tossing three coins simultaneously. Event Probability 3 heads, 0 tails 2 heads, 1 tail 1 head, 2 tails 0 heads, 3 tails

Solution Steps

We are tasked with finding the probability distribution for the number of heads when tossing three coins simultaneously. The possible outcomes for the number of heads \( X \) can be 0, 1, 2, or 3.

Using the binomial probability formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

where:

• \( n = 3 \) (the number of trials),
• \( p = 0.5 \) (the probability of getting heads),
• \( q = 1 - p = 0.5 \) (the probability of getting tails).

We calculate the probabilities for each event:

1. For 3 heads, 0 tails (\( x = 3 \)): \[ P(X = 3) = \binom{3}{3} \cdot (0.5)^3 \cdot (0.5)^{0} = 1 \cdot 0.125 \cdot 1 = 0.125 \]

2. For 2 heads, 1 tail (\( x = 2 \)): \[ P(X = 2) = \binom{3}{2} \cdot (0.5)^2 \cdot (0.5)^{1} = 3 \cdot 0.25 \cdot 0.5 = 0.375 \]

3. For 1 head, 2 tails (\( x = 1 \)): \[ P(X = 1) = \binom{3}{1} \cdot (0.5)^1 \cdot (0.5)^{2} = 3 \cdot 0.5 \cdot 0.25 = 0.375 \]

4. For 0 heads, 3 tails (\( x = 0 \)): \[ P(X = 0) = \binom{3}{0} \cdot (0.5)^{0} \cdot (0.5)^{3} = 1 \cdot 1 \cdot 0.125 = 0.125 \]

The probabilities calculated for each event are summarized in the following table:

\[ \begin{array}{|c|c|} \hline \text{Event} & \text{Probability} \\ \hline 3 \text{ heads, } 0 \text{ tails} & 0.125 \\ 2 \text{ heads, } 1 \text{ tail} & 0.375 \\ 1 \text{ head, } 2 \text{ tails} & 0.375 \\ 0 \text{ heads, } 3 \text{ tails} & 0.125 \\ \hline \end{array} \]

Final Answer