Probability that A speaks truth is `4/5` . A coin is tossed A reports that a head appears. The probability that actually there was head is

Options

A. `4/5`
B. `3/5`
C. `2/5`
D. `1/2`

Answer

To solve the problem, we will use Bayes' theorem. Let's break down the solution step by step.

### Step 1: Define the Events
Let:
- \( E_1 \): Event that A speaks the truth.
- \( E_2 \): Event that A speaks falsely.
- \( X \): Event that a head appears when the coin is tossed.

### Step 2: Identify the Given Probabilities
From the problem statement, we have:
- \( P(E_1) = \frac{4}{5} \) (Probability that A speaks the truth)
- \( P(E_2) = 1 - P(E_1) = 1 - \frac{4}{5} = \frac{1}{5} \) (Probability that A speaks falsely)

### Step 3: Determine the Probability of Getting Heads
When the coin is tossed, the probability of getting heads, regardless of whether A speaks the truth or not, is:
- \( P(X | E_1) = \frac{1}{2} \) (Probability of heads given A speaks the truth)
- \( P(X | E_2) = \frac{1}{2} \) (Probability of heads given A speaks falsely)

### Step 4: Apply Bayes' Theorem
We want to find the probability that there was actually a head given that A reports a head, which can be expressed as:
\[ P(E_1 | X) = \frac{P(E_1) \cdot P(X | E_1)}{P(E_1) \cdot P(X | E_1) + P(E_2) \cdot P(X | E_2)} \]

### Step 5: Substitute the Values
Now, substituting the known values into the equation:
\[ P(E_1 | X) = \frac{\left(\frac{4}{5}\right) \cdot \left(\frac{1}{2}\right)}{\left(\frac{4}{5} \cdot \frac{1}{2}\right) + \left(\frac{1}{5} \cdot \frac{1}{2}\right)} \]

### Step 6: Simplify the Expression
Calculating the numerator:
\[ \text{Numerator} = \frac{4}{5} \cdot \frac{1}{2} = \frac{4}{10} \]

Calculating the denominator:
\[ \text{Denominator} = \left(\frac{4}{5} \cdot \frac{1}{2}\right) + \left(\frac{1}{5} \cdot \frac{1}{2}\right) = \frac{4}{10} + \frac{1}{10} = \frac{5}{10} \]

### Step 7: Final Calculation
Now substituting back into the equation:
\[ P(E_1 | X) = \frac{\frac{4}{10}}{\frac{5}{10}} = \frac{4}{5} \]

### Conclusion
Thus, the probability that actually there was a head when A reports a head is:
\[ \boxed{\frac{4}{5}} \]