A box contains 3 coins `B_(1), B_(2),B_(3)` and the probability of getting heads on the coins are `(1)/(2), (1)/(4),(1)/(8)` respectively. If one of the coins is selected at random and tossed for 3 times and exactly 3 times and exactly 3 heads appeared, then the probability that it was coin `B_(1)` is

Options

A. `(9)/(73)`
B. `(10)/(73)`
C. `(36)/(73)`
D. `(64)/(73)`

Answer

To solve the problem step by step, we will use Bayes' theorem to find the probability that coin \( B_1 \) was selected given that we obtained exactly 3 heads after tossing the coin 3 times.

### Step 1: Define the Events
Let:
- \( A_1 \): The event that coin \( B_1 \) is selected.
- \( A_2 \): The event that coin \( B_2 \) is selected.
- \( A_3 \): The event that coin \( B_3 \) is selected.
- \( H \): The event of getting exactly 3 heads in 3 tosses.

### Step 2: Calculate Prior Probabilities
Since one of the coins is selected at random, the prior probabilities are:
\[ P(A_1) = P(A_2) = P(A_3) = \frac{1}{3} \]

### Step 3: Calculate the Probability of Getting 3 Heads for Each Coin
1. For coin \( B_1 \) (probability of heads = \( \frac{1}{2} \)):
\[ P(H | A_1) = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \]

2. For coin \( B_2 \) (probability of heads = \( \frac{1}{4} \)):
\[ P(H | A_2) = \left(\frac{1}{4}\right)^3 = \frac{1}{64} \]

3. For coin \( B_3 \) (probability of heads = \( \frac{1}{8} \)):
\[ P(H | A_3) = \left(\frac{1}{8}\right)^3 = \frac{1}{512} \]

### Step 4: Calculate the Total Probability of Getting 3 Heads
Using the law of total probability:
\[ P(H) = P(H | A_1) P(A_1) + P(H | A_2) P(A_2) + P(H | A_3) P(A_3) \]
Substituting the values:
\[ P(H) = \left(\frac{1}{8} \cdot \frac{1}{3}\right) + \left(\frac{1}{64} \cdot \frac{1}{3}\right) + \left(\frac{1}{512} \cdot \frac{1}{3}\right) \]
\[ P(H) = \frac{1}{24} + \frac{1}{192} + \frac{1}{1536} \]

To add these fractions, we find a common denominator, which is 1536:
\[ P(H) = \frac{64}{1536} + \frac{8}{1536} + \frac{1}{1536} = \frac{73}{1536} \]

### Step 5: Apply Bayes' Theorem
We want to find \( P(A_1 | H) \):
\[ P(A_1 | H) = \frac{P(H | A_1) P(A_1)}{P(H)} \]
Substituting the values:
\[ P(A_1 | H) = \frac{\left(\frac{1}{8}\right) \left(\frac{1}{3}\right)}{\frac{73}{1536}} \]
\[ P(A_1 | H) = \frac{\frac{1}{24}}{\frac{73}{1536}} = \frac{1536}{24 \cdot 73} = \frac{64}{73} \]

### Final Answer
Thus, the probability that the selected coin was \( B_1 \) given that exactly 3 heads appeared is:
\[ \frac{64}{73} \]