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, we will use Bayes' Theorem. We need to find the probability that coin \( B_1 \) was selected given that we observed 3 heads in 3 tosses.

### Step 1: Define the events
Let:
- \( A \) be the event that we selected coin \( B_1 \).
- \( B \) be the event that we got 3 heads in 3 tosses.

### Step 2: Find the prior probabilities
The probability of selecting each coin is:
- \( P(A) = P(B_1) = \frac{1}{3} \)
- \( P(B_2) = \frac{1}{3} \)
- \( P(B_3) = \frac{1}{3} \)

### Step 3: Calculate the probability of getting 3 heads for each coin
- For coin \( B_1 \): The probability of heads is \( \frac{1}{2} \).
\[ P(B | A) = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \]

- For coin \( B_2 \): The probability of heads is \( \frac{1}{4} \).
\[ P(B | B_2) = \left(\frac{1}{4}\right)^3 = \frac{1}{64} \]

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

### Step 4: Calculate the total probability of getting 3 heads
Using the law of total probability:
\[ P(B) = P(B | A) P(A) + P(B | B_2) P(B_2) + P(B | B_3) P(B_3) \]
Substituting the values:
\[ P(B) = \left(\frac{1}{8} \cdot \frac{1}{3}\right) + \left(\frac{1}{64} \cdot \frac{1}{3}\right) + \left(\frac{1}{512} \cdot \frac{1}{3}\right) \]
Calculating each term:
\[ P(B) = \frac{1}{24} + \frac{1}{192} + \frac{1}{1536} \]

Finding a common denominator (which is 1536):
\[ P(B) = \frac{64}{1536} + \frac{8}{1536} + \frac{1}{1536} = \frac{73}{1536} \]

### Step 5: Apply Bayes' Theorem
Now we apply Bayes' Theorem to find \( P(A | B) \):
\[ P(A | B) = \frac{P(B | A) P(A)}{P(B)} \]
Substituting the known values:
\[ P(A | B) = \frac{\left(\frac{1}{8}\right) \left(\frac{1}{3}\right)}{\frac{73}{1536}} \]
Calculating the numerator:
\[ P(A | B) = \frac{\frac{1}{24}}{\frac{73}{1536}} = \frac{1536}{24 \cdot 73} = \frac{64}{73} \]

### Final Answer
Thus, the probability that it was coin \( B_1 \) given that we got 3 heads is:
\[ \boxed{\frac{64}{73}} \]