15 coins are tossed , the probability of getting heads will be

Options

A. `(511)/(32768)`
B. `(1001)/(32768)`
C. `(3003)/(32768)`
D. `(3005)/(32768)`

Explanation

To solve the problem of finding the probability of getting heads when 15 coins are tossed, we can use the concept of binomial distribution. Here’s a step-by-step solution:

### Step 1: Understand the scenario
When we toss a coin, there are two possible outcomes: heads (H) or tails (T). The probability of getting heads (p) is 1/2, and the probability of getting tails (q) is also 1/2.

### Step 2: Define the parameters
In this case:
- Number of trials (n) = 15 (since we are tossing 15 coins)
- Probability of getting heads (p) = 1/2
- Probability of getting tails (q) = 1/2

### Step 3: Use the binomial probability formula
The probability of getting exactly k heads in n tosses is given by the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k q^{(n-k)} \]
where:
- \( \binom{n}{k} \) is the binomial coefficient, which calculates the number of ways to choose k successes (heads) from n trials (coin tosses).

### Step 4: Calculate the probability for a specific number of heads
Let’s calculate the probability of getting exactly 10 heads (k = 10):
- Here, n = 15 and k = 10.
- The binomial coefficient \( \binom{15}{10} \) can be calculated as:
\[ \binom{15}{10} = \frac{15!}{10!(15-10)!} = \frac{15!}{10!5!} \]

### Step 5: Calculate the binomial coefficient
Calculating \( \binom{15}{10} \):
\[ \binom{15}{10} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} \]
Now, calculate the numerator:
\[ 15 \times 14 \times 13 \times 12 \times 11 = 360360 \]
And the denominator:
\[ 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
Thus,
\[ \binom{15}{10} = \frac{360360}{120} = 3003 \]

### Step 6: Calculate the probability
Now substitute back into the formula:
\[ P(X = 10) = \binom{15}{10} \left(\frac{1}{2}\right)^{10} \left(\frac{1}{2}\right)^{5} \]
This simplifies to:
\[ P(X = 10) = 3003 \left(\frac{1}{2}\right)^{15} \]
\[ P(X = 10) = 3003 \times \frac{1}{32768} \]
\[ P(X = 10) = \frac{3003}{32768} \]

### Step 7: Final answer
Thus, the probability of getting exactly 10 heads when 15 coins are tossed is:
\[ P(X = 10) = \frac{3003}{32768} \]