If 10 coins are tossed simultaneoiusly , then probability of getting at least 7 heads is

Options

A. `(120)/(1024)`
B. `(45)/(1024)`
C. `(10)/(1024)`
D. `(176)/(1024)`

Answer

To solve the problem of finding the probability of getting at least 7 heads when tossing 10 coins simultaneously, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Binomial Distribution Parameters**:
- We are tossing 10 coins, so \( n = 10 \).
- The probability of getting heads (success) in a fair coin toss is \( p = \frac{1}{2} \).
- The probability of getting tails (failure) is also \( q = \frac{1}{2} \).

2. **Define the Event**:
- We want to find the probability of getting at least 7 heads. This can be expressed as:
\[ P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \]

3. **Use the Binomial Probability Formula**:
- The probability of getting exactly \( r \) successes (heads) in \( n \) trials (coin tosses) is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
- Here, \( \binom{n}{r} \) is the binomial coefficient, which can be calculated as:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

4. **Calculate Each Probability**:
- For \( P(X = 7) \):
\[ P(X = 7) = \binom{10}{7} \left(\frac{1}{2}\right)^7 \left(\frac{1}{2}\right)^{3} = \binom{10}{7} \left(\frac{1}{2}\right)^{10} \]
- For \( P(X = 8) \):
\[ P(X = 8) = \binom{10}{8} \left(\frac{1}{2}\right)^8 \left(\frac{1}{2}\right)^{2} = \binom{10}{8} \left(\frac{1}{2}\right)^{10} \]
- For \( P(X = 9) \):
\[ P(X = 9) = \binom{10}{9} \left(\frac{1}{2}\right)^9 \left(\frac{1}{2}\right)^{1} = \binom{10}{9} \left(\frac{1}{2}\right)^{10} \]
- For \( P(X = 10) \):
\[ P(X = 10) = \binom{10}{10} \left(\frac{1}{2}\right)^{10} \left(\frac{1}{2}\right)^{0} = \binom{10}{10} \left(\frac{1}{2}\right)^{10} \]

5. **Calculate the Binomial Coefficients**:
- \( \binom{10}{7} = \frac{10!}{7!3!} = 120 \)
- \( \binom{10}{8} = \frac{10!}{8!2!} = 45 \)
- \( \binom{10}{9} = \frac{10!}{9!1!} = 10 \)
- \( \binom{10}{10} = 1 \)

6. **Combine the Probabilities**:
- Now we can sum these probabilities:
\[ P(X \geq 7) = \left(120 + 45 + 10 + 1\right) \left(\frac{1}{2}\right)^{10} = 176 \left(\frac{1}{1024}\right) = \frac{176}{1024} \]

7. **Simplify the Result**:
- The final probability can be simplified:
\[ P(X \geq 7) = \frac{176}{1024} = \frac{11}{64} \]

### Final Answer:
The probability of getting at least 7 heads when tossing 10 coins is \( \frac{11}{64} \).