Ten unbiased coins are thrown simultaneously. The probability of getting at least seven heads is

Options

A. `3/64`
B. `5/64`
C. `7/64`
D. `11/64`

Answer

To find the probability of getting at least 7 heads when 10 unbiased coins are thrown, we can use the binomial probability formula. Let's break down the solution step by step.

### Step 1: Understand the Problem
We need to find the probability of getting at least 7 heads when 10 coins are tossed. This means we need to calculate the probabilities for getting 7, 8, 9, and 10 heads.

### Step 2: Define the Variables
- Let \( n = 10 \) (the number of coins).
- The probability of getting heads (success) on a single toss, \( p = \frac{1}{2} \).
- The probability of getting tails (failure), \( q = 1 - p = \frac{1}{2} \).

### Step 3: Use the Binomial Probability Formula
The binomial probability formula is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
where \( \binom{n}{r} \) is the binomial coefficient, which represents the number of ways to choose \( r \) successes in \( n \) trials.

### Step 4: Calculate the Required Probabilities
We need to calculate:
\[ P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \]

#### Calculate \( P(X = 7) \):
\[ P(X = 7) = \binom{10}{7} \left(\frac{1}{2}\right)^7 \left(\frac{1}{2}\right)^{3} = \binom{10}{7} \left(\frac{1}{2}\right)^{10} \]
\[ \binom{10}{7} = \frac{10!}{7!(10-7)!} = \frac{10!}{7!3!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]
Thus,
\[ P(X = 7) = 120 \cdot \left(\frac{1}{2}\right)^{10} = \frac{120}{1024} \]

#### Calculate \( P(X = 8) \):
\[ P(X = 8) = \binom{10}{8} \left(\frac{1}{2}\right)^8 \left(\frac{1}{2}\right)^{2} = \binom{10}{8} \left(\frac{1}{2}\right)^{10} \]
\[ \binom{10}{8} = \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \]
Thus,
\[ P(X = 8) = 45 \cdot \left(\frac{1}{2}\right)^{10} = \frac{45}{1024} \]

#### Calculate \( P(X = 9) \):
\[ P(X = 9) = \binom{10}{9} \left(\frac{1}{2}\right)^9 \left(\frac{1}{2}\right)^{1} = \binom{10}{9} \left(\frac{1}{2}\right)^{10} \]
\[ \binom{10}{9} = 10 \]
Thus,
\[ P(X = 9) = 10 \cdot \left(\frac{1}{2}\right)^{10} = \frac{10}{1024} \]

#### Calculate \( P(X = 10) \):
\[ P(X = 10) = \binom{10}{10} \left(\frac{1}{2}\right)^{10} = 1 \cdot \left(\frac{1}{2}\right)^{10} = \frac{1}{1024} \]

### Step 5: Sum the Probabilities
Now, we add all these probabilities together:
\[ P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \]
\[ P(X \geq 7) = \frac{120}{1024} + \frac{45}{1024} + \frac{10}{1024} + \frac{1}{1024} \]
\[ P(X \geq 7) = \frac{120 + 45 + 10 + 1}{1024} = \frac{176}{1024} \]

### Step 6: Simplify the Result
Now, we simplify \( \frac{176}{1024} \):
\[ \frac{176 \div 16}{1024 \div 16} = \frac{11}{64} \]

### Final Answer
Thus, the probability of getting at least 7 heads when 10 unbiased coins are thrown is:
\[ \boxed{\frac{11}{64}} \]