In tossing 10 coins, the probability of getting exactly 5 heads is

Options

A. `(9)/(128)`
B. `(63)/(256)`
C. `(1)/(2)`
D. `(193)/(256)`

Answer

To find the probability of getting exactly 5 heads when tossing 10 coins, we can use the binomial probability formula. The steps are as follows:

### Step 1: Identify the parameters
In this problem:
- The number of trials (n) = 10 (since we are tossing 10 coins)
- The number of successes (k) = 5 (we want exactly 5 heads)
- The probability of success on a single trial (p) = 1/2 (the probability of getting heads in a single toss)
- The probability of failure (q) = 1/2 (the probability of getting tails in a single toss)

### Step 2: Use the binomial probability formula
The binomial probability formula is given by:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
Where:
- \( \binom{n}{k} \) is the binomial coefficient, which can be calculated as \( \frac{n!}{k!(n-k)!} \)

### Step 3: Calculate the binomial coefficient
We need to calculate \( \binom{10}{5} \):
\[ \binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!} \]
Calculating \( 10! \), \( 5! \):
\[ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5! \quad \text{(we can cancel } 5! \text{)} \]
Thus,
\[ \binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]

### Step 4: Calculate \( p^k \) and \( q^{n-k} \)
Now, we calculate \( p^k \) and \( q^{n-k} \):
\[ p^k = \left(\frac{1}{2}\right)^5 = \frac{1}{32} \]
\[ q^{n-k} = \left(\frac{1}{2}\right)^{10-5} = \left(\frac{1}{2}\right)^5 = \frac{1}{32} \]

### Step 5: Substitute values into the formula
Now we can substitute these values into the binomial probability formula:
\[ P(X = 5) = \binom{10}{5} \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^5 = 252 \times \frac{1}{32} \times \frac{1}{32} \]
\[ P(X = 5) = 252 \times \frac{1}{1024} \]
\[ P(X = 5) = \frac{252}{1024} \]

### Step 6: Simplify the fraction
Now we can simplify \( \frac{252}{1024} \):
- Divide both the numerator and the denominator by 4:
\[ \frac{252 \div 4}{1024 \div 4} = \frac{63}{256} \]

### Final Answer
Thus, the probability of getting exactly 5 heads when tossing 10 coins is:
\[ \frac{63}{256} \]

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