One hundred identical coins, each with probability 'p' of showing heads are tossed once. If `0 lt p lt 1` and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, then the value of p is

Options

A. `1/2`
B. `51/101`
C. `49/101`
D. `47/101`

Answer

To solve the problem, we need to find the value of \( p \) such that the probability of getting heads on 50 coins is equal to the probability of getting heads on 51 coins when 100 identical coins are tossed. We will use the binomial distribution for this purpose.

### Step-by-Step Solution:

1. **Understanding the Binomial Distribution**:
The probability of getting exactly \( k \) heads in \( n \) tosses of a coin with probability \( p \) of heads is given by:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Here, \( n = 100 \), and we need to set up equations for \( k = 50 \) and \( k = 51 \).

2. **Setting Up the Equation**:
We know that:
\[ P(X = 50) = P(X = 51) \]
This translates to:
\[ \binom{100}{50} p^{50} (1-p)^{50} = \binom{100}{51} p^{51} (1-p)^{49} \]

3. **Using Binomial Coefficients**:
We can express the binomial coefficients:
\[ \binom{100}{51} = \frac{100!}{51! \cdot 49!} \]
and
\[ \binom{100}{50} = \frac{100!}{50! \cdot 50!} \]

4. **Substituting the Coefficients**:
Substituting these into our equation gives:
\[ \frac{100!}{50! \cdot 50!} p^{50} (1-p)^{50} = \frac{100!}{51! \cdot 49!} p^{51} (1-p)^{49} \]

5. **Canceling Common Terms**:
We can cancel \( 100! \) from both sides:
\[ \frac{1}{50! \cdot 50!} p^{50} (1-p)^{50} = \frac{1}{51! \cdot 49!} p^{51} (1-p)^{49} \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{p^{50} (1-p)^{50}}{50!} = \frac{p^{51} (1-p)^{49}}{51 \cdot 50!} \]
We can cancel \( 50! \) from both sides:
\[ p^{50} (1-p)^{50} = \frac{p^{51} (1-p)^{49}}{51} \]

7. **Dividing Both Sides**:
Dividing both sides by \( p^{50} (1-p)^{49} \) (assuming \( p \neq 0 \) and \( 1-p \neq 0 \)):
\[ (1-p) = \frac{p}{51} \]

8. **Solving for \( p \)**:
Rearranging gives:
\[ 51(1-p) = p \]
Expanding and rearranging:
\[ 51 - 51p = p \implies 51 = 52p \implies p = \frac{51}{52} \]

### Final Answer:
The value of \( p \) is:
\[ \boxed{\frac{51}{101}} \]