Question

A fair coin is tossed 100 times. The probability of getting a head for an even number of times is

• A. $\frac{1}{2}$
• B. $\frac{3}{8}$
• C. $\frac{1}{8}$
• D. $\frac{3}{4}$

Hint:

For any fair coin tossed $n$ times ($n \ge 1$), the symmetry of outcomes ensures that the probability of getting an even number of heads is always exactly $\frac{1}{2}$, regardless of whether $n$ is 10, 100, or 10,000!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A fair coin is flipped 100 times. We need to find the probability that the total count of heads obtained turns out to be an even number.

Step 2: Key Formula or Approach:
This experiment follows a Binomial Distribution $B(n, p)$ where $n = 100$, and the probability of getting a head is $p = \frac{1}{2}$, which means $q = 1 - p = \frac{1}{2}$.
The probability of getting exactly $r$ successes is given by ${^{n}\text{C}_r} \cdot p^r \cdot q^{n-r}$. We need to sum this expression for all even values of $r$ ($0, 2, 4, \dots, 100$).

Step 3: Detailed Explanation:
The sum of probabilities for obtaining an even number of heads is:
$$P(\text{Even}) = {^{100}\text{C}_0} p^0 q^{100} + {^{100}\text{C}_2} p^2 q^{98} + \dots + {^{100}\text{C}_{100}} p^{100} q^0$$ Since $p = q = \frac{1}{2}$, each product term simplifies to $\left(\frac{1}{2}\right)^{100}$. We can factor this out from the entire expression:
$$P(\text{Even}) = \left(\frac{1}{2}\right)^{100} \left[ {^{100}\text{C}_0} + {^{100}\text{C}_2} + {^{100}\text{C}_4} + \dots + {^{100}\text{C}_{100}} \right]$$ By standard binomial coefficient identities, the sum of all even-indexed combinatoric coefficients is exactly half of the total sum of coefficients:
$${^{n}\text{C}_0} + {^{n}\text{C}_2} + {^{n}\text{C}_4} + \dots = 2^{n-1}$$ Substituting $n = 100$ gives:
$$\text{Sum of even coefficients} = 2^{100-1} = 2^{99}$$ Now, plug this sum back into our probability equation:
$$P(\text{Even}) = \frac{1}{2^{100}} \times 2^{99} = 2^{99 - 100} = 2^{-1} = \frac{1}{2}$$ This matches option (A).

Step 4: Final Answer:
The probability of getting a head an even number of times is $\frac{1}{2}$, which corresponds to option (A).