Multiple choice

A coin is tossed '$2n$' times. The chance that the number of times one gets head is not equal to the number of times one gets tail is

  1. $\displaystyle \frac{(2n!)}{(n!)^{2}}. \left(\frac{1}{2} \right)^{2n}$
  2. $1-\displaystyle \frac{(2n!)}{(n!)^{2}}$
  3. $1-\displaystyle \frac{(2n!)}{(n!)^{2}}.\frac{1}{4^{n}}$
  4. $\displaystyle \frac{(2n!)}{(n!)^{2}}\frac{1}{4^{n}}$
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C Correct answer
Explanation

Total outcomes = 2^(2n). Outcomes with equal heads and tails = 2nCn. Probability of equal heads and tails = (2nCn) / 2^(2n). Probability of not equal = 1 - (2nCn) / 4^n.

AI explanation

When a coin is tossed 2n times, the probability of getting exactly n heads and n tails is found using the binomial distribution formula, given by 2nCn multiplied by (1/2) raised to the power of 2n. This simplifies to the factorial expression (2n!) divided by (n!)^2, all multiplied by 1/4^n. The probability that the number of heads is not equal to the number of tails is the complement of this event, calculated as 1 minus this value.

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