Question

A coin is tossed \(2 n\) times. The probability that the number of times one get head is not equal to number of times one gats tail is (a) \(1-\left(2 / 4^{n}\right)\) (b) \(1-\left[\\{(2 n) !\\} /\left\\{(n !)^{2}\right\\}\right] \cdot\left(1 / 4^{n}\right)\) (c) \(1-\left[\\{(2 n) !\\} /\left\\{(n !)^{2}\right\\}\right]\) (d) \(\left[\\{(2 n) !\\} /\left\\{(n !)^{2}\right\\}\right] \cdot\left(1 / 4^{n}\right)\)

Short answer

None of the given options is correct, as the calculated probability expression does not match any of them. The correct probability expression is \(P = \frac{C(2n, n+1)}{2^{2n}} + \frac{C(2n, n-1)}{2^{2n}}\).

Step-by-step solution

List out the possibilities

If a coin is tossed \(2n\) times, there will be a total of \(2^{2n}\) possible outcomes. We need to find the probability of the cases where the number of Heads is not equal to the number of Tails.

Use the combination formula

To find the number of ways we can get \(n+1\) Heads and \(n-1\) Tails or \(n-1\) Heads and \(n+1\) Tails, we can use the combination formula: \[ C(n, r) = \frac{n!}{r!(n-r)!} \] In our case, the number of ways we can get \(n+1\) Heads and \(n-1\) Tails is \[ C(2n, n+1) = \frac{(2n)!}{(n+1)!(2n-n-1)!} \] and the number of ways we can get \(n-1\) Heads and \(n+1\) Tails is \[ C(2n, n-1) = \frac{(2n)!}{(n-1)!(2n-n+1)!} \]

Calculate probabilities

Now that we have the number of ways we can get these scenarios, we can calculate the probability for each scenario using the formula: Probability = \(\frac{\text{Number of successful outcomes}}{\text{Total number of outcomes}}\) For \(n+1\) Heads and \(n-1\) Tails, the probability is: \[ P_{n+1} = \frac{C(2n, n+1)}{2^{2n}} \] For \(n-1\) Heads and \(n+1\) Tails, the probability is: \[ P_{n-1} = \frac{C(2n, n-1)}{2^{2n}} \]

Sum of probabilities

Now, we need to find the probability that the number of Heads is not equal to the number of Tails. This is the sum of the probabilities \(P_{n+1}\) and\(P_{n-1}\) : \[ P = P_{n+1} + P_{n-1} \] Substitute the probabilities found in step 3: \[ P = \frac{C(2n, n+1)}{2^{2n}} + \frac{C(2n, n-1)}{2^{2n}} \]

Compare with the given options

Comparing our final probability expression with the given options, none of the given options matches our calculated probability expression. Therefore, none of the given options is correct.

Key concepts

Combinatorics

Combinatorics is a branch of mathematics dealing with combinations, permutations, and counting problems of discrete structures. It's a key component when analyzing games of chance, such as tossing a coin multiple times. In our exercise, combinatorics comes into play when determining the number of possible outcomes from tossing a coin 2n times.

Each coin toss has two possible outcomes - head (H) or tail (T) - and when a coin is tossed 2n times, the number of all possible sequences of Hs and Ts (such as HHTT or TTHH) is 2²ⁿ, as each toss is independent and has two outcomes. To understand how the combinations work, let's review the basic formula for combinations:

• \(C(n, k) = \frac{n!}{k!(n-k)!}\) where \(n\) is the total number of items, \(k\) is the number of items to choose, and \(!\) denotes the factorial function. This formula helps calculate the number of ways to choose \(k\) items from a set of \(n\) items, without regard to the order.

Applying combinatorics to our problem, we can identify how many distinct ways we can achieve an unequal number of heads and tails in 2n tosses, which is crucial before we can calculate the probability of such an event.

Probability Theory

Probability theory is the mathematical framework for quantifying the likelihood of events, ranging from flipping coins to more complex phenomena. It forms the foundation for statistical inference and decision-making under uncertainty. In probability theory, we often refer to the classical probability definition, which assumes all outcomes are equally likely.

The probability of an event is given by the formula:

• Probability = \(\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}\)

For our coin tossing exercise, we are provided with a scenario where we have to calculate the probability of getting an unequal number of heads and tails. We use the concepts of combinatorics to find the favorable outcomes, and then, with probability theory, we put them over the total possibilities (\(2^{2n}\)) to find the likelihood of our event. It's important to note that each combination of heads and tails is an independent event, and their probabilities must be considered separately before summing them up to get the total probability.

Binomial Theorem

The binomial theorem provides a quick way to expand binomial expressions raised to any power. It's expressed as:

• \((a + b)^n = \sum_{k=0}^{n} C(n, k) a^{n-k} b^k\)

This theorem can also be applied in probability, especially when dealing with binomial distributions, which describe the number of successes in a fixed number of independent Bernoulli trials. Each trial has two possible outcomes with fixed probabilities that sum to one. The formula obtained from the binomial theorem relates to the probabilities of obtaining exactly \(k\) successes (or heads in the context of our coin toss) in \(n\) trials:

• Probability of \(k\) heads = \(C(n, k) p^k (1-p)^{n-k}\)

Where \(p\) represents the probability of having head in one toss, and \(1-p\) represents the probability of a tail. In our problem, each coin toss is considered a Bernoulli trial with a success probability \(p=1/2\), reflecting the equal likelihood of heads or tails. To solve the problem, we use the binomial distribution formula to calculate the probabilities of having \(n+1\) or \(n-1\) heads (or tails), which represent the uneven distributions of outcomes we're interested in.