Coin Toss Probability Explained

This problem asks us to determine the probability of a specific outcome when a fair coin is tossed n times. The outcome of interest is when the difference between the number of heads (H) and the number of tails (T) is exactly \(n - 3\).

Defining Coin Toss Variables and Conditions

Let's define the variables involved:

• \(n\): The total number of times a fair coin is tossed.
• \(H\): The number of heads obtained in the \(n\) tosses.
• \(T\): The number of tails obtained in the \(n\) tosses.

Since each toss results in either a head or a tail, the sum of the number of heads and tails must equal the total number of tosses:

$$H + T = n \quad (1)$$

The condition given in the problem is that the difference between the number of heads and tails is \(n - 3\). We can express this using the absolute difference:

$$|H - T| = n - 3 \quad (2)$$

This absolute difference implies two possible scenarios:

1. The number of heads is greater than the number of tails: \(H - T = n - 3\)
2. The number of tails is greater than the number of heads: \(T - H = n - 3\), which is equivalent to \(H - T = -(n - 3) = 3 - n\)

Parity Analysis for Heads-Tails Difference

We can analyze the possibility of these conditions being met by considering the parity (whether a number is even or odd) of the counts.

We have the following two equations:

• \(H + T = n\)
• \(H - T = k\) (where \(k\) represents the difference)

Adding these two equations together, we get:

$$ (H + T) + (H - T) = n + k $$

$$ 2H = n + k $$

For \(H\) to be a valid number of heads, it must be an integer. This means that \(2H\) must be an even number. Consequently, the sum \(n + k\) must also be an even number.

For the sum \(n + k\) to be even, both \(n\) and \(k\) must have the same parity: either both must be even, or both must be odd.

Applying Parity to Coin Toss Difference \(n-3\)

In this specific problem, the difference required is \(k = n - 3\). Let's check the parity relationship between \(n\) and \(k = n - 3\):

• Scenario 1: \(n\) is even. If \(n\) is an even number, then \(n - 3\) (even minus odd) results in an odd number. So, \(n\) is even and \(k\) is odd.
• Scenario 2: \(n\) is odd. If \(n\) is an odd number, then \(n - 3\) (odd minus odd) results in an even number. So, \(n\) is odd and \(k\) is even.

In both scenarios, \(n\) and \(k = n - 3\) always have different parities.

Because \(n\) and \(k\) have different parities, their sum \(n + k\) will always be an odd number. Substituting \(k = n - 3\), we get \(n + k = n + (n - 3) = 2n - 3\), which is always odd.

Since \(2H = 2n - 3\) (an odd number), there is no integer value for \(H\) that satisfies this equation. The same logic applies if we consider \(T - H = n - 3\), which leads to \(2T = 2n - 3\), again yielding no integer solution for \(T\).

Possibility Conclusion for Coin Toss Difference

The analysis shows that it is mathematically impossible for the difference between the number of heads and tails to be exactly \(n - 3\) in \(n\) coin tosses, regardless of the value of \(n\). This means the event described in the question is an impossible event.

Calculating Probability for Coin Toss Scenario

The probability of an impossible event occurring is always 0.

Therefore, the probability that the difference between the number of heads and tails is \(n - 3\) is 0.

Matching Probability Result to Options

We compare our result to the options provided:

• Option 1: \(\frac{1}{2^n}\)
• Option 2: 0
• Option 3: \({}^nC_{n - 3} \cdot \frac{1}{2^n}\)
• Option 4: \(\frac{1}{2^{n - 3}}\)

Our calculated probability of 0 matches Option 2.