Understanding Coin Toss Probability

The problem asks for the probability that a tail appears an odd number of times when a coin is tossed 5 times. A single coin toss is a simple probability experiment with two possible outcomes: Head (H) or Tail (T). Since the coin is not mentioned as biased, we assume it is a fair coin.

• Probability of getting a Head, \(P(H) = \frac{1}{2}\)
• Probability of getting a Tail, \(P(T) = \frac{1}{2}\)

We are tossing the coin 5 times, and each toss is an independent event. This scenario fits the description of a binomial distribution.

Binomial Probability Calculation

In a binomial distribution, if an experiment is performed \(n\) times, and the probability of success in a single trial is \(p\) (and probability of failure is \(q = 1-p\)), the probability of getting exactly \(k\) successes in \(n\) trials is given by the formula:

In this case:

• Number of trials, \(n = 5\)
• Success is defined as getting a Tail.
• Probability of success (getting a Tail), \(p = \frac{1}{2}\)
• Probability of failure (getting a Head), \(q = 1 - \frac{1}{2} = \frac{1}{2}\)

We want to find the probability that the number of tails is odd. When a coin is tossed 5 times, the number of tails can be 0, 1, 2, 3, 4, or 5. The odd numbers of tails are 1, 3, or 5.

So, we need to calculate \(P(X=1) + P(X=3) + P(X=5)\).

Calculate \(P(X=k)\) for Odd \(k\)

Let's calculate the probability for each odd number of tails:

• For exactly 1 Tail (\(k=1\)): \[P(X=1) = \binom{5}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{5-1} = 5 \times \left(\frac{1}{2}\right)^1 \times \left(\frac{1}{2}\right)^4 = 5 \times \left(\frac{1}{2}\right)^5\] \[P(X=1) = 5 \times \frac{1}{32} = \frac{5}{32}\]
• For exactly 3 Tails (\(k=3\)): \[P(X=3) = \binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{5-3} = \binom{5}{2} \times \left(\frac{1}{2}\right)^3 \times \left(\frac{1}{2}\right)^2\] \[\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10\] \[P(X=3) = 10 \times \left(\frac{1}{2}\right)^5 = 10 \times \frac{1}{32} = \frac{10}{32}\]
• For exactly 5 Tails (\(k=5\)): \[P(X=5) = \binom{5}{5} \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^{5-5} = 1 \times \left(\frac{1}{2}\right)^5 \times \left(\frac{1}{2}\right)^0\] \[\binom{5}{5} = 1\] \[P(X=5) = 1 \times \left(\frac{1}{2}\right)^5 = 1 \times \frac{1}{32} = \frac{1}{32}\]

Summing Probabilities for Odd Tails

The total probability of getting an odd number of tails is the sum of these probabilities:

So, the probability that tail appears an odd number of times is \(\frac{1}{2}\).

Alternative Approach: Symmetry in Binomial Distribution (\(p=1/2\))

When the probability of success \(p\) is equal to the probability of failure \(q\) (i.e., \(p=q=1/2\)), the binomial distribution is symmetric. For any number of trials \(n\), the sum of probabilities for an odd number of successes is equal to the sum of probabilities for an even number of successes (excluding \(k=0\) if \(n\) is odd, including \(k=0\) if \(n\) is even, but the principle holds for the sum of odds vs sum of evens).

The sum of all probabilities is 1: \[\sum_{k=0}^{n} P(X=k) = 1\] When \(p=q=1/2\), the probability of getting an odd number of successes (tails) is always \(1/2\) for \(n \ge 1\).

This can be shown using the property that \(\sum_{k \text{ odd}} \binom{n}{k} = 2^{n-1}\) for \(n \ge 1\). \[P(\text{Odd Tails}) = \sum_{k \text{ odd}} \binom{n}{k} \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^n \sum_{k \text{ odd}} \binom{n}{k} = \left(\frac{1}{2}\right)^n \times 2^{n-1} = \frac{2^{n-1}}{2^n} = \frac{1}{2}\] For \(n=5\), this property gives the probability as \(\frac{1}{2}\) directly.

Sum of probabilities for odd tails: \(P(X=1) + P(X=3) + P(X=5) = \frac{5}{32} + \frac{10}{32} + \frac{1}{32} = \frac{16}{32} = \frac{1}{2}\).

Sum of probabilities for even tails: \(P(X=0) + P(X=2) + P(X=4) = \frac{1}{32} + \frac{10}{32} + \frac{5}{32} = \frac{16}{32} = \frac{1}{2}\).

The probability that tail appears an odd number of times is \(\frac{1}{2}\).

Probability Concepts Revision Table

Additional Probability Information

Understanding binomial probability is key when dealing with repeated independent trials, like coin tosses. Here are some related concepts:

• Bernoulli Trial: A single experiment with only two possible outcomes (success or failure). A single coin toss is a Bernoulli trial.
• Probability of Even Number of Tails: In this problem, we found the probability of an odd number of tails is 1/2. The probability of getting an even number of tails would be \(P(X=0) + P(X=2) + P(X=4) = \frac{1}{32} + \frac{10}{32} + \frac{5}{32} = \frac{16}{32} = \frac{1}{2}\). This confirms the symmetry for \(p=1/2\).
• Expected Value: The expected number of tails in 5 tosses is \(E(X) = np = 5 \times \frac{1}{2} = 2.5\). This doesn't have to be an integer, as it's an average over many repetitions.
• Sum of Binomial Coefficients: The sum of all binomial coefficients for a given \(n\) is \(\sum_{k=0}^{n} \binom{n}{k} = 2^n\). The sum of coefficients for odd indices equals the sum for even indices, and each sum is \(2^{n-1}\) (for \(n \ge 1\)). This property is very useful for probabilities when \(p=1/2\).