Question

If a coin be tossed \(n\) times then probability that the head comes odd number of times (1) \(\frac{1}{2}\) (2) \(\frac{1}{2^{n}}\) (3) \(\frac{1}{2^{n-1}}\) (4) \(\frac{1}{n^{2}}\)

Short answer

The probability is \(\frac{1}{2}\).

Step-by-step solution

Understand the problem

We need to find the probability that a coin tossed \(n\) times results in an odd number of heads.

Determine the nature of outcomes

Each toss of a coin has two outcomes: heads (H) and tails (T). The total number of possible outcomes for \(n\) tosses is \(2^n\).

Calculate the probabilities

The probability of obtaining a specific sequence of heads and tails is \(1/2^n\) since each outcome is independent and there are \(2^n\) possible sequences.

Use symmetry

The total number of possible outcomes where heads appear an even number of times is equal to the number of outcomes where heads appear an odd number of times. Therefore, each of these events has a probability of \(1/2\) given the symmetry.

Conclusion

The probability that heads occur an odd number of times in \(n\) tosses is \(\frac{1}{2}\).

Key concepts

probability theory

Probability theory helps us understand and measure how likely an event is to occur. When tossing a coin, each outcome (heads or tails) is equally likely. The chance of getting heads is \( \frac{1}{2} \) and the same goes for tails. When dealing with multiple coin tosses, we consider each outcome to be independent of the others. This means the result of one toss doesn't affect another. To calculate the overall probability, we multiply the individual probabilities. For instance, the probability of getting two heads in two tosses is \( \left( \frac{1}{2} \right) \times \left( \frac{1}{2} \right) = \frac{1}{4} \). Probability theory is handy for situations involving random outcomes, like dice rolls or drawing cards, helping us to predict possible results and their likelihood.

binomial distribution

Binomial distribution describes the number of successes in a fixed number of trials, with each trial having the same probability of success. In our coin toss example, 'success' could be defined as getting heads. The distribution helps us find the probability of getting an exact number of heads when tossing the coin multiple times. Suppose we flip a coin three times. We want to know the probability of getting exactly two heads. Here, the number of trials \(n = 3\) and the probability of success \(p = \frac{1}{2}\). We will use the binomial formula: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \), where \( \binom{n}{k} \) is the binomial coefficient. For two heads \(k = 2\), \( P(X = 2) = \binom{3}{2} \left( \frac{1}{2} \right)^2 \left( \frac{1}{2} \right)^{1} = 3 \times \frac{1}{4} \times \frac{1}{2} = \frac{3}{8}\). This formula can be used for any number of tosses and heads.

symmetry in probability

Symmetry in probability is a valuable concept that simplifies problems. When tossing a fair coin, the chance of getting heads is always the same as getting tails. This symmetry extends to multiple tosses. If we consider getting any number of heads, the outcomes for an even or odd number of heads are symmetric. This means the probability of getting an even number of heads is equal to getting an odd number of heads. For example, when we toss a coin four times, there are equal chances to get 0, 2, or 4 heads (even) as there are to get 1 or 3 heads (odd). Therefore, the probability that heads occur an odd number of times in any number of tosses is \(\frac{1}{2}\). Symmetry makes it easier to calculate probabilities without counting every possible outcome.