Question

If a fair coin is tossed four times, what is the probability of getting tails exactly twice? A. \(\frac{1}{4}\) B. \(\frac{3}{8}\) C. \(\frac{1}{2}\) D. \(\frac{3}{4}\)

Short answer

\( \frac{3}{8} \)

Step-by-step solution

Understand the Scenario

Consider a fair coin which has two outcomes each time it is tossed: heads (H) or tails (T). The problem is concerned with four tosses of the coin, and we want to find the probability of getting tails exactly twice.

Determine Total Outcomes

For four coin tosses, the total number of possible outcomes is calculated as \(2^4 = 16\). This is because each toss has 2 possible outcomes and there are 4 tosses.

Determine the Specific Outcomes

We need to find how many ways we can get exactly two tails in four coin tosses. This scenario is a combination problem where we select 2 positions out of 4 for the tails. The number of ways to do this is given by the combination formula: \[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = 6 \] where \( \binom{n}{k} \) is the number of ways to choose \( k \) successes (tails) from \( n \) trials (tosses).

Calculate the Probability

The probability of getting exactly two tails is the ratio of the successful outcomes to the total outcomes. Using the values from previous steps: \[ \text{Probability} = \frac{\text{Number of successful outcomes}}{\text{Total number of outcomes}} = \frac{6}{16} = \frac{3}{8} \]

Key concepts

Combinations

Combinations are a way to calculate the number of ways to choose items from a larger set, where the order does not matter. In the context of our coin toss problem, we need to find out how many ways we can get exactly two tails in four tosses. This is done using the combination formula: \ \ \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \ \ \] Here, \ n \ is the total number of tosses, and \ k \ is the number of tails we want. In our problem, \ n = 4\ and \ k = 2 \ so we calculate: \ \ \[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1} = 6 \ \ \] This means there are 6 unique ways to get exactly two tails in four coin tosses. \ \ The key takeaway is that combinations calculate the number of ways to select a subset of items from a larger set without considering the order. This is crucial in probability calculations where order does not matter, like in our coin toss exercise.

Binomial Probability

The binomial probability formula helps to find the probability of getting a fixed number of successes (like tails) in a specific number of trials (like coin tosses), where each trial is independent and has only two possible outcomes. In our case, we are interested in the probability of getting exactly two tails in four tosses of a fair coin. The formula for binomial probability is: \ \ \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \ \ \] Where: \ - \ P(X = k) \ is the probability of getting \ k \ successes (tails), \ - \ \binom{n}{k} \ is the number of combinations (as calculated in the previous section), \ - \ p \ is the probability of a single success (getting tails, which is 0.5 for a fair coin), \ - \ n \ is the total number of trials (4 tosses) \ - \k \ is the number of successes (2 tails).\ \ \ Plugging the values into the formula: \ \ \[ P(X = 2) = \binom{4}{2} (0.5)^2 (0.5)^{4-2} = 6 \times 0.25 \times 0.25 = \frac{6}{16} = \frac{3}{8} \ \ \] So the probability of getting exactly two tails in four tosses of a fair coin is \ \frac{3}{8} \. The binomial probability formula simplifies calculating such probabilities by considering both the number of possible outcomes and the likelihood of each.

Independent Events

Independent events are events where the outcome of one event does not affect the outcome of another. This concept is crucial in understanding our coin toss problem. \ \ Each toss of a fair coin is an independent event. This means the result of the first toss (heads or tails) does not influence the result of any subsequent toss. The independence of coin tosses allows us to multiply individual probabilities to find a combined probability for multiple events. In mathematical terms, for two independent events A and B, the probability of both events occurring is: \ \ \[ P(A \text{ and } B) = P(A) \times P(B) \ \ \] Since each coin toss is independent and has a 0.5 probability of landing either heads or tails, the probability of any specific sequence of four tosses (like HHTT or TTHH) is: \ \ \[ P(\text{HHTT}) = 0.5 \times 0.5 \times 0.5 \times 0.5 = (0.5)^4 = \frac{1}{16} \ \ \] When considering exactly two tails in four tosses, knowing each individual outcome doesn’t affect the others simplifies the calculation using combinations and probabilities presented before. Understanding that events are independent is essential for correctly applying various probability rules and formulas, especially in scenarios involving multiple trials or repetitions.