Question

A coin is tossed two independent times, each resulting in a tail \((\mathrm{T})\) or a head (H). The sample space consists of four ordered pairs: TT, TH, HT, HH. Making certain assumptions, compute the probability of each of these ordered pairs. What is the probability of at least one head?

Short answer

The probability of each ordered pair (TT, TH, HT, HH) is 0.25. The probability of getting at least one head is 0.75.

Step-by-step solution

Compute Probability of Each Ordered Pair

To compute the probability of an ordered pair is as simple as multiplying the probabilities of the individual outcomes. Since these are fair coin tosses, the probability of getting a head (H) or a tail (T) in each toss is 0.5. For example, to compute the probability of TT (both tosses yield tails), we do: \[ \text{P(TT)} = P(T) * P(T) = 0.5 * 0.5 = 0.25 \] Similarly, we can compute the probabilities for other three outcomes (TH, HT, HH). They are all 0.25.

Determine the Probability of at Least One Head

To determine the probability of getting at least one head in two coin tosses, we can sum up probabilities of the outcomes that include at least one head. That means, we account for TH, HT, HH outcomes. So, the probability of getting at least one head is: \[ \text{P(at least one H)} = P(TH) + P(HT) + P(HH) = 0.25 + 0.25 + 0.25 = 0.75 \]