Question

A coin is tossed two independent times, each resulting in a tail (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 probabilities for outcomes TT, TH, HT, HH are all \( \frac{1}{4} \) each. The probability of getting at least one head in two tosses is \( \frac{3}{4} \).

Step-by-step solution

Compute the individual probabilities

As there are two possible outcomes (tail or head) for each coin toss, and the tosses are independent, the probability for each outcome (TT, TH, HT, HH) is simply the product of the individual coin toss results. As each toss is an independent and unbiased event, the probability of a head (H) or a tail (T) is equal, i.e.,\( \frac{1}{2} \). Therefore, each outcome (TT, TH, HT, HH) has the probability of \( \frac{1}{2} \) * \( \frac{1}{2} \) = \( \frac{1}{4} \).

Calculate the probability of at least one head

To get at least one head (H) in two tosses, we could have the pairs HT, TH, or HH. As we calculated in the previous step, the probability for each pair is \( \frac{1}{4} \). So the total probability of getting at least one head is \( \frac{1}{4} \) (HT) + \( \frac{1}{4} \) (TH) + \( \frac{1}{4} \) (HH) = \( \frac{3}{4} \)

Key concepts

Independent Events

Understanding independent events is crucial for studying probability theory. Independent events are those where the occurrence of one event does not affect the probability of the other happening. For example, when tossing a fair coin, whether you get a head or tail on the first toss doesn't influence the outcome of the second toss; they are independent events. In our coin toss exercise, we assume both coin tosses are independent. From this assumption, to find the probability of two independent events both occurring, you would multiply the probability of each event. If the probability of getting heads on a single coin toss is \( \frac{1}{2} \), then the probability of getting heads twice in a row, being independent events, would be \( \frac{1}{2} \) multiplied by \( \frac{1}{2} \) which equals \( \frac{1}{4} \).

This principle applies no matter how many independent events you are considering. The key takeaway is that independence equates to the lack of influence between events, making probability calculations straightforward when dealing with multiple coin tosses or other similar scenarios.

Sample Space

The sample space in probability theory is the set of all possible outcomes of a random experiment. It's the 'universe' of all the events that could potentially occur. In the coin toss example provided, the sample space consists of all the ordered pairs of results you could see from tossing a coin twice: TT, TH, HT, and HH. The term 'ordered pairs' means that the order matters—HT is different from TH because the first toss resulted in heads for HT, while for TH, it's the second toss that was heads.

Visualizing or listing the sample space is often a helpful step in solving probability problems, as it allows you to see all the scenarios you need to consider. When each outcome in the sample space is equally likely, as with a fair coin, you can determine the probability of an event by dividing the number of ways an event can occur by the total number of outcomes in the sample space.

Coin Toss Probability

Now, let's dive into coin toss probability more specifically. For a fair coin, the probability of getting heads (H) or tails (T) on a single toss is equal, each with a probability of \( \frac{1}{2} \). But how do we handle the coin toss probability when multiple coin tosses are involved, as in our exercise? It's simple when we remember that coin tosses are independent events.

Computing Single Toss Probabilities

For one coin toss, the probability is \( \frac{1}{2} \) for heads and \( \frac{1}{2} \) for tails.

Probability of Sequential Outcomes

When throwing the coin twice, the probability of each ordered pair (TT, TH, HT, HH) is the product of individual outcomes, so \( \frac{1}{2} \) times \( \frac{1}{2} \) equals \( \frac{1}{4} \).

To calculate the probability of getting at least one head in two tosses, we sum up the probabilities of getting heads in any of the tosses (HT, TH, or HH). Each has a probability of \( \frac{1}{4} \), so altogether, the probability is \( \frac{1}{4} + \frac{1}{4} + \frac{1}{4} \) which simplifies to \( \frac{3}{4} \) or 75%. This approach can be extended to more complex scenarios, and it's all about understanding how to use the basics of sample spaces and independent events probability calculations.