Problem 14 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the probability for the experiment of tossing a coin three times. Use the sample space $$S=\\{H H H, H H T, H T H, H T T, T H H, T H T, T T H, T T T\\}$$ The probability of getting a head on the first toss

Step-by-Step Solution

Answer

The probability of getting a 'Head' on the first toss when a coin is flipped three times is \(\frac{1}{2}\).

Step 1: Understand the sample space and identify required outcomes

The sample space presented consists of all possible outcomes when a coin is flipped three times. For each flip, there are two possible outcomes: Head (H) or Tail (T). Thus, for three flips, there are \(2^3 = 8\) possible outcomes. In terms of the first toss resulting in 'Head' (H), the outcomes are: \{H H H, H H T, H T H, H T T\}. So there are 4 outcomes in which the first toss resulted in 'Head' (H).

Step 2: Apply the formula of probability

The general formula for probability is given by the ratio of the number of favourable outcomes to the total number of outcomes. In this case, the number of favourable outcomes are the ones where the first toss results in 'Head'. This counts to 4 out of a total of 8 possible outcomes. This means the probability (\(P\)) of getting a 'Head' on the first toss is \(\frac{4}{8}\)

Step 3: Simplify the result

The calculated probability value can be simplified by dividing the numerator and denominator by their greatest common divisor (GCD), which in this case is 4. Hence the probability simplifies to \(\frac{1}{2}\).

Key Concepts

Sample Space

When dealing with experiments in probability, like tossing a coin multiple times, we first need to understand the concept of a sample space. Sample space is a term used to describe all the possible outcomes of a probabilistic experiment. In the case of tossing a coin three times, there are two outcomes for each toss, either 'Head' (H) or 'Tail' (T).

To construct the sample space for three coin tosses, we consider all combinations of these outcomes. For the first toss, we have two possibilities, and for each of those, the second toss also has two possibilities, leading to four outcomes by the second toss. Extending this to the third toss, each of those four outcomes can again result in either 'H' or 'T', which gives us a total of eight unique outcomes for the sample space: \(\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}\).

Understanding the sample space is crucial as it forms the foundation for determining the probability of various events within that experiment.

Favorable Outcomes

Once the sample space is established, we look at favorable outcomes to find the probability of a specific event. Favorable outcomes are those that align with the event we’re interested in. In the example of a coin tossed three times, if we are to find the probability of getting a 'Head' on the first toss, we only consider those outcomes where the first coin toss resulted in 'Head'.

In the solution provided, the favorable outcomes are \(\{HHH, HHT, HTH, HTT\}\), which all show 'Head' on the first toss. There are four such outcomes out of the total eight possible outcomes in the sample space. It's important to correctly identify these favorable outcomes, as the probability we calculate relies on whether or not we've accurately captured the correct set of outcomes that fit our event criteria.

Probability Formula

With the favorable outcomes and sample space defined, we utilize the probability formula to calculate the likelihood of an event. The formula is a straightforward ratio: \( P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes in the sample space}} \), where \( P(E) \) represents the probability of event \( E \).

In our coin toss example, the event \( E \) is getting a 'Head' on the first toss. There are four favorable outcomes for this event: \(\{HHH, HHT, HTH, HTT\}\), and there are eight possible outcomes in total. Plugging these numbers into our formula gives us: \( P(\text{'Head' on first toss}) = \frac{4}{8} \). To simplify, we divide both the numerator and denominator by the greatest common divisor, which is 4 in this case, resulting in a probability of \( \frac{1}{2} \). This reflects the intuitive fact that for a fair coin, the chance of getting a 'Head' on any single toss is 50%.