Problem 6 - A First Course in Probability

Question

In Problem \(4.5,\) for \(n=3,\) if the coin is assumed fair, what are the probabilities associated with the values that \(X\) can take on?

Step-by-Step Solution

Answer

For a fair coin with \(n=3\) tosses, the random variable \(X\) represents the number of heads obtained. The probabilities associated with the values that \(X\) can take on are: \(P(X=0) = 0.125\) \(P(X=1) = 0.375\) \(P(X=2) = 0.375\) \(P(X=3) = 0.125\)

Step 1: List all possible outcomes after 3 tosses

We can obtain a total of 2^3 = 8 possible outcomes after 3 tosses: HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT.

Step 2: Determine the random variable X

Let the random variable \(X\) represent the number of heads obtained in 3 coin tosses. Thus, \(X\) can take on the values of 0, 1, 2, or 3.

Step 3: Calculate the probabilities for each value of X

We need to find the probabilities for each possible value of \(X\) from the set of outcomes: 1. \(P(X=0)\): There's only one possible outcome with 0 heads – TTT, and since each toss is independent and the coin is fair, the probability is given by \(P(X=0) = P(T) \times P(T) \times P(T) = 0.5 \times 0.5 \times 0.5 = 0.125\). 2. \(P(X=1)\): There are three possible outcomes with 1 head – HTT, THT, and TTH. The probability of each is given by \(P(H) \times P(T) \times P(T) = 0.5 \times 0.5 \times 0.5 = 0.125\). So, the total probability of getting 1 head is \(P(X=1) = 3 \times 0.125 = 0.375\). 3. \(P(X=2)\): There are also three possible outcomes with 2 heads – HHT, HTH, and THH. The probability of each is given by \(P(H) \times P(H) \times P(T) = 0.5 \times 0.5 \times 0.5 = 0.125\). So, the total probability of getting 2 heads is \(P(X=2) = 3 \times 0.125 = 0.375\). 4. \(P(X=3)\): There's only one possible outcome with 3 heads – HHH, and its probability is given by \(P(X=3) = P(H) \times P(H) \times P(H) = 0.5 \times 0.5 \times 0.5 = 0.125\). Now, we have found the probabilities associated with the values that \(X\) can take on: \(P(X=0) = 0.125\) \(P(X=1) = 0.375\) \(P(X=2) = 0.375\) \(P(X=3) = 0.125\)

Key Concepts

Probabilities of Outcomes

Understanding the probabilities of outcomes is crucial when tackling problems involving random events, such as coin tosses. In the given exercise, we analyze a sequence of three coin tosses and aim to determine the likelihood of various results. Coin tosses are classic examples of a binary outcome: the result is either 'heads' (H) or 'tails' (T) each time the coin is flipped. The fair coin assumption indicates that the probability of getting a head (H) or a tail (T) is the same with each individual toss, which is \(0.5\) or 50%.

When listing all the possible outcomes of three coin tosses, we observe there are eight possible sequences. These sequences can be categorized based on the number of heads that appear. For instance, the sequence TTT represents zero heads, while HHT represents two heads. To find the probability of obtaining exactly a certain number of heads, we sum the probabilities of obtaining that outcome for each distinct sequence.

• The calculation for no heads is straightforward; it's the probability of tails coming up three times in a row.
• With one head, we encounter three distinct sequences that can achieve this result, so we multiply the probability of one such sequence by three.
• The same logic applies to the case of two heads.
• Finally, for three heads, we only have one sequence: HHH.

It's essential in such problems to systematically account for all possible outcomes and to understand that the sum of the probabilities of all mutually exclusive outcomes always equals 1.

Random Variable

In probabilistic terms, a random variable is a numerical description of the outcome of a statistical experiment. In our coin toss scenario, the random variable \(X\) represents the number of heads resulting from three coin tosses. \(X\) is a discrete random variable because it takes on a limited number of distinct values, specifically 0, 1, 2, or 3 in this case.

Characterizing each outcome of the experiment with a numeric value allows us to calculate various probabilities associated with these outcomes. It provides a systematic framework for tallying outcomes and simplifies the communication of probabilistic findings. As you improve your understanding, take away the practical use of random variables—they allow us to transform random outcomes into measurable quantities upon which we can perform mathematical operations and make predictions or informed decisions.

Independent Events

The concept of independent events plays a pivotal role in calculating probabilities in scenarios like coin tosses. Two events are said to be independent if the occurrence of one does not influence the occurrence of the other. Each coin toss in our question is an independent event—what happened in previous tosses has no effect on the outcome of the next toss.

Independence is profoundly useful because it simplifies calculations. We can compute the probability of multiple independent events occurring together by multiplying the probabilities of each individual event. For instance,

• to determine the probability of getting three heads in a row (\(P(X=3)\)), we multiply the probability of getting a head on each independent toss (\(P(H)\)),
• the probability of getting two heads and one tail in any order for \(P(X=2)\) gets computed similarly.

Grasping the independence of events helps avoid misconceptions, such as the gambler's fallacy—the incorrect belief that past events can influence the randomness of future events.