Problem 2 - Calculus for Biology and Medicine

Question

Toss a fair coin four times. Let \(X\) be the random variable that counts the number of heads. Find the probability mass function describing the distribution of \(X\).

Step-by-Step Solution

Answer

The probability mass function is: \(P(X=0)=\frac{1}{16}, P(X=1)=\frac{1}{4}, P(X=2)=\frac{3}{8}, P(X=3)=\frac{1}{4}, P(X=4)=\frac{1}{16}\).

Step 1: Understand the experiment

The experiment involves tossing a fair coin four times. Each toss results in either a Head (H) or Tail (T). The random variable \(X\) counts the number of Heads obtained in these four tosses.

Step 2: Identify possible values for X

The random variable \(X\) can take values from 0 to 4, corresponding to 0, 1, 2, 3, or 4 Heads in four tosses.

Step 3: Use the binomial distribution formula

The probability of getting exactly \(k\) Heads out of \(n\) coin tosses (where each toss is independent) is given by the binomial distribution: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]Here, \(n = 4\) (four tosses), \(k\) is the number of heads, and \(p = 0.5\) (probability of getting a Head in a toss).

Step 4: Calculate probabilities for each X value

We calculate the probability for each possible number of Heads:- \(X = 0\): \[ P(X = 0) = \binom{4}{0}(0.5)^0 (0.5)^4 = 1 \times 1 \times \frac{1}{16} = \frac{1}{16} \]- \(X = 1\): \[ P(X = 1) = \binom{4}{1} (0.5)^1(0.5)^3 = 4 \times \frac{1}{2} \times \frac{1}{8} = \frac{4}{16} = \frac{1}{4} \]- \(X = 2\): \[ P(X = 2) = \binom{4}{2} (0.5)^2(0.5)^2 = 6 \times \frac{1}{4} \times \frac{1}{4} = \frac{6}{16} = \frac{3}{8} \]- \(X = 3\): \[ P(X = 3) = \binom{4}{3} (0.5)^3(0.5)^1 = 4 \times \frac{1}{8} \times \frac{1}{2} = \frac{4}{16} = \frac{1}{4} \]- \(X = 4\): \[ P(X = 4) = \binom{4}{4} (0.5)^4 (0.5)^0 = 1 \times \frac{1}{16} \times 1 = \frac{1}{16} \]

Step 5: Present the probability mass function

The probability mass function for \(X\) is:- \(P(X = 0) = \frac{1}{16}\)- \(P(X = 1) = \frac{1}{4}\)- \(P(X = 2) = \frac{3}{8}\)- \(P(X = 3) = \frac{1}{4}\)- \(P(X = 4) = \frac{1}{16}\)

Key Concepts

Binomial Distribution

The binomial distribution is crucial when dealing with scenarios where you have a fixed number of repetitive trials, like coin tosses. It helps determine the probability of achieving a certain number of successful outcomes. In our exercise, tossing a fair coin four times and counting the number of heads is an example of a binomial distribution.

Here's why:

• The number of trials (\(n\)) is fixed, which is four tosses in this case.
• There are only two possible outcomes for each trial: Head or Tail. These binary outcomes make the scenario a classic example of a binomial experiment.
• The probability of getting a head (\(p\)) remains constant at 0.5 for each toss, assuming a fair coin.

The binomial distribution formula helps us compute the probability of getting exactly 'k' heads out of 'n' coin tosses. This is given by:
\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]
Where \(\binom{n}{k}\) is the binomial coefficient determining how many ways you can achieve 'k' successes. This beautiful formula is central when analyzing probability in repetitive trials.

Random Variable

A random variable is a mathematical concept used to quantify outcomes of a random phenomenon. It is essentially a function that assigns numerical values to the results of a random experiment. In the context of our coin toss experiment, we introduce the random variable \(X\).

Here's how \(X\) is set up:

• \(X\) counts the number of heads obtained when tossing the coin four times.
• \(X\) can take any integer value from 0 to 4 in this experiment — ranging from getting no heads at all to getting all four heads.

Random variables like \(X\) help to structure and analyze data outcomes effectively. They bridge the gap between qualitative ("how many heads did we get?") and quantitative ("assigning numerical probabilities") analysis, allowing for a more comprehensive understanding of the situation.

Coin Toss Experiment

A coin toss is a simple and common experiment often used in probability exercises due to its straightforward nature. This experiment perfectly demonstrates the key features of a binomial distribution, as each toss of a fair coin has an independent and consistent chance of resulting in a head or a tail.

Let's break down the process:

• Each toss acts as a single trial, with two potential outcomes: Head or Tail.
• The probability of landing a head is 0.5, and similarly, the probability for a tail is also 0.5.
• In our four-toss experiment, these independent and identical trials make calculating probabilities efficient using binomial distribution principles.

Such experiments are foundational, not only in classroom settings but in real-world applications where binary events occur. The simplicity and clarity of a coin toss make it an ideal element for probability studies.

Probability Calculation

Understanding how to calculate probabilities in a binomial setting requires applying the appropriate formulas and grasping the underlying mechanics of probability. For the coin toss experiment, we are tasked with finding the probability of achieving a specific number of heads.

Here's a step-by-step approach to calculating these probabilities:

• Start by recognizing the number of experiments or trials, which is four in our scenario.
• Identify the probability of success (\(p\)), which is 0.5 for a head per toss.
• Use the binomial distribution formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\).

For instance, when calculating \(P(X = 2)\), you'll find 2 heads in 4 tosses:
- Calculate \(\binom{4}{2}\), resulting in 6 ways to get two heads.
- Multiply by the probability of getting heads twice: \((0.5)^2 = 0.25\).
- Multiply by the probability of getting two tails (or non-heads): \((0.5)^2 = 0.25\).
The result is \[P(X = 2) = 6 \times 0.25 \times 0.25 = 0.375\] or \(\frac{3}{8}\).
Knowing how to apply these basics allows accurate probability predictions and enriches your understanding of statistical methods.