Problem 33 - Calculus for Biology and Medicine

Question

33\. Toss a fair coin 10 times. Let \(X\) be the number of heads. Find. (a) \(P(X=5)\). (b) \(P(X \geq 8)\). (c) \(P(X \leq 9)\).

Step-by-Step Solution

Answer

(a) 0.246, (b) 0.0547, (c) 0.99902.

Step 1: Define the Experiment

The problem involves tossing a fair coin 10 times, where each toss results in either a head (H) or a tail (T). The random variable \(X\) represents the number of heads obtained in these 10 tosses.

Step 2: Identify the Distribution

Since each coin toss is independent and has the same probability of resulting in a head, \(X\) follows a binomial distribution with parameters \(n = 10\) (number of trials) and \(p = 0.5\) (probability of getting a head on one toss).

Step 3: Calculate P(X=5)

The probability of getting exactly 5 heads, \(P(X=5)\), can be calculated using the binomial probability formula: \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]Plug in \(n = 10\), \(k = 5\), and \(p = 0.5\):\[P(X=5) = \binom{10}{5} (0.5)^5 (0.5)^5 = \binom{10}{5} (0.5)^{10}\]Calculating further:\[P(X=5) = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \times (0.5)^{10} = 252 \times \frac{1}{1024} \approx 0.246\]

Step 4: Calculate P(X ≥ 8)

To find \(P(X \geq 8)\), we sum the probabilities of getting 8, 9, and 10 heads using the binomial formula:\[P(X \geq 8) = P(X=8) + P(X=9) + P(X=10)\]\[P(X=k) = \binom{10}{k} (0.5)^{k} (0.5)^{10-k} = \binom{10}{k} (0.5)^{10}\]Calculating each term:\[P(X=8) = \binom{10}{8} (0.5)^{10} = 45 \cdot \frac{1}{1024} = 0.0439\]\[P(X=9) = \binom{10}{9} (0.5)^{10} = 10 \cdot \frac{1}{1024} = 0.0098\]\[P(X=10) = \binom{10}{10} (0.5)^{10} = 1 \cdot \frac{1}{1024} = 0.00098\]Adding these probabilities:\[P(X \geq 8) = 0.0439 + 0.0098 + 0.00098 = 0.0547\]

Step 5: Calculate P(X ≤ 9)

Finding \(P(X \leq 9)\) directly is easier using the concept of complementary probability:\[P(X \leq 9) = 1 - P(X=10)\]Since \(P(X=10) = 0.00098\), we have:\[P(X \leq 9) = 1 - 0.00098 = 0.99902\]

Key Concepts

Probability Theory

Probability theory is a branch of mathematics that deals with the likelihood of different outcomes occurring in experiments or situations involving uncertainty. The probability of an event is a number between 0 and 1, where 0 means the event cannot happen, and 1 means it will certainly happen. To understand this better, let’s consider a coin toss experiment. Each toss has two possible outcomes: heads or tails.
- The probability of getting heads in a fair coin toss is 0.5 (since there are two possible outcomes, each equally likely).
- If we toss the coin multiple times, like in our exercise with 10 coin tosses, we use probability theory to determine the likelihood of observing a certain number of heads. This is calculated using the binomial distribution formula, which takes into account the number of trials and the probability of success on each trial.

Random Variables

A random variable is a variable that takes on different numerical values based on the outcomes of a random process. In this context, a random variable is often used to quantify outcomes. For example, when tossing a coin 10 times, the number of heads that appear can be represented as a random variable, denoted by \(X\).
- In our exercise, random variable \(X\) represents the number of heads obtained in 10 tosses. It can take on integer values from 0 to 10, inclusive.
- Random variables can be either discrete or continuous, but in our case, all possible results (like 2, 5, or 8 heads) are distinct values, making \(X\) a discrete random variable.
By modeling the problem with a random variable \(X\), we can apply mathematical tools, such as probability distributions, to analyze and predict the outcomes.

Combinatorics

Combinatorics is a field of mathematics that studies the counting, arrangement, and combination of objects. It's particularly useful when determining the number of ways an event can occur, which is essential in calculating probabilities. In our exercise, we used combinatorics to find how many ways we can get a certain number of heads in 10 coin tosses.
- The binomial coefficient \(\binom{n}{k}\) is fundamental in this calculation. It represents the number of combinations of \(n\) items taken \(k\) at a time, regardless of order.
- For instance, to find the probability of getting exactly 5 heads in 10 tosses, we calculate \(\binom{10}{5}\). This gives us the number of possible sequences that result in exactly 5 heads.
Using combinatorics simplifies the process of breaking down complex probability problems into manageable calculations.

Independent Events

Independent events are events where the occurrence or outcome of one event does not affect the outcome of another. This concept is crucial when dealing with multiple events or actions, such as successive coin tosses.
- Each coin toss in the exercise is considered independent; the result of one toss does not influence the result of the next.
- This independence means that the probability of getting a head remains constant at 0.5 for each toss, regardless of previous outcomes.
Understanding the independence of events helps us use the binomial distribution effectively, as it requires each trial to have the same probability of success and to be independent from one another.