Question

In Australia, $16\mathrm{\%}$ of the adult population is near- sighted. 17 If three Australians are chosen at random, what is the probability that (a) two are nearsighted and one is not? (b) exactly one is nearsighted? (c) at most one is nearsighted? (d) none of them are nearsighted?

Short answer

The probabilities are (a) 0.064512, (b) 0.338688, (c) 0.931392, (d) 0.592704.

Step-by-step solution

Understand the problem

The problem is asking us to find the probability of various combinations of nearsighted people within a group of three randomly chosen Australians. Since 16% of the adult population is near-sighted, the probability of a randomly chosen adult being nearsighted is 0.16, and the probability of them not being nearsighted is 0.84.

Calculate probability for part (a)

To find the probability that two are nearsighted and one is not, we can use the binomial probability formula, where n=3 (number of trials), k=2 (number of successes, i.e. nearsighted individuals), p=0.16 (probability of success), and q=0.84 (probability of failure). The formula for binomial probability is $P(X=k)=(\genfrac{}{}{0ex}{}{n}{k})\cdot p^k\cdot q^{n-k}$. We calculate $P(2)=(\genfrac{}{}{0ex}{}{3}{2})\cdot (0.16{)}^2\cdot (0.84{)}^1$.

Calculate probability for part (b)

Similarly, to find the probability that exactly one person is nearsighted, we apply the binomial formula with k=1: $P(1)=(\genfrac{}{}{0ex}{}{3}{1})\cdot (0.16{)}^1\cdot (0.84{)}^2$.

Calculate probability for part (c)

To find the probability that at most one person is nearsighted, we need to add the probability of having exactly one nearsighted person and the probability of having none. So we calculate $P(0)=(\genfrac{}{}{0ex}{}{3}{0})\cdot (0.16{)}^0\cdot (0.84{)}^3$ and add it to our result from Step 3.

Calculate probability for part (d)

To find the probability that none of them are nearsighted, we use the formula with k=0: $P(0)=(\genfrac{}{}{0ex}{}{3}{0})\cdot (0.16{)}^0\cdot (0.84{)}^3$. This gives us the probability of all three being not nearsighted.

Perform the calculations

Now we will perform the actual calculations for each part. Using a calculator or performing the computations by hand, we compute the binomial coefficients and probabilities:(a) $P(2)=(\genfrac{}{}{0ex}{}{3}{2})\cdot {0.16}^2\cdot 0.84=3\cdot 0.0256\cdot 0.84=0.064512$(b) $P(1)=(\genfrac{}{}{0ex}{}{3}{1})\cdot 0.16\cdot {0.84}^2=3\cdot 0.16\cdot 0.7056=0.338688$(c) $P(\text{at most one})=P(1)+P(0)=0.338688+(\genfrac{}{}{0ex}{}{3}{0})\cdot (0.84{)}^3=0.338688+{0.84}^3=0.338688+0.592704=0.931392$(d) $P(0)=(0.84{)}^3=0.592704$

Key concepts

Binomial Probability Formula

The binomial probability formula is a fundamental concept in probability theory, especially when dealing with discrete random variables.
This formula provides a way to calculate the probability of achieving exactly 'k' successes in 'n' independent trials, where the result of each trial can be classified as a success or failure. The general formula is given by:
$$P(X=k)=(\genfrac{}{}{0ex}{}{n}{k})\cdot p^k\cdot q^{n-k}$$
where $(\genfrac{}{}{0ex}{}{n}{k})$ is the binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$, representing the number of ways to choose 'k' successes from 'n' trials. The term 'p' stands for the probability of success on any given trial, and 'q' is the probability of failure, with $q=1-p$.

Applying the Formula

To apply this formula, we need to identify the probability of success 'p' and understand that each trial is independent, meaning the outcome of one trial does not affect the others. In the exercise given, the success is defined as finding a nearsighted individual, with a success probability of 0.16. Then, we calculate the scenario of interest—such as two successes out of three trials—to find the specific probability.

Probability of Nearsightedness

In probability, when we are tasked to deal with real-world statistics such as the probability of nearsightedness within a population, we can use the principles of probability to make quantifiable predictions.
In our example involving the percentage of nearsighted individuals in Australia, understanding the probability of a single event helps us to calculate more complex scenarios involving multiple individuals. The probability of an adult being nearsighted in this scenario is 0.16, which we will utilize in our binomial probability calculations as 'p'.

Contextual Understanding

It's important to note that these probabilities are based on past data and assume that the selection of individuals is random and independent. The assumption of independence is critical; in real life, factors such as genetics, environment, and social interactions could influence the likelihood of nearsightedness, but for the purpose of our calculations, we simplify and consider each choice as independent.

Combinatorics in Probability

Combinatorics plays an integral role in probability, especially when we need to determine the number of ways in which events can occur. It allows us to calculate the binomial coefficients that are essential in the binomial probability formula.
The field of combinatorics provides the tools to count the combinations and permutations of sets, which is essential in calculating the likelihood of specific outcomes.

Combinations vs. Permutations

In the context of probability, combinations are used when the order of events does not matter, while permutations are used when the order is significant. For calculating the probabilities in our example, we use combinations because the order of selecting nearsighted individuals does not change the outcome. The binomial coefficient represents the number of combinations of 'k' successes in 'n' trials and is a key component in determining the probability of those successes.