Problem 59 - A First Course in Probability

Question

If you buy a lottery ticket in 50 lotteries, in each of which your chance of winning a prize is \(\frac{1}{100},\) what is the (approximate) probability that you will win a prize (a) at least once? (b) exactly once? (c) at least twice?

Step-by-Step Solution

Answer

The approximate probabilities are: (a) winning at least once: \(1 - \left(\frac{99}{100}\right)^{50}\), (b) winning exactly once: \(\binom{50}{1} \left(\frac{1}{100}\right)^1 \left(\frac{99}{100}\right)^{49}\), and (c) winning at least twice: \(1 - \left(\frac{99}{100}\right)^{50} - \binom{50}{1} \left(\frac{1}{100}\right)^1 \left(\frac{99}{100}\right)^{49}\).

Step 1: (a) Probability of winning at least once

We can solve this problem by finding the probability of not winning any prize and then subtracting it from 1 to get the probability of winning at least one prize. The probability of not winning a single lottery is \(1-\frac{1}{100} = \frac{99}{100}\), so the probability of not winning any prize in all 50 lotteries is the product of the probabilities of not winning each lottery: $$q^n = \left(\frac{99}{100}\right)^{50}$$ Now, we can find the probability of winning at least one prize by subtracting this value from 1: $$P(X\geq 1) = 1-q^n = 1 - \left(\frac{99}{100}\right)^{50}$$

Step 2: (b) Probability of winning exactly once

To calculate the probability of winning exactly one prize, we can use the Binomial Probability Formula and let \(k=1\): $$P(X=1) = {n\choose 1}p^1q^{(n-1)} = \binom{50}{1} \left(\frac{1}{100}\right)^1 \left(\frac{99}{100}\right)^{49}$$

Step 3: (c) Probability of winning at least twice

One way to find the probability of winning at least twice can be to find the probabilities of winning exactly 0 and 1 times and then subtracting their sum from 1. We already computed the probability of winning 0 times which is \(q^n\) and the probability of winning 1 time which is \(P(X=1)\). So the probability of winning at least twice is: $$P(X\geq 2) = 1 - P(X=0) - P(X=1) = 1 - q^n - P(X=1) = 1 - \left(\frac{99}{100}\right)^{50} - \binom{50}{1} \left(\frac{1}{100}\right)^1 \left(\frac{99}{100}\right)^{49}$$

Key Concepts

Binomial Probability Formula

Understanding the binomial probability formula is crucial for solving problems involving a fixed number of independent trials, like drawing lottery tickets. In our lottery example, you buy a ticket for multiple lotteries, each with a set chance of winning. To determine the probability of winning exactly once, the binomial probability formula is used, which is expressed as:

\[\begin{equation}P(X=k) = {n \choose k}p^kq^{(n-k)}\end{equation}\]
where:

• \( n \) is the number of trials, or lotteries
• \( k \) is the number of desired successes, in this case, winning
• \( p \) is the probability of success on a single trial
• \( q \) is the probability of failure on a single trial, which equals \( 1-p \)

For example, to find the probability of winning exactly once when you play 50 lotteries, you'd use the formula with \( n=50 \), \( k=1 \), \( p=1/100 \), and \( q=99/100 \). The formula calculates all the different ways one win can occur across all trials and the probability of each occurrence.

Complementary Probability

When we speak of complementary probability, we refer to the probability that an event does not occur. It's a critical concept because sometimes it's easier to calculate the probability of an event not happening and subtract it from 1 to find the probability that it does happen, as in our lottery scenario.

The complementary probability of winning at least once involves first finding the likelihood of not winning at all. In our example, this was computed by raising the chance of losing a single lottery, \( q = 99/100 \), to the power of the number of lotteries played, or \( q^{n} \). The complement, the probability of at least one win, is then \( 1 - q^n \). This approach simplifies calculations, especially when dealing with a large number of trials.

Binomial Distribution

Binomial distribution describes the behavior of a series of independent trials, each with the same probability of success. In our exercise, purchasing lottery tickets and looking for a winning one is an example of such trials. Binomial distribution helps to find the probabilities of having various numbers of wins. It centers around two parameters: the number of trials (\( n \)) and the probability of success for an individual trial (\( p \)).

With this distribution, you can calculate the probabilities of all possible numbers of wins—whether it's none, exactly one, or more than one. In the case of wanting at least two wins in 50 lotteries, it helps to use the cumulative property of binomial distribution, summarizing the probability of one or fewer wins and subtracting from 1 to find the probability of two or more wins.