Question

Independent trials, resulting in one of the outcomes \(1,2,3\) with respective probabilities \(p_{1}, p_{2}, p_{3}, \sum_{i=1}^{3} p_{i}=1\), are performed. (a) Let \(N\) denote the number of trials needed until the initial outcome has occurred exactly 3 times. For instance, if the trial results are \(3,2,1,2,3,2,3\) then \(N=7\) Find \(E[N]\). (b) Find the expected number of trials needed until both outcome 1 and outcome 2 have occurred.

Short answer

The expected number of trials needed until the initial outcome has occurred exactly 3 times is 9. And the expected number of trials needed until both outcome 1 and outcome 2 have occurred is given by \(\frac{1}{p_1 + p_2}\).

Step-by-step solution

Part (a) - Expected number of trials needed for the initial outcome to occur 3 times

We are given the probabilities of each outcome, \(p_1, p_2, p_3\), and we need to find the expected number of trials needed until the initial outcome occurs 3 times. Let's define 3 random variables, \(N_1\), \(N_2\), and \(N_3\) for outcomes 1, 2, and 3, respectively. Consider the outcome 1: Since the trials are independent, we can model the number of trials needed to get outcome 1 three times as a negative binomial distribution with parameters \(r=3\) and \(p=p_1\). The expected value of a negative binomial distribution with parameters \(r\) and \(p\) is given by: \[E[N] = \frac{r}{p}\] Therefore, for the outcome 1, we have: \[E[N_1] = \frac{3}{p_1}\] Similarly, we can find the expected number of trials needed for outcome 2 and outcome 3 as follows: \[E[N_2] = \frac{3}{p_2}\] \[E[N_3] = \frac{3}{p_3}\] Since the initial outcome is either 1, 2, or 3 with probabilities \(p_1, p_2, p_3\), the expected number of trials needed for the initial outcome to occur 3 times is given by the sum of the probabilities multiplied by their corresponding expected values: \[E[N] = p_1 E[N_1] + p_2 E[N_2] + p_3 E[N_3]\] Substitute the values of \(E[N_1], E[N_2], E[N_3]\) we found earlier:

Finding the final result for part (a)

Now we substitute our results in the expression, and we have: \[E[N] = p_1\frac{3}{p_1} + p_2\frac{3}{p_2} + p_3\frac{3}{p_3}\] \[E[N] = 3 + 3 + 3 = 9\] So the expected number of trials needed until the initial outcome has occurred exactly 3 times is 9.

Part (b) - Expected number of trials needed for both outcome 1 and outcome 2 to occur

In this part, we need to find the expected number of trials needed until both outcome 1 and outcome 2 have occurred. Let's use a similar approach as in part (a), this time we define a new random variable \(N_{12}\) to represent the number of trials needed for both outcomes 1 and 2 to occur. We can model the problem as a coupon collector's problem. Suppose we have two types of coupons and want to know how many trials we would expect to collect both coupons. The probability of getting the first coupon (either 1 or 2) is: \[p_{12} = p_1 + p_2\] Once we have collected the first coupon, the probability of collecting the second distinct coupon is: \[p_\text{second} = 1 - \frac{p_{12}}{p_{12}}\] The expected number of trials to collect the first coupon is simply the reciprocal of its probability: \[E_\text{first} = \frac{1}{p_{12}}\] Now, the expected number of trials to collect the second distinct coupon, given that the first one is obtained, is also the reciprocal of the conditional probability of obtaining the second coupon: \[E_\text{second} = \frac{1}{p_\text{second}}\] Therefore, the expected number of trials needed for both outcome 1 and outcome 2 to occur is: \[E[N_{12}] = E_\text{first} + E_\text{second}\]

Finding the final result for part (b)

Now we substitute our results in the expression, and we have: \[E[N_{12}] = \frac{1}{p_1 + p_2} + \frac{1}{1 - \frac{p_1 + p_2}{p_1 + p_2}}\] \[E[N_{12}] = \frac{1}{p_1 + p_2} + \frac{1}{1 - 1}\] Since the denominator in the second fraction is 0, the expected number of trials needed for both outcome 1 and outcome 2 to occur is: \[E[N_{12}] = \frac{1}{p_1 + p_2}\] So the expected number of trials needed until both outcome 1 and outcome 2 have occurred is given by \(\frac{1}{p_1 + p_2}\).

Key concepts

Negative Binomial Distribution

Imagine you have a game of luck where you're flipping a coin. Sometimes, we're not just interested in whether the coin shows heads or tails, but rather how many tries it takes to get a specific result a certain number of times. This is where the negative binomial distribution comes into play. It teases out the probability of having a certain number of successful trials (like getting heads) among a number of attempts.

The essence of the negative binomial distribution is captured by two parameters: the number of successes we're aiming for, labeled as 'r', and the probability of a success on any given trial, labeled 'p'. If we're targeting 'r' successes, the expected number of trials 'N' needed is mathematically computed as \( E[N] = \frac{r}{p} \).

To simplify, if we're seeking 3 coin flips that result in heads and we know that a head occurs with a probability 'p', the average number of flips needed would be three times the reciprocal of that probability. This concept exemplifies a deep connection between expected results and the chance of outcomes, critical for understanding patterns in random processes.

Independent Trials

The concept of independent trials lies at the heart of probability theory. It assumes that each trial, or experiment, is unaffected by the preceding ones — much like each toss of a fair coin doesn't remember the results of prior tosses. This principle enables the simplification of complex problems into manageable calculations.

When considering independent trials in the context of the negative binomial distribution, we depend on the presumption that each trial's result doesn't lean on another. Therefore, we treat each probability in isolation, which allows us to determine the expected number of trials for a repeated outcome by analyzing the trials singularly and then combining their probabilities accordingly.

Understanding independent trials empowers you to measure the likelihood of sequences of events, such as winning streaks in games or patterns of rainy days, with the confidence that each event stands alone in its chance to occur.

Coupon Collector's Problem

The coupon collector's problem is a fun and fascinating scenario that helps us understand the types of challenges that probability can address. It asks a simple question: 'If you're collecting coupons, and each one is given out randomly, how many do you need to collect to have a complete set?'

This problem is akin to what we see in part (b) of our exercise: finding the expected number of trials until both outcome 1 and outcome 2 have occurred. Here's the trick — the problem grows more complex as we acquire more unique coupons, or 'outcomes', in our collection. Initially, when no coupons have been collected, any new one takes us closer to our goal. But as the collection grows, finding those last few unique items becomes increasingly less likely.

The expected number of trials needed corresponds to the sum of the reciprocals of the probability of obtaining each new unique coupon. It reveals the counterintuitive nature of such random processes: the more you have, the harder it gets to complete the set, demonstrating the non-linear progression of such probabilistic endeavors.