Question

There are three coins in a barrel. These coins, when flipped, will come up heads with respective probabilities \(0.3,0.5,0.7 .\) A coin is randomly selected from among these three and is then flipped ten times. Let \(N\) be the number of heads obtained on the ten flips. (a) Find \(P[N=0\\}\). (b) Find \(P[N=n\\}, n=0,1, \ldots, 10\) (c) Does \(N\) have a binomial distribution? (d) If you win \(\$ 1\) each time a head appears and you lose \(\$ 1\) each time a tail appears, is this a fair game? Explain.

Short answer

\(P(N=0) = \frac{1}{3} [(0.7)^{10} + (0.5)^{10} + (0.3)^{10}]\) \(P(N=n) = \frac{1}{3} [\binom{10}{n} \times (P(N=n|C_1) + P(N=n|C_2) + P(N=n|C_3) )]\) N does not have a binomial distribution because the probability of success is not constant for each trial. This game is not fair overall because the expected value is not the same for each coin. Coin 2 is fair, but coins 1 and 3 are not.

Step-by-step solution

1. Determine the probability of no heads for each coin

First, we need to find the probability of obtaining a tail for each coin. This can be calculated as the complement of each coin's head probability. Coin 1: \(P(T) = 1 - P(H) = 1 - 0.3 = 0.7\) Coin 2: \(P(T) = 1 - P(H) = 1 - 0.5 = 0.5\) Coin 3: \(P(T) = 1 - P(H) = 1 - 0.7 = 0.3\) Now, we need to find the probability of obtaining 0 heads in 10 flips for each coin. Coin 1: \(P(N=0|C_1) = (0.7)^{10}\) Coin 2: \(P(N=0|C_2) = (0.5)^{10}\) Coin 3: \(P(N=0|C_3) = (0.3)^{10}\)

2. Compute P[N=0] using the Law of Total Probability

The probability of obtaining 0 heads is the sum of the probabilities of getting 0 heads for each coin, weighted by the probability of choosing each coin (which is 1/3, since we're selecting randomly from 3 coins). \(P(N=0) = \frac{1}{3} [P(N=0|C_1) + P(N=0|C_2) + P(N=0|C_3)]\) #b. Find P[N=n], n=0,1,…,10#

1. Find the probability of obtaining n heads for each coin

We will use the binomial probability formula: \(P(N=n|C_i) = \binom{10}{n} \times \text{(probability of H for coin }i)^n \times (1 - \text{probability of H for coin }i)^{10-n}\)

2. Compute P[N=n] using the Law of Total Probability

Just as in part (a), we will sum the probabilities of obtaining n heads for each coin, weighted by the probability of choosing each coin (1/3). \(P(N=n) = \frac{1}{3} [P(N=n|C_1) + P(N=n|C_2) + P(N=n|C_3)]\) #c. Does N have a binomial distribution?#

1. Observe the probabilities and parameters of the problem.

A binomial distribution has two parameters: the number of trials (n) and the probability of success (p). However, in this case, the probability of success is not constant, since each coin has a different probability of obtaining a head. This indicates that N does not have a simple binomial distribution because the probability of success is not the same for each trial. #d. If you win \(1 each time a head appears and you lose \)1 each time a tail appears, is this a fair game? Explain.#

1. Calculate the expected value of each coin flip.

For this game to be fair, the expected value (EV) of each flip should be 0. Let's calculate the EV for each coin: EV1 = \(0.3 \times 1 + 0.7 \times (-1) = -0.4\) EV2 = \(0.5 \times 1 + 0.5 \times (-1) = 0\) EV3 = \(0.7 \times 1 + 0.3 \times (-1) = 0.4\)

2. Determine if the game is fair or not.

Since the expected value for each coin is not 0, this game is not fair overall. Coin 2 has an expected value of 0 which makes it a fair game, but coins 1 and 3 are not. This means that, overall, this game is biased as the expected value is not the same for each coin.

Key concepts

Binomial Distribution

The binomial distribution is a probability model that describes the number of successes in a fixed number of independent Bernoulli trials. A Bernoulli trial is an experiment where there are only two possible outcomes, commonly termed as 'success' and 'failure'. For the binomial distribution to be applicable, each trial must have the same probability of success (p), and the trials must be independent, meaning the outcome of one trial does not affect the outcome of another.

In the context of the given exercise, to determine if the number of heads (N) has a binomial distribution, we need to consider whether the conditions for binomial distribution are met. The problem is not simple because we're dealing with different coins having different probabilities of coming up heads—0.3, 0.5, and 0.7—as a result, the probability of success (p) is not constant. This variability in p indicates that the trials (coin flips) are not identical, therefore, N does not follow a traditional binomial distribution.

Law of Total Probability

The law of total probability is a fundamental rule that provides a way to calculate the probability of an event based on several different and exhaustive partitions of the sample space. It is particularly useful when dealing with the probability of outcomes resulting from several different scenarios or conditions.

In layman's terms, it helps us break down complex probability problems into simpler parts that are easier to manage. To apply this law, as seen in the step-by-step solution, we considered all the ways in which we could obtain zero heads (N=0) after ten flips. Since we could randomly select any of the three coins, each with a different head-probability, we used the law of total probability to combine the individual scenarios. Consequently, the probability of obtaining zero heads is the weighted average of the probabilities of each coin not showing heads in 10 flips.

Expected Value

The expected value, often denoted as EV, of a random variable is a measure of the central tendency of its probability distribution and represents the average outcome we would expect if we were to repeat the experiment an infinite number of times. To calculate the expected value, we multiply each outcome by its probability and then sum these products.

From the exercise, the expected value is used to evaluate whether a game is fair. In a fair game, over time, you would neither lose nor gain money, so the EV should be zero. However, the expected value calculations for each coin show that only the second coin with a 0.5 probability results in an EV of 0, while the other coins have a positive or negative expected value. Therefore, the game is not fair since the overall EV is not equal for each coin, and a fair game requires an expected value of zero for any single bet or flip. This is crucial for understanding games of chance and whether or not to engage in them.