Probability #5 - Questions and Answers

Question 1 Consider a Poisson distribution with a mean of two occurrences per time period. Compute the probability of two occurrences in one time period. Answer Given: μ = 2 per time period Formula Poisson probability: f(k) = (μ^k * e^(-μ)) / k! f(2) = (2² * e^(-2)) / 2! ≈ 0.2707 Question 2 Two fair dice are rolled. Find the joint probability mass function of X and Y when (a) X is the largest value obtained on any die and Y is the sum of the values; (b) X is the value on the first die and Y is the larger of the two values; (c) X is the smallest and Y is the largest value obtained on the dice. Answer 1 Define N₁ and N₂ as random variables that mark numbers obtained on the first and the second die. We know that N₁ and N₂ are independent and that N₁,N₂ a) Here we have that X = max (N₁,N₂) and Y = N₁ + N₂. Thus X ∈ {1, ...,6} and Y ∈ {2, ..., 12}. Also we have that X < Y almost certainly. So take any k < l, where k and l are from the ranges given above. Consider event X = k, Y = l. That means that the maximum value on any die is k and that the sum on both dice is l. Observe that the only possible pairs of (N₁,N₂) corresponding to that event are (k, l - k) and (l - k, k) if l < 2k. If l = 2k, the only possible pair is (k,k). Hence the required PMF is P(X = k, Y = l) = {2/36, k < l < 2k {1/36, l = 2k Answer 2 Here we have that X = N₁ and Y = max (N₁, N₂). Observe that both variables are in {1,..., 6} and that X ≤ Y almost certainly. Take any k ≤ l from the range given above. We have that P(X = k, Y = l) = P(Y = l | X = k)P(X = k) Suppose that k < l and that we are given X = k. In that case, N₂ can be any number from the range 1,..., k to obtain the required Y = l. Hence P(Y = l | X = k)P(X = k) = k/36 if k=l If k < l and we are given that X = k, N₂ must be equal to l to obtain Y = l. So, in that case P(Y = l | X = k)P(X = k) = 1/36 Answer 3 Here we have that X = min(N₁, N₂) and Y = max(N₁, N₂). We also have that X ≤ Y almost certainly. So, take any k ≤ l Suppose that k < l. In this case we have to have N₁ = k, N₂ = l or N₁ = l, N₂ = k. So, there are only two possibilities, hence P(X = k, Y = l) = 2/36 if k=l, the only possibility is (N₁, N₂) = (k, k), thus P(X = k, Y = l) = 1/36 Question 3 A bag has 9 red marbles and 6 green marbles. How do you find the probability of drawing a green marble, replacing it, and drawing another green marble? Answer First, we have the probability of drawing a green marble as 6 out of (6 + 9), which is 6/15 After that, we replace the green marble into the bag. At this moment, the bag becomes the same as it is before a green marble is drawn, i.e. there are still 6 green marbles and 9 red ones. If we are to draw another green marble, the same probability, 6/15 will be resulted. As both requirements have to be fulfilled, the required probability is the multiplication of both, (6/15) * (6/15) = 36/225 = 4/25 Question 4 Consider the following discrete probability distribution. What is the probability that x equals 7? x 3 4 7 9 P(X= x) 0.22 0.17 ? 0.18 Answer Explanation: The Probability Distribution of the Random Variable X, is x 3 4 7 9 P(X= x) 0.22 0.17 ? 0.18 We know that for any Probability Distribution: ∑ P(X = x) = 1 Hence: P(X = 3) + P(X = 4) + P(X = 7) + P(X = 9) = 1 ∴ 0.22 + 0.17 + P(X = 7) + 0.18 = 1 ∴ 0.57 + P(X = 7) = 1 ∴ P(X = 7) = 0.43 Question 5 The spinner is spun twice. Two possible outcomes are AC and CA. a. Create a sample space that lists all of the equally likely outcomes. b. Refer to the sample space to find the probability of spinning A one or more times in two spins. Answer a.Any of the three letters can be an outcome in each spin so the sample space is: AA,AB,AC,BA,BB,BC,CA,CB,CC b.Check for outcomes that have one or two A's. There are 5 outcomes that have one or two A's out of the sample space of 9. So, the probability is: 5/9