Inkerval

Joint distributions practice problems

You work with two random variables from one experiment. You add table cells to get marginals. You divide a cell by a row total to get a conditional. You multiply marginals to test independence. For continuous variables you integrate a density over a region. A problem belongs here when it tracks two quantities at once, for example a table of p(x,y) values or a density f(x,y). You will practice: Joint PMF and marginals, Conditional distributions, Independence of random variables, Densities on rectangles.

11 problems, each with a worked answer. Work them on paper, in order — the difficulty ramps gently.

1.

A food truck logs each order's X = number of tacos (1 or 2) and Y = whether a drink was added (0 = no, 1 = yes). Joint PMF: p(1,0) = 0.20, p(1,1) = 0.30, p(2,0) = 0.25, p(2,1) = 0.25. Find p_X(1), the probability an order has exactly one taco.

Show answer

Answer:

Marginalize over y by summing the row for x equal to 1. Summing the row removes Y from the picture.

2.

A two-question quiz records X = 1 if question 1 was answered correctly (else 0) and Y likewise for question 2. Known joint values: p(0,0) = 0.15, p(0,1) = 0.25, p(1,0) = 0.35. Find p(1,1).

Show answer

Answer:

All cells of a joint PMF sum to 1. The three known cells total 0.75.

3.

On a random day, X = 1 if it rains (else 0) and Y = 1 if it is windy (else 0), with p(0,0) = 0.4, p(0,1) = 0.2, p(1,0) = 0.1, p(1,1) = 0.3. Find P(Y = 1 | X = 1).

Show answer

Answer:

First find the marginal of X at 1. Then renormalize the row where X is 1.

4.

A dart lands uniformly at random on a rectangular board 0 ≤ X ≤ 4, 0 ≤ Y ≤ 2. Find P(X ≤ 1 and Y ≤ 1).

Show answer

Answer:

Uniform on a rectangle means probability equals the area of the event over the area of the board. The event is a 1 by 1 square.

5.

At a gym, a random visit is recorded with X = 1 if the member uses weights (else 0) and Y = 1 if the member uses cardio machines (else 0). The joint PMF is p(0,0) = 0.1, p(0,1) = 0.3, p(1,0) = 0.4, p(1,1) = 0.2. Find P(X = 1).

Show answer

Answer:

Add the joint probabilities across the row X = 1.

6.

A pet store records each customer with X = 1 if they buy food (else 0) and Y = 1 if they buy a toy (else 0). Three joint values are known: p(0,0) = 0.35, p(0,1) = 0.15, p(1,0) = 0.30. Find p(1,1).

Show answer

Answer:

All four joint probabilities must sum to 1.

7.

A commuter's bus is late with probability 0.2 and her train is late with probability 0.15, independently of each other. Let X and Y be the lateness indicators (1 = late, 0 = on time). Find P(X = 1 and Y = 0).

Show answer

Answer:

By independence, multiply P(X = 1) by P(Y = 0).

8.

Two judges taste the same dish and each gives it a score of 1 or 2. Let X and Y be their scores, with joint PMF p(1,1) = 0.35, p(1,2) = 0.15, p(2,1) = 0.05, p(2,2) = 0.45. Find the probability that the two judges give the same score.

Show answer

Answer:

P(X = Y) is the sum of the diagonal entries of the table.

9.

A delivery drone lands at a point (X, Y) uniformly distributed over a rectangular field given by 0 ≤ X ≤ 10 and 0 ≤ Y ≤ 4, in meters. Find P(X > 8 and Y < 2).

Show answer

Answer:

For a uniform density on a rectangle, probability equals favorable area over total area.

10.

A coffee shop logs each order with X = 1 if it is an espresso drink (else 0) and Y = 1 if it uses oat milk (else 0). The joint PMF is p(0,0) = 0.30, p(0,1) = 0.20, p(1,0) = 0.35, p(1,1) = 0.15. Find P(Y = 1 | X = 1).

Show answer

Answer:

First get the marginal P(X = 1), then divide the joint value by it.

11.

Let p(x,y) = c(x + y) for x ∈ {1, 2} and y ∈ {1, 2}, and p(x,y) = 0 otherwise. Find the constant c that makes p a valid joint PMF.

Show answer

Answer:

The four values of x + y are 2, 3, 3, 4. The total probability must equal 1.

Inkerval is handwritten math practice for iPad and Apple Pencil. You solve by hand; it remembers what you know and brings it back when it should.

$29 a month, after a 14-day free trial. The waitlist is open now.

Be first in when the doors open. No spam, ever.

More probability practice