GOZUNTA
Try it now
All courses Geometry · S-MD.7 50 of 57
Analyze decisions and strategies with probability concepts in applied settings.

Find the expected value of a game from its outcomes and probabilities

Problem
Compute the expected value as \(0.20(\$10)+0.80(-\$2)\), treating the loss as a negative outcome.
Your answer
Show answer choicesHide answer choices Work through the mathematics first, then compare your reasoning.Select the answer that matches your work.
With a free account

Read the steps or watch the work.

Every problem included with your free account keeps its hints, complete written solution, and step-by-step learning video available for review.

Create a free account
With paid access

Take the session to paper.

Export learning sessions as spacious printable PDFs when pencil-and-paper practice works better.

Compare plans

Hint

Multiply each outcome by its probability, then add the results.

Expected value is a probability-weighted average: \(E=\sum (\text{outcome})(\text{probability})\).

Solution walkthrough

01

Use net signed outcomes

\[x_1=+10\text{ dollars},\ p_1=0.20;\qquad~x_2=-2\text{ dollars},\ p_2=0.80\]

Winning contributes positive ten dollars, while losing contributes negative two dollars. Their probabilities sum to 1.

02

Write the expected-value model

\[E=\sum~\text{xp}=(0.20)(10)+(0.80)(-2)\]

Expected value is the probability-weighted average of all net outcomes, so each signed dollar result is multiplied by its probability.

03

Evaluate both contributions

\[(0.20)(10)=2.00;\qquad(0.80)(-2)=-1.60\]

The winning contribution is positive two dollars and the losing contribution is negative one dollar sixty cents.

04

Combine and interpret

\[E=2.00-1.60=0.40\text{ dollars}\]

The game has a positive expected net result of forty cents per play in the long run; this is an average, not a guaranteed single-play payoff.

+

Another way

  1. Use cents: 0.20(1000)+0.80(-200)=200-160=40 cents.

!

Common mistake

Do not treat the two-dollar loss as positive; its negative sign is essential in the weighted average.

Solution walkthrough video