Expected distinct colours in a three-candy grab

Question

A bag holds $10$ candies: $1$ green, $1$ yellow, $2$ red and $6$ blue. You take $3$ candies at random without replacement. What is the expected number of different colours among them?

Short answer

$2.1$ . Sum indicators that each colour appears. A colour with $k$ candies is missed with probability $\binom{10-k}{3}/\binom{10}{3}$ , so the four colours appear with probabilities $\tfrac{116}{120},\tfrac{64}{120},\tfrac{36}{120},\tfrac{36}{120}$ , totalling $\tfrac{252}{120}=2.1$ .

Adapted from QuantQuestions.io; question and answer rewritten by SelfMadeQuant.com.

See the hint and the full worked solution in the question bank (free account).

Open the worked solution

Lessons behind this question

Expectation, Mean, and Variance
Conditional Expectation and Variance Revisited
Sample Spaces, Events, and the Probability Axioms

Start with a free lesson: Expectation, Variance and Covariance is part of a free topic. Create a free account, no card needed, to read it in full with worked solutions.

Read it free