Suppose we have an urn containing 5 yellow balls and 7 green balls. We draw 3 balls, without replacement, from the urn. Find the expected number of yellow balls drawn.

References

Doyle, Peter G. (2006). "Grinstead and Snell's Introduction to Probability" (PDF). Retrieved June 6, 2024.

Solution: E

The number of yellows balls drawn has probability distribution:

[[math]] P(N=k) = \frac{\binom{5}{k} \binom{7}{3-k}}{\binom{12}{3}}. [[/math]]

Calculating the above gives:

Probability distribution for the number of yellow balls drawn

k	P(N=k)
1	0.4773
2	0.3182
3	0.0455

And therefore the expected value equals:

0.4773 + 0.3182*2 + 0.0455*3 = 1.2502