Compound probability drills

Compute probabilities of compound and independent events.

Free, timed SHSAT practice sets on compound probability— every question includes a full worked explanation, and there's no account required.

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How to solve compound probability problems

For "this and that", multiply the probabilities of each step.

Quick check: an "and" probability is smaller than each single probability. Common traps: adding instead of multiplying, keeping the same denominator on the second draw without replacement, and finding "both are X" when the question asks for "neither is X".

Worked examples

Example 1.
Bag A contains 6 yellow, 11 white, 8 purple, 8 red marbles, and Bag B contains 12 black, 10 blue, 11 orange marbles. One marble is drawn at random from each bag. What is the probability that the marble from Bag A is yellow AND the marble from Bag B is blue?
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Answer:
A

The two draws are independent, so multiply: P(A is yellow) \(\times\) P(B is blue) = \(\dfrac{2}{11}\) \(\times\) \(\dfrac{10}{33}\) = \(\dfrac{20}{363}\). (Adding the two probabilities instead of multiplying gives \(\dfrac{16}{33}\). Putting both bags together (\((6 + 10) \div (33 + 33)\)) gives \(\dfrac{8}{33}\). Stopping after the Bag A probability gives \(\dfrac{2}{11}\).)

Example 2.
A jar contains 8 green, 9 blue, 2 black marbles. Two marbles are drawn at random, one at a time, WITHOUT replacement. What is the probability that both marbles are green?
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Answer:
D

First draw: P(green) = \(\dfrac{8}{19}\). Since the marble is NOT replaced, the second draw has only 7 green left out of 18 total: P(green) = \(\dfrac{7}{18}\). Multiply: \(\dfrac{28}{171}\). (Putting the marble back (\(\dfrac{8}{19} \times \dfrac{8}{19}\)) gives \(\dfrac{64}{361}\). Adding the two probabilities instead of multiplying gives \(\dfrac{277}{342}\). Taking 1 away from the green count but not from the total (\(\dfrac{8}{19} \times \dfrac{7}{19}\)) gives \(\dfrac{56}{361}\).)

Example 3.
A basket contains 7 blue, 2 white, 10 black marbles. Two marbles are drawn at random, one at a time, WITHOUT replacement. What is the probability that NEITHER marble is blue?
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Answer:
A

Non-blue marbles: 12 of 19. First draw: \(\dfrac{12}{19}\). Without replacement, second draw: \(\dfrac{11}{18}\). Multiply: \(\dfrac{22}{57}\). (Finding the probability that both ARE blue gives \(\dfrac{7}{57}\). Putting the first marble back (\(\dfrac{12}{19} \times \dfrac{12}{19}\)) gives \(\dfrac{144}{361}\). Taking 1 minus the probability that both are blue gives \(\dfrac{50}{57}\).)