Simple probability drills

Compute the probability of a single event.

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

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How to solve simple probability problems

For equally likely outcomes:

\(P(\text{event}) = \dfrac{\text{favorable outcomes}}{\text{total outcomes}}\)

Example: a bag has 3 red and 5 blue marbles. \(P(\text{red}) = \dfrac{3}{3 + 5} = \dfrac{3}{8}\).

Common traps: putting the total on top (the answer to a probability question is never more than 1), dividing by the "other" count (3 out of 5) instead of the total (3 out of 8), and giving P(red) when the question asks for P(not red).

Worked examples

Example 1.
A bag contains 10 orange, 2 black, 8 white, 12 blue marbles. One marble is drawn at random. What is the probability that it is orange?
Show answer and explanation
Answer:
C

P(orange) = (favorable) \(\div\) (total) = 10 \(\div\) 32 = \(\dfrac{5}{16}\). (Turning the fraction upside down (\(32 \div 10\)) gives \(\dfrac{16}{5}\). Finding the probability of NOT orange (\(22 \div 32\)) gives \(\dfrac{11}{16}\). Comparing orange to the other marbles (\(10 \div 22\)) gives \(\dfrac{5}{11}\).)

Example 2.
A spinner has 18 equal sections, and 4 of them are red. What is the probability that one spin lands on a section that is NOT red?
Show answer and explanation
Answer:
A

Non-red sections = 18 \(-\) 4 = 14. P(not red) = 14 \(\div\) 18 = \(\dfrac{7}{9}\). (Finding P(red) instead (\(4 \div 18\)) gives \(\dfrac{2}{9}\). Turning the fraction upside down (\(18 \div 14\)) gives \(\dfrac{9}{7}\). Comparing non-red to red sections (\(14 \div 4\)) gives \(\dfrac{7}{2}\).)

Example 3.
A bag holds 8 white marbles and 12 black marbles. How many white marbles must be added to the bag so that the probability of drawing a white marble at random is \(\dfrac{3}{5}\)?
Show answer and explanation
Answer:
C

After adding \(k\) white marbles there are \(8 + k\) white out of \(20 + k\) in all. Set \(\dfrac{8 + k}{20 + k} = \dfrac{3}{5}\). Cross-multiply: \(5(8 + k) = 3(20 + k)\), so \(k = 10\). Check: \(\dfrac{18}{30} = \dfrac{3}{5}\). (Keeping the total at 20 (\(\dfrac{8 + k}{20} = \dfrac{3}{5}\)) gives 4. Stopping at the new total (\(20 + 10\)) gives 30. Stopping at the new number of white marbles (\(8 + 10\)) gives 18.)