Practice Probability practice: 6 questions with answers Six probability practice questions, from easy to hard, with worked explanations. Pick an answer to check it instantly.
These questions come from our SHSAT, TACHS and SSAT practice apps. For parents
Probability measures how likely something is, from 0 (impossible) to 1 (certain). For equally likely outcomes, it comes down to one fraction.
Favorable over total
Probability = number of favorable outcomes ÷ total number of outcomes.
For example, a bag holds 3 red, 5 blue and 2 green marbles, 10 in all. The probability of drawing a blue marble is 5/10, or 1/2.
Two follow-up rules
“Not” means subtract from 1. The probability of not drawing blue is 1 − 1/2 = 1/2.Two independent events: multiply. Flipping heads twice in a row is 1/2 × 1/2 = 1/4. If the first draw isn’t put back, the total changes for the second draw, so recount before you multiply.
The questions start easy and get harder. Try each one before you look at the explanation.
Practice questions Each question and explanation was reviewed by a person. Pick an answer to see if you're right.
Question 1 A jar contains
3 3 3 white marbles and
7 7 7 green marbles, and no other marbles. If one marble is drawn from the jar at random, what is the probability that it is white?
A 3 10 \frac{3}{10} 10 3 B 7 10 \frac{7}{10} 10 7 C 3 7 \frac{3}{7} 7 3 D 1 10 \frac{1}{10} 10 1
The jar holds
3 + 7 = 10 3 + 7 = 10 3 + 7 = 10 marbles in all, and
3 3 3 of them are white, so
P ( white ) = 3 10 P(\text{white}) = \frac{3}{10} P ( white ) = 10 3 . Check:
P ( green ) = 7 10 P(\text{green}) = \frac{7}{10} P ( green ) = 10 7 , and
3 10 + 7 10 = 1 \frac{3}{10} + \frac{7}{10} = 1 10 3 + 10 7 = 1 , as it must.
Question 2 A jar contains 12 marbles: 5 blue, 4 green, and 3 yellow. Priya draws one marble at random, records its color, and puts it back in the jar. She then draws a second marble at random. What is the probability that both marbles she draws are blue?
A 25 144 \dfrac{25}{144} 144 25 B 5 6 \dfrac{5}{6} 6 5 C 25 132 \dfrac{25}{132} 132 25 D 20 132 \dfrac{20}{132} 132 20 E 5 12 \dfrac{5}{12} 12 5
Because the marble is replaced, the two draws are independent and the jar has 12 marbles each time. P(blue on a draw) =
5 12 \frac{5}{12} 12 5 . P(both blue) =
5 12 × 5 12 = 25 144 \frac{5}{12}\times\frac{5}{12}=\frac{25}{144} 12 5 × 12 5 = 144 25 .
Question 3 A quality checker at a factory tested a random sample of 60 flashlights from one day's production and found that 3 of them were defective. The factory produced 2,400 flashlights that day. Based on the sample, about how many of the day's flashlights should be expected to be defective?
A 3 B 40 C 72 D 120 E 800
The experimental probability of a defective flashlight is
3 60 = 1 20 \frac{3}{60} = \frac{1}{20} 60 3 = 20 1 . Applying this rate to the full production:
1 20 × 2400 = 120 \frac{1}{20} \times 2400 = 120 20 1 × 2400 = 120 . Second method: the full production of 2,400 is
2400 ÷ 60 = 40 2400 \div 60 = 40 2400 ÷ 60 = 40 groups the size of the sample, and each group is expected to contain about 3 defective flashlights, giving
40 × 3 = 120 40 \times 3 = 120 40 × 3 = 120 . Sanity check: 1 out of every 20 flashlights is defective, and
120 × 20 = 2400 120 \times 20 = 2400 120 × 20 = 2400 , so an estimate of 120 defective flashlights is consistent.
Question 4 Two fair number cubes, each with faces numbered 1 through 6, are rolled. What is the probability that the product of the two numbers rolled is exactly 12?
A 1 18 \frac{1}{18} 18 1 B 1 12 \frac{1}{12} 12 1 C 1 9 \frac{1}{9} 9 1 D 1 6 \frac{1}{6} 6 1 E 1 3 \frac{1}{3} 3 1
There are
6 × 6 = 36 6 \times 6 = 36 6 × 6 = 36 equally likely ordered outcomes. Factor pairs of 12 using numbers 1-6 are
2 × 6 2 \times 6 2 × 6 and
3 × 4 3 \times 4 3 × 4 ; the pair
1 × 12 1 \times 12 1 × 12 is impossible because 12 is not on a cube. As ordered outcomes these are
( 2 , 6 ) , ( 6 , 2 ) , ( 3 , 4 ) , ( 4 , 3 ) (2,6), (6,2), (3,4), (4,3) ( 2 , 6 ) , ( 6 , 2 ) , ( 3 , 4 ) , ( 4 , 3 ) , which is 4 outcomes, so the probability is
4 36 = 1 9 \frac{4}{36} = \frac{1}{9} 36 4 = 9 1 . Sanity check: scanning the 36-cell multiplication table, exactly the four cells listed hold a 12, confirming
1 9 \frac{1}{9} 9 1 .
Question 5 A team of
3 3 3 students is chosen at random from a group of
5 5 5 boys and
4 4 4 girls. What is the probability that the team has exactly
2 2 2 girls?
A 1 6 \frac{1}{6} 6 1 B 5 42 \frac{5}{42} 42 5 C 10 21 \frac{10}{21} 21 10 D 1 14 \frac{1}{14} 14 1 E 5 14 \frac{5}{14} 14 5
Total ways to choose
3 3 3 of
9 9 9 students:
( 9 3 ) = 84 \binom{9}{3}=84 ( 3 9 ) = 84 . Ways to choose exactly
2 2 2 girls and
1 1 1 boy:
( 4 2 ) ( 5 1 ) = 6 ⋅ 5 = 30 \binom{4}{2}\binom{5}{1}=6\cdot5=30 ( 2 4 ) ( 1 5 ) = 6 ⋅ 5 = 30 . Probability
= 30 84 = 5 14 =\tfrac{30}{84}=\tfrac{5}{14} = 84 30 = 14 5 .
Question 6 On Maria's walk to school she passes
3 3 3 traffic lights. Each light is independently green when she arrives with probability
0.7 0.7 0.7 . What is the probability that she has to stop (light not green) at least once?
A 0.027 0.027 0.027 B 0.343 0.343 0.343 C 0.657 0.657 0.657 D 0.9 0.9 0.9 E 0.973 0.973 0.973
Use the complement. P(never stops) = P(all three green)
= 0.7 3 = 0.343 =0.7^3=0.343 = 0. 7 3 = 0.343 . So P(stops at least once)
= 1 − 0.343 = 0.657 =1-0.343=0.657 = 1 − 0.343 = 0.657 .
For parents
These practice questions come from our SHSAT, TACHS and SSAT practice apps. Each app has more than 1,200 practice questions, every one with a worked explanation.
Checked against official sources. How we make this site Wednesday, September 30, 2026