Probability and counting — SHSAT practice questions
Probability and counting on the SHSAT Mathematics exam covers finding chances for one or more events and enumerating favorable outcomes, including a spinner paired with a die, marbles left after some are removed, and numbered tiles that meet one condition or another. A student has to write a ratio of desired outcomes to possible outcomes, multiply for independent and events, add for or events while avoiding double-counted overlap, and recompute the total after items are set aside. Items are multiple choice, with answers usually written as simplified fractions, and common traps reuse the original bag total or treat overlapping cases as disjoint.
Section: Mathematics 57 questions on the paper
241 practice questions in the app
Sample questions
Question 1
A box contains 20 tiles numbered 1 through 20. If one tile is chosen at random, what is the probability that the number on the tile is divisible by 2 or divisible by 5?
A21
B53
C32
D43
Show answer
Answer: B — 53
There are 10 multiples of 2 and 4 multiples of 5 from 1 through 20. The 2 multiples of 10 were counted twice, so there are 10+4−2=12 favorable tiles. The probability is 2012=53.
Question 2
A spinner is divided into 5 equal-sized sections: 3 blue, 1 red, and 1 green. Elena spins the spinner once and then separately rolls a standard six-sided die once. What is the probability that the spinner lands on blue AND the die shows a number greater than 4?
A101
B152
C51
D1514
Show answer
Answer: C — 51
P(blue) = 3/5 and P(die greater than 4) = P(5 or 6) = 2/6 = 1/3. Since the spin and the roll are independent, multiply the probabilities: 3/5 x 1/3 = 1/5.
Question 3
A bag contains 45 marbles, 9 of which are red. Without looking, Malik removes 15 marbles from the bag, 4 of which turn out to be red, and sets them aside. The remaining marbles stay in the bag. What is the probability that a marble drawn at random from the marbles remaining in the bag will be red?
A91
B152
C61
D103
Show answer
Answer: C — 61
After removing 15 marbles (4 of them red), 45 - 15 = 30 marbles remain in the bag, and 9 - 4 = 5 of those are red, so P(red) = 5/30 = 1/6.
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