Study on the flySHSATMathematicsProbability and counting
Mathematics
Probability and counting questions ask how likely an outcome is: favorable outcomes divided by all equally likely outcomes.
Most errors come from miscounting, not from the formula. When two conditions are joined by the word or, outcomes meeting both are counted once, so among multiples of three or multiples of four, twelve is not added twice. Independent events, such as a coin toss and a die roll, combine by multiplying their probabilities.
Some SHSAT Mathematics items take objects out before a draw, so totals must be updated first. Answer choices are usually fractions, several built from uncorrected counts.
Pick an answer to see whether it is right — nothing is saved, and nothing needs an account.
Question 1Mid
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?
Answer: B —
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 favorable tiles. The probability is .
Question 2Mid
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?
Answer: C —
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 3Mid
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?
Answer: C —
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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