One-event probability word problems. Game on.

One-event probability word problems is a grade 7 math skill aligned to Common Core standard 7.SP.C.7: develop a probability model and use it to find probabilities of events. Below are 8 practice questions with answers and step-by-step explanations, drawn from the 10 one-event probability word problems problems our math games drill.

CCSS 7.SP.C.710 questions in the bank
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Warm-upeasy

A jar contains 55 red tokens and 77 blue tokens. One token is chosen at random. What is the probability that the token is red?

There are 5+7=125 + 7 = 12 tokens in all and 55 are red, so P(red)=512P(\text{red}) = \dfrac{5}{12}.

Mid-gameeasy

A game spinner is divided into 1212 equal sections. Exactly 44 sections are gold and the rest are silver. What is the probability that the spinner lands on gold on one spin?

P(gold)=412=13P(\text{gold}) = \dfrac{4}{12} = \dfrac{1}{3}.

Mid-gameeasy

One letter is chosen at random from the word STUDENT. What is the probability that the letter is a vowel?

STUDENT has 77 letters. The vowels are U and E, so P(vowel)=27P(\text{vowel}) = \dfrac{2}{7}.

Mid-gameeasy

A bowl contains 44 lemons and 99 limes. One piece of fruit is chosen at random. What is the probability that the fruit is a lime?

There are 4+9=134 + 9 = 13 pieces of fruit and 99 are limes, so P(lime)=913P(\text{lime}) = \dfrac{9}{13}.

Mid-gameeasy

A standard six-sided number cube is rolled once. What is the probability of rolling an odd number?

The odd faces are 11, 33, and 55, so P(odd)=36=12P(\text{odd}) = \dfrac{3}{6} = \dfrac{1}{2}.

Mid-gamemedium

A club has 2424 members. Of these members, 99 play soccer. If one member is selected at random, what is the probability that the member plays soccer?

P(soccer)=924=38P(\text{soccer}) = \dfrac{9}{24} = \dfrac{3}{8}.

Mid-gamemedium

A school raffle has 5050 identical tickets sold. Maya buys 55 of the tickets. One winning ticket is drawn at random. What is the probability that Maya wins?

Maya holds 55 of the 5050 tickets, so P(Maya wins)=550=110P(\text{Maya wins}) = \dfrac{5}{50} = \dfrac{1}{10}.

Buzzer beatermedium

A whole number from 11 through 2525 is chosen at random. What is the probability that the number is a multiple of 44?

The multiples of 44 from 11 to 2525 are 4,8,12,16,20,4, 8, 12, 16, 20, and 2424, so P=625P = \dfrac{6}{25}.

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