Expected value from probability (ACT). Game on.

Expected value from probability (ACT) is a grade 9 math skill aligned to Common Core standard HSS.MD.A.2. Below are 8 practice questions with answers and step-by-step explanations, drawn from the 10 expected value from probability (act) problems our math games drill.

CCSS HSS.MD.A.210 questions in the bank
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Warm-upeasy

A raffle ticket pays $10\text{\char36}10, $5\text{\char36}5, or nothing. Winning $10\text{\char36}10 is twice as likely as winning $5\text{\char36}5, and winning nothing is equally likely as winning $5\text{\char36}5. What is the expected value of one ticket?

Weights are 22, 11, and 11 (total 44), so probabilities are 24\dfrac{2}{4}, 14\dfrac{1}{4}, and 14\dfrac{1}{4}. Then E=∑piviE = \sum p_i v_i, which means the sum of each probability times its value: E=24(10)+14(5)+14(0)=$6.25E = \dfrac{2}{4}(10) + \dfrac{1}{4}(5) + \dfrac{1}{4}(0) = \text{\char36}6.25.

Mid-gameeasy

A school store prize is $12\text{\char36}12, $4\text{\char36}4, or nothing. The $12\text{\char36}12 prize is twice as likely as the $4\text{\char36}4 prize, and winning nothing is equally likely as the $4\text{\char36}4 prize. What is the expected value?

Weights are 22, 11, and 11 (total 44), so probabilities are 24\dfrac{2}{4}, 14\dfrac{1}{4}, and 14\dfrac{1}{4}. Then E=∑piviE = \sum p_i v_i, which means the sum of each probability times its value: E=24(12)+14(4)+14(0)=$7E = \dfrac{2}{4}(12) + \dfrac{1}{4}(4) + \dfrac{1}{4}(0) = \text{\char36}7.

Mid-gameeasy

A lemonade stand earns $15\text{\char36}15 on a busy day, $5\text{\char36}5 on a slow day, or $0\text{\char36}0 if it rains. A busy day is equally likely as a slow day, and a rainy day is twice as likely as a slow day. What is the expected profit?

Weights are 11, 11, and 22 (total 44), so probabilities are 14\dfrac{1}{4}, 14\dfrac{1}{4}, and 24\dfrac{2}{4}. Then E=∑piviE = \sum p_i v_i, which means the sum of each probability times its value: E=14(15)+14(5)+24(0)=$5E = \dfrac{1}{4}(15) + \dfrac{1}{4}(5) + \dfrac{2}{4}(0) = \text{\char36}5.

Mid-gameeasy

On a bonus quiz question, a student earns 44 points, 11 point, or 00 points. Earning 44 points is twice as likely as earning 11 point, and earning 00 points is 33 times as likely as earning 11 point. What is the expected number of bonus points?

Weights are 22, 11, and 33 (total 66), so probabilities are 26\dfrac{2}{6}, 16\dfrac{1}{6}, and 36\dfrac{3}{6}. Then E=∑piviE = \sum p_i v_i, which means the sum of each probability times its value: E=26(4)+16(1)+36(0)=96=1.5E = \dfrac{2}{6}(4) + \dfrac{1}{6}(1) + \dfrac{3}{6}(0) = \dfrac{9}{6} = 1.5 points.

Mid-gameeasy

One spin of a prize wheel pays $6\text{\char36}6, $2\text{\char36}2, or nothing. Winning $6\text{\char36}6 is 55 times as likely as winning $2\text{\char36}2, and winning nothing is equally likely as winning $2\text{\char36}2. What is the expected payout?

Weights are 55, 11, and 11 (total 77), so probabilities are 57\dfrac{5}{7}, 17\dfrac{1}{7}, and 17\dfrac{1}{7}. Then E=∑piviE = \sum p_i v_i, which means the sum of each probability times its value: E=57(6)+17(2)+17(0)=327≈$4.57E = \dfrac{5}{7}(6) + \dfrac{1}{7}(2) + \dfrac{1}{7}(0) = \dfrac{32}{7} \approx \text{\char36}4.57.

Mid-gameeasy

A scratch card pays $10\text{\char36}10, $5\text{\char36}5, or nothing. Winning $10\text{\char36}10 is twice as likely as winning $5\text{\char36}5, and winning nothing is 33 times as likely as winning $5\text{\char36}5. What is the expected value of one card?

Weights are 22, 11, and 33 (total 66), so probabilities are 26\dfrac{2}{6}, 16\dfrac{1}{6}, and 36\dfrac{3}{6}. Then E=∑piviE = \sum p_i v_i, which means the sum of each probability times its value: E=26(10)+16(5)+36(0)=256≈$4.17E = \dfrac{2}{6}(10) + \dfrac{1}{6}(5) + \dfrac{3}{6}(0) = \dfrac{25}{6} \approx \text{\char36}4.17.

Mid-gameeasy

A prize drawing awards $20\text{\char36}20, $10\text{\char36}10, or nothing. Winning $20\text{\char36}20 is equally likely as winning $10\text{\char36}10, and winning nothing is 33 times as likely as winning $10\text{\char36}10. What is the expected value of the prize?

Weights are 11, 11, and 33 (total 55), so probabilities are 15\dfrac{1}{5}, 15\dfrac{1}{5}, and 35\dfrac{3}{5}. Then E=∑piviE = \sum p_i v_i, which means the sum of each probability times its value: E=15(20)+15(10)+35(0)=$6E = \dfrac{1}{5}(20) + \dfrac{1}{5}(10) + \dfrac{3}{5}(0) = \text{\char36}6.

Buzzer beatereasy

A game chest gives $9\text{\char36}9, $3\text{\char36}3, or nothing. Winning $9\text{\char36}9 is twice as likely as winning $3\text{\char36}3, and winning nothing is 44 times as likely as winning $3\text{\char36}3. What is the expected value of one opening?

Weights are 22, 11, and 44 (total 77), so probabilities are 27\dfrac{2}{7}, 17\dfrac{1}{7}, and 47\dfrac{4}{7}. Then E=∑piviE = \sum p_i v_i, which means the sum of each probability times its value: E=27(9)+17(3)+47(0)=217=$3E = \dfrac{2}{7}(9) + \dfrac{1}{7}(3) + \dfrac{4}{7}(0) = \dfrac{21}{7} = \text{\char36}3.

Tests that cover expected value from probability (act):
ACTPARCCNJSLAMaine MTA
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