NJSLA Algebra I Using Probability to Make Decisions Practice

Practice playeasy

A scratch card pays $\text{\char36}10$ , $\text{\char36}5$ , or nothing. Winning $\text{\char36}10$ is twice as likely as winning $\text{\char36}5$ , and winning nothing is $3$ times as likely as winning $\text{\char36}5$ . What is the expected value of one card?

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

Practice playeasy

A prize drawing awards $\text{\char36}20$ , $\text{\char36}10$ , or nothing. Winning $\text{\char36}20$ is equally likely as winning $\text{\char36}10$ , and winning nothing is $3$ times as likely as winning $\text{\char36}10$ . What is the expected value of the prize?

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

Practice playeasy

A game chest gives $\text{\char36}9$ , $\text{\char36}3$ , or nothing. Winning $\text{\char36}9$ is twice as likely as winning $\text{\char36}3$ , and winning nothing is $4$ times as likely as winning $\text{\char36}3$ . What is the expected value of one opening?

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

Practice playeasy

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

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

Practice playeasy

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

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

Practice playeasy

A lemonade stand earns $\text{\char36}15$ on a busy day, $\text{\char36}5$ on a slow day, or $\text{\char36}0$ 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 $1$ , $1$ , and $2$ (total $4$ ), so probabilities are $\dfrac{1}{4}$ , $\dfrac{1}{4}$ , and $\dfrac{2}{4}$ . Then $E = \sum p_i v_i$ , which means the sum of each probability times its value: $E = \dfrac{1}{4}(15) + \dfrac{1}{4}(5) + \dfrac{2}{4}(0) = \text{\char36}5$ .

Practice playeasy

On a bonus quiz question, a student earns $4$ points, $1$ point, or $0$ points. Earning $4$ points is twice as likely as earning $1$ point, and earning $0$ points is $3$ times as likely as earning $1$ point. What is the expected number of bonus points?

Weights are $2$ , $1$ , and $3$ (total $6$ ), so probabilities are $\dfrac{2}{6}$ , $\dfrac{1}{6}$ , and $\dfrac{3}{6}$ . Then $E = \sum p_i v_i$ , which means the sum of each probability times its value: $E = \dfrac{2}{6}(4) + \dfrac{1}{6}(1) + \dfrac{3}{6}(0) = \dfrac{9}{6} = 1.5$ points.

Practice playeasy

One spin of a prize wheel pays $\text{\char36}6$ , $\text{\char36}2$ , or nothing. Winning $\text{\char36}6$ is $5$ times as likely as winning $\text{\char36}2$ , and winning nothing is equally likely as winning $\text{\char36}2$ . What is the expected payout?

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