g.pr.10.6 (mc)

florence wants to play a game at a casino where she will roll a fair die and spin a spinner with the colors red, blue, green, and yellow evenly spread out. the game costs \$5 to play and there are 3 ways to win.

* roll a 1 or 2 and spin red or yellow-win \$2
* roll a 3 and spin blue or red-win \$20
* roll an even number and spin a yellow-win \$5

find the probability of not winning a prize.

$\bigcirc$ $\frac{1}{3}$
$\bigcirc$ $\frac{5}{8}$
$\bigcirc$ $\frac{2}{3}$
$\bigcirc$ $\frac{3}{4}$

Question

g.pr.10.6 (mc)

florence wants to play a game at a casino where she will roll a fair die and spin a spinner with the colors red, blue, green, and yellow evenly spread out. the game costs \$5 to play and there are 3 ways to win.

* roll a 1 or 2 and spin red or yellow-win \$2
* roll a 3 and spin blue or red-win \$20
* roll an even number and spin a yellow-win \$5

find the probability of not winning a prize.

$\bigcirc$ $\frac{1}{3}$
$\bigcirc$ $\frac{5}{8}$
$\bigcirc$ $\frac{2}{3}$
$\bigcirc$ $\frac{3}{4}$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Independent Events", "Probability Formulas", "Complementary Events" ], "new_concepts": [], "current_concepts": [ "Independent Events", "Probability Formulas", "Complementary Events" ] } ## Calculate the probability of each winning event $$ \begin{aligned} P(\text{Win 1}) &= P(\text{Roll } 1 \text{ or } 2) \times P(\text{Spin red or yellow}) = \frac{2}{6} \times \frac{2}{4} = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \ P(\text{Win 2}) &= P(\text{Roll } 3) \times P(\text{Spin blue or red}) = \frac{1}{6} \times \frac{2}{4} = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12} \ P(\text{Win 3}) &= P(\text{Roll even}) \times P(\text{Spin yellow}) = \frac{3}{6} \times \frac{1}{4} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} \end{aligned} $$ ## Sum the probabilities of winning any prize $$ \begin{aligned} P(\text{Win}) &= P(\text{Win 1}) + P(\text{Win 2}) + P(\text{Win 3}) \ &= \frac{1}{6} + \frac{1}{12} + \frac{1}{8} \ &= \frac{4}{24} + \frac{2}{24} + \frac{3}{24} = \frac{9}{24} = \frac{3}{8} \end{aligned} $$ ## Calculate the probability of not winning a prize $$ P(\text{Not Win}) = 1 - P(\text{Win}) = 1 - \frac{3}{8} = \frac{5}{8} $$

(A) $\frac{1}{3}$

(B) $\frac{5}{8}$

(C) $\frac{2}{3}$

(D) $\frac{3}{4}$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Statistics", "Probability Formulas" ] }

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Question

Explanation:

Calculate the probability of each winning event

Sum the probabilities of winning any prize

Calculate the probability of not winning a prize

Answer: