Which event is most likely to occur? $\newline$ Rolling an odd number on a twelve-sided die, numbered from $1$ to $12$ . $\newline$ Spinning a spinner divided into five equal-sized sections colored red/green/yellow/blue/purple and landing on red or blue or green. $\newline$ Winning a raffle that sold a total of $100$ tickets, if you buy $41$ tickets. $\newline$ Reaching into a bag full of $33$ strawberry chews and $7$ cherry chews without looking and pulling out a strawberry chew.

View step-by-step help

Home

Math Problems

Precalculus

Mean, variance, and standard deviation of binomial distributions

Full solution

Q. Which event is most likely to occur?

\newline

Rolling an odd number on a twelve-sided die, numbered from

1

12

\newline

Spinning a spinner divided into five equal-sized sections colored red/green/yellow/blue/purple and landing on red or blue or green.

\newline

Winning a raffle that sold a total of

100

tickets, if you buy

41

tickets.

\newline

Reaching into a bag full of

33

strawberry chews and

7

cherry chews without looking and pulling out a strawberry chew.

Event $1$ Analysis: Let's analyze each event to determine its probability. $\newline$ Event $1$ : Rolling an odd number on a twelve-sided die. $\newline$ There are $6$ odd numbers ( $1$ , $3$ , $5$ , $7$ , $9$ , $11$ ) on a twelve-sided die. $\newline$ Probability of rolling an odd number = Number of odd outcomes / Total number of outcomes $\newline$ = $6 / 12$ $\newline$ = $1 / 2$
Event $2$ Analysis: Event $2$ : Spinning a spinner divided into five equal-sized sections and landing on red or blue or green. $\newline$ There are $3$ favorable outcomes (red, blue, green) out of $5$ possible outcomes. $\newline$ Probability of landing on red or blue or green = Number of favorable outcomes / Total number of outcomes $\newline$ = $\frac{3}{5}$
Event $3$ Analysis: Event $3$ : Winning a raffle that sold a total of $100$ tickets, if you buy $41$ tickets. $\newline$ Probability of winning the raffle = Number of tickets you have / Total number of tickets $\newline$ = $\frac{41}{100}$
Event $4$ Analysis: Event $4$ : Reaching into a bag with $33$ strawberry chews and $7$ cherry chews and pulling out a strawberry chew. $\newline$ Probability of pulling out a strawberry chew $= \frac{\text{Number of strawberry chews}}{\text{Total number of chews}}$ $\newline$ $= \frac{33}{(33 + 7)}$ $\newline$ $= \frac{33}{40}$
Comparison: Now, let's compare the probabilities to determine which event is most likely to occur. $\newline$ Event $1$ : Probability = $\frac{1}{2} = 0.5$ $\newline$ Event $2$ : Probability = $\frac{3}{5} = 0.6$ $\newline$ Event $3$ : Probability = $\frac{41}{100} = 0.41$ $\newline$ Event $4$ : Probability = $\frac{33}{40} = 0.825$ $\newline$ The event with the highest probability is Event $4$ , pulling out a strawberry chew.