A glass jar contains a total of $46$ marbles. The jar has yellow, blue, and green marbles. There are $4$ blue marbles, $5$ green marbles, and the rest are yellow. Determine which event has a probability closest to $1$ . $\newline$ A. Picking a green marble $\newline$ B. Picking a yellow marble $\newline$ C. Picking a red marble $\newline$ D. Picking a blue marble

View step-by-step help

Home

Math Problems

Algebra 2

Identify independent events

Full solution

Q. A glass jar contains a total of

46

marbles. The jar has yellow, blue, and green marbles. There are

4

blue marbles,

5

green marbles, and the rest are yellow. Determine which event has a probability closest to

1

\newline

A. Picking a green marble

\newline

B. Picking a yellow marble

\newline

C. Picking a red marble

\newline

D. Picking a blue marble

Find Yellow Marbles: First, let's find out the total number of yellow marbles since we know the total number of marbles and the number of blue and green marbles. $\newline$ Total marbles = $46$ $\newline$ Blue marbles = $4$ $\newline$ Green marbles = $5$ $\newline$ Yellow marbles = Total marbles - (Blue marbles + Green marbles) $\newline$ Yellow marbles = $46 - (4 + 5)$ $\newline$ Yellow marbles = $46 - 9$ $\newline$ Yellow marbles = $37$
Calculate Probabilities: Now, let's calculate the probability of picking a marble of each color. $\newline$ The probability of picking a green marble $P(\text{Green})$ is the number of green marbles divided by the total number of marbles. $\newline$ $P(\text{Green}) = \frac{\text{Number of green marbles}}{\text{Total number of marbles}}$ $\newline$ $P(\text{Green}) = \frac{5}{46}$
Compare Probabilities: The probability of picking a yellow marble $P(\text{Yellow})$ is the number of yellow marbles divided by the total number of marbles. $P(\text{Yellow}) = \frac{\text{Number of yellow marbles}}{\text{Total number of marbles}}$ $P(\text{Yellow}) = \frac{37}{46}$
Event with Closest Probability to $1$ : The probability of picking a blue marble $P(\text{Blue})$ is the number of blue marbles divided by the total number of marbles. $P(\text{Blue}) = \frac{\text{Number of blue marbles}}{\text{Total number of marbles}}$ $P(\text{Blue}) = \frac{4}{46}$
Event with Closest Probability to $1$ : The probability of picking a blue marble $P(\text{Blue})$ is the number of blue marbles divided by the total number of marbles. $P(\text{Blue}) = \frac{\text{Number of blue marbles}}{\text{Total number of marbles}}$ $P(\text{Blue}) = \frac{4}{46}$ Since there are no red marbles mentioned in the problem, the probability of picking a red marble $P(\text{Red})$ is zero. $P(\text{Red}) = \frac{0}{46}$ $P(\text{Red}) = 0$
Event with Closest Probability to $1$ : The probability of picking a blue marble $P(\text{Blue})$ is the number of blue marbles divided by the total number of marbles. $P(\text{Blue}) = \frac{\text{Number of blue marbles}}{\text{Total number of marbles}}$ $P(\text{Blue}) = \frac{4}{46}$ Since there are no red marbles mentioned in the problem, the probability of picking a red marble $P(\text{Red})$ is zero. $P(\text{Red}) = \frac{0}{46}$ $P(\text{Red}) = 0$ To find the event with the probability closest to $1$ , we compare the probabilities calculated for green, yellow, blue, and red marbles. $P(\text{Green}) = \frac{5}{46}$ $P(\text{Yellow}) = \frac{37}{46}$ $P(\text{Blue}) = \frac{4}{46}$ $P(\text{Red}) = 0$ The probability closest to $1$ is $P(\text{Yellow}) = \frac{37}{46}$ .