Question

lesson 20.3 checkpoint
once you have completed the above problems and checked your solutions, complete the lesson checkpoint
below.
complete the lesson reflection above by circling your current understanding of the learning goal.
there are 4 green, 10 red, and 6 yellow marbles in a bag. each time you randomly choose a marble, you
put it aside before choosing another marble at random.

1. find the probability that you choose three red marbles. choose the option that demonstrates the correct

work to find the probability.
a. \\( \frac { 10 } { 20 } \cdot \frac { 9 } { 20 } \cdot \frac { 8 } { 20 } \\)
b. \\( \frac { 10 } { 20 } \cdot \frac { 10 } { 20 } \cdot \frac { 10 } { 20 } \\)
c. \\( \frac { 10 } { 20 } \cdot \frac { 9 } { 19 } \cdot \frac { 8 } { 18 } \\)

1. based on the work below, what probability is trying to be found?

\\( \frac { 6 } { 20 } \cdot \frac { 5 } { 19 } \\)
a. the probability that the 1st 2 marbles are yellow and green.
b. the probability that the 1st 2 marbles are all yellow.
c. the probability that the 1st 2 marbles are all different.

Explanation:

1. Probability of choosing three red marbles

Step1: Calculate total number of marbles

Total marbles \(=4 + 10+6=20\).

Step2: Probability of first - red marble

Probability of choosing first red marble \(P(R_1)=\frac{10}{20}\) (10 red out of 20 total).

Step3: Probability of second - red marble

After one red is chosen, there are 9 red and 19 total marbles. So \(P(R_2)=\frac{9}{19}\).

Step4: Probability of third - red marble

After two reds are chosen, there are 8 red and 18 total marbles. So \(P(R_3)=\frac{8}{18}\).

Step5: Use multiplication rule for dependent events

\(P(R_1\cap R_2\cap R_3)=\frac{10}{20}\times\frac{9}{19}\times\frac{8}{18}\)

2. Analyze \(\frac{6}{20}\cdot\frac{5}{19}\)

Step1: First - marble probability

There are 6 yellow marbles out of 20. So \(P(Y_1)=\frac{6}{20}\).

Step2: Second - marble probability

After one yellow is chosen, there are 5 yellow marbles out of 19. So \(P(Y_2)=\frac{5}{19}\).

Step3: Use multiplication rule for dependent events

\(P(Y_1\cap Y_2)=\frac{6}{20}\times\frac{5}{19}\) (probability that first two marbles are yellow)

Answer:

1. C. \(\frac{10}{20}\cdot\frac{9}{19}\cdot\frac{8}{18}\)
2. B. The probability that the \(1^{st}\) 2 marbles are all yellow.