Question

round all answers to 4 decimal places.
a. a bag contains 5 green marbles, 9 blue marbles, and 10 white marbles. if a marble is drawn from the bag, replaced, and another marble is drawn, what is the probability of drawing first a green marble and then a white marble?
b. a bag contains 9 black marbles, 10 white marbles, and 5 blue marbles. if a marble is drawn from the bag, set aside, and another marble is drawn, what is the probability of drawing first a black marble and then a blue marble?

Explanation:

Step1: Calculate total marbles in part a

The total number of marbles in the first - bag is $5 + 9+10=24$.

Step2: Calculate probability of drawing a green marble first

The probability of drawing a green marble on the first draw is $P(G)=\frac{5}{24}$ since there are 5 green marbles out of 24.

Step3: Calculate probability of drawing a white marble second (with replacement)

Since the marble is replaced, the total number of marbles remains 24. The probability of drawing a white marble on the second draw is $P(W)=\frac{10}{24}$.

Step4: Calculate the joint - probability for part a

Since the two draws are independent events (because of replacement), the probability of drawing a green marble first and then a white marble is $P(G)\times P(W)=\frac{5}{24}\times\frac{10}{24}=\frac{50}{576}\approx0.0868$.

Step5: Calculate total marbles in part b

The total number of marbles in the second - bag is $9 + 10+5 = 24$.

Step6: Calculate probability of drawing a black marble first

The probability of drawing a black marble on the first draw is $P(B_1)=\frac{9}{24}$.

Step7: Calculate probability of drawing a blue marble second (without replacement)

After drawing a black marble and setting it aside, there are 23 marbles left. The probability of drawing a blue marble on the second draw is $P(B_2)=\frac{5}{23}$.

Step8: Calculate the joint - probability for part b

Since the two draws are dependent events (because of no replacement), the probability of drawing a black marble first and then a blue marble is $P(B_1)\times P(B_2)=\frac{9}{24}\times\frac{5}{23}=\frac{45}{552}\approx0.0815$.

Answer:

a. $0.0868$
b. $0.0815$