Question

round all answers to 4 decimal places.
a. a bag contains 6 yellow marbles, 7 blue marbles, and 9 green marbles. if a marble is drawn from the bag, replaced, and another marble is drawn, what is the probability of drawing first a yellow marble and then a green marble?
b. a bag contains 7 black marbles, 9 green marbles, and 6 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

Total marbles in part a = 6 + 7+ 9 = 22.

Step2: Calculate probability of drawing yellow first in part a

The probability of drawing a yellow marble first, $P(Y_1)=\frac{6}{22}$.

Step3: Calculate probability of drawing green second in part a

Since the marble is replaced, the probability of drawing a green marble second, $P(G_2)=\frac{9}{22}$.

Step4: Calculate combined probability in part a

The probability of drawing a yellow then a green (independent - with replacement) is $P = P(Y_1)\times P(G_2)=\frac{6}{22}\times\frac{9}{22}=\frac{54}{484}\approx0.1116$.

Step5: Calculate total marbles in part b

Total marbles in part b = 7 + 9+ 6 = 22.

Step6: Calculate probability of drawing black first in part b

The probability of drawing a black marble first, $P(B_1)=\frac{7}{22}$.

Step7: Calculate probability of drawing blue second in part b

Since the first marble is set - aside, there are 21 marbles left. So the probability of drawing a blue marble second, $P(B_2)=\frac{6}{21}$.

Step8: Calculate combined probability in part b

The probability of drawing a black then a blue (dependent - without replacement) is $P = P(B_1)\times P(B_2)=\frac{7}{22}\times\frac{6}{21}=\frac{42}{462}\approx0.0909$.

Answer:

a. 0.1116
b. 0.0909