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QuestionRhianna has 1 blue, 1 white, and 1 red marble. Find the probabilities of drawing marbles with replacement: a) Blue, then red = b) Red, then white = c) Blue, then blue, then blue =

Studdy Solution

STEP 1

Assumptions1. The bag contains1 blue marble,1 white marble, and1 red marble. . After a marble is drawn, it is returned to the bag before the next draw.
3. Each marble has an equal chance of being drawn.

STEP 2

First, we need to find the total number of marbles in the bag. We can do this by adding the number of each color of marble.
Total marbles=Blue marbles+White marbles+Red marblesTotal\, marbles = Blue\, marbles + White\, marbles + Red\, marbles

STEP 3

Now, plug in the given values for the number of each color of marble to calculate the total number of marbles.
Total marbles=1+1+1Total\, marbles =1 +1 +1

STEP 4

Calculate the total number of marbles in the bag.
Total marbles=1+1+1=3Total\, marbles =1 +1 +1 =3

STEP 5

The probability of drawing a particular color of marble is the number of that color of marble divided by the total number of marbles. Since each color of marble has the same number (1), the probability is the same for each color.
Probability=Number of a particular color of marbleTotal marblesProbability = \frac{Number\, of\, a\, particular\, color\, of\, marble}{Total\, marbles}

STEP 6

Plug in the values for the number of a particular color of marble and the total number of marbles to calculate the probability.
Probability=13Probability = \frac{1}{3}

STEP 7

For part a), we need to find the probability of drawing a blue marble and then a red marble. Since the marbles are returned to the bag after each draw, these are independent events, so we multiply the probabilities.
(A Blue,thenared)=(Blue)×(Red)(A\, Blue, then a red) =(Blue) \times(Red)

STEP 8

Plug in the values for the probabilities of drawing a blue marble and a red marble.
(A Blue,thenared)=13×13(A\, Blue, then a red) = \frac{1}{3} \times \frac{1}{3}

STEP 9

Calculate the probability of drawing a blue marble and then a red marble.
(A Blue,thenared)=3×3=9(A\, Blue, then a red) = \frac{}{3} \times \frac{}{3} = \frac{}{9}

STEP 10

For part b), we need to find the probability of drawing a red marble and then a white marble. Again, these are independent events, so we multiply the probabilities.
(A red,thenawhite)=(Red)×(White)(A\, red, then a white) =(Red) \times(White)

STEP 11

Plug in the values for the probabilities of drawing a red marble and a white marble.
(A red,thenawhite)=3×3(A\, red, then a white) = \frac{}{3} \times \frac{}{3}

STEP 12

Calculate the probability of drawing a red marble and then a white marble.
(A red,thenawhite)=×=9(A\, red, then a white) = \frac{}{} \times \frac{}{} = \frac{}{9}

STEP 13

For part c), we need to find the probability of drawing a blue marble three times in a row. Since the marbles are returned to the bag after each draw, these are independent events, so we multiply the probabilities.
(A Blue,thenaBlue,thenaBlue)=(Blue)×(Blue)×(Blue)(A\, Blue, then a Blue, then a Blue) =(Blue) \times(Blue) \times(Blue)

STEP 14

Plug in the values for the probabilities of drawing a blue marble.
(A Blue,thenaBlue,thenaBlue)=3×3×3(A\, Blue, then a Blue, then a Blue) = \frac{}{3} \times \frac{}{3} \times \frac{}{3}

STEP 15

Calculate the probability of drawing a blue marble three times in a row.
(A Blue,thenaBlue,thenaBlue)=3×3×3=27(A\, Blue, then a Blue, then a Blue) = \frac{}{3} \times \frac{}{3} \times \frac{}{3} = \frac{}{27}So, the probabilities area) A Blue, then a red =/9b) A red, then a white =/9c) A Blue, then a Blue, then a Blue =/27

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