What is the probability of a blue marble?

Question

Mikes teacher has three boxes and she puts two red marbles in box A, two blue marbles in box B, and one blue and one red marble in box C.

Part 1 - What is the probability of Mike picking a box so that he gets at least one blue marble?

Part 2 - His teacher now removes one marble from each of the box. What is the probability of Mike getting a blue marble?

Answer 1

PART 1:

1/3 of the time he will get box A with a 0% probability of blue

1/3 of the time he will get box B with a 100% probability of blue

1/3 of the time he will get box C with a 50% probability of blue

1/3 * 0 + 1/3 * 1 + 1/3 * 1/2

= 1/3 + 1/6

= 2/6 + 1/6

= 3/6

= 1/2

Answer:

50%

PART 2:

The probability of box A and B will not change.

1/3 of the time he will get box A with a 0% probability of blue

1/3 of the time he will get box B with a 100% probability of blue

1/3 of the time he will get box C

--- 1/2 of the time box C will have 1 red marble (0%)

--- 1/2 of the time box C will have 1 blue marble (100%)

1/3 * 0 + 1/3 * 1 + 1/3 (1/2 * 0 + 1/2 * 1)

= 0 + 1/3 + 1/3(0 + 1/2)

= 0 + 1/3 + 1/6

= 2/6 + 1/6

= 3/6

= 1/2

Answer:

Still 50%

Answer 2

Part 1: There are two boxes out of three containing blue marbles, so a random pick would have a 2/3 probability of getting at least one blue marble.

Part 2:

There are two possibilities now, depending on whether the teacher removed the blue or the red marble in box C. Without knowing which, we treat each case as equally probable.

There's a probability of 1/2 that Mike's chance of getting a blue marble hasn't changed because there are still two boxes with blue marbles. There's a probability of 1/2 that Mike's chance has changed to 1/3, because boxes A and C now each contain one red marble.

The overall probability of getting a blue marble is now

1/2 * 2/3 + 1/2 * 1/3 = 1/2 * (2/3 + 1/3) = 1/2

Answer 3

Each Box has the same chance of being selected, so each has a probability of 1/3

2 of the boxes have a blue marble, and the choices are independent, so simply add the probabilities....

1/3 + 1/3 = 2/3

The probability is 2/3 of drawing at least one blue marble

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Part 2 -

Assuming that the removal of the marble from Box C is at random, the probability that Box C has a blue marble is 1/2.

So the probability of drawing a Blue marble is now

0 + 1/3 + 1/3*1/2

or

1/2

If you know which marble was removed from Box C, you can calculate the probability of drawing a Blue marble using the same method.  It will be either 1/3  or 2/3 based on which marble was removed.

Answer 4

part 1... 3/6 also 1/2

part 2... it all depends whether she takes a blue or a red marble from box c

if she takes the red marble 2/3

if she takes the blue marble 1/3

Answer 5

Part 1 - Box A does not contain at least one blue marble, but Box B and Box C do.  Your probability is 2/3.

Part 2 -

If Mike picks box A, there is a 0 probability that he gets a blue marble.

If Mike picks box C, there is a 1/2 probability that he gets a blue marble (assuming the teacher took the marbles out randomly, there's a 50% chance that the marble she took was the blue one).

If Mike picks box B, there is a 1 probability that he gets a blue marble.

Assuming he picks the box randomly, you can multiply each of the three possibilities by 1/3 and add them together to get your total probability:

(1/3 x 0) + (1/3 x 1/2) + (1/3 x 1) = 3/6 or 1/2.  There is a 1/2 or 50% probability that Mike gets a blue marble.

Answer 6

Part 1- two of the three boxes have blue marbles in them, so the probability is 2/3

Part2- now one box surely has a blue marble, another surely doesn't and the last has a blue marble with probability 1/2, so the total probability is 1/3*1+1/3*0+1/3*1/2=1/2

Answer 7

1) That he picks Box B or C

P(B)=1/3

P(C)=1/3

P(B or C) = 1/3+1/3=2/6=1/3

2)Blue marble from A =0

Blue marble from B= 1

Blue marble from C= 1  or 0

(1/3)(0)+(1/3)(1)+(1/3)1

+ (1/3)(0)(1/3)(1)+(1/3)(0)

=2/3