probability problem

[kmor2107]

My question is: A bag contains several marbles. Each marble is either red, white, or blue. The probability of choosing a red marble is 1/3, and the probability of choosing a white marble is 1/6. 1. What is the probability of choosing a blue marble? Explain. 2. What is the least number of marbles that can be in the bag? Explain. Suppose the bag contains the least number of marbles. How many of each color does the bag contain? 3. Can the bag contain 48 marbles? If so, how many of each color would it contain?

[graphica]

Honestly Probability and P&C are my weakest points!
Still will try (don't shoot me if I am wrong) ;)

The probability of choosing a blue marble
1- (1/6+1/3) = 1- (1+2/6=3/6=1/2) = 1/2 blue marbles.

Lowest number = 6 (3 blue, 2 Red and 1 White)

If the bag had 48 marbles then the total will be:
3x+2x+x=48
6x=48
x=8

blue=24 red=16 and white= 8

Apologize in advance if I am wrong!

[mschwrtz]

Looks good to me.

The probability of choosing a blue marble
1- (1/6+1/3) = 1- (1+2/6=3/6=1/2) = 1/2 blue marbles.

Yes, the sum of the probabilities must be 1.

Lowest number = 6 (3 blue, 2 Red and 1 White)

Yes, 6 is the least number divisible by 2, 3, and 6.

If the bag had 48 marbles then the total will be:
3x+2x+x=48
6x=48
x=8

blue=24 red=16 and white= 8

All correct.