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Question

A bag contains four marbles. One marble is blue, one is red, one is white, and one is yellow. Two marbles are draw from the bag by removing one marble and then, without replacing the first marble, removing a second marble. In how many ways can the two marbles be drawn from the bag so that at least one of the marbles is red?

 

NOTE: I'm stuck between answers 3 or 6?

A. 3

B. 4

C. 6

D. 12

Answer 1

There are two marbles drawn.  The problem is asking about combinations ("how many ways"), not about probability.

 

With one blue, one red, one white, one yellow marbles in the bag -- and without replacement (which is important) -- the possible results are (PLZ list then all for better understanding):

  B R

  B W

  B Y

  R B   (note:  this "way" is different from B R)

  R W

  R Y

  W B

  W R

  W Y

  Y B

  Y R

  Y W

 

How many of these "ways" have "at least one marble that is red" (another note:  without replacement, it is not possible to have 2 red marbles;  also, we don't count "ways" that have no red marbles)?

 

Now, start thinking about drawing two marbles from a bag of four different-colored marbles, ignoring all outcomes that did not have one red marble (remember, two is impossible without replacement).  You have 3 outcomes that have a red marble as the first of the set.  Then (remember, without replacement), if you did not draw a red marble for the first one, but one of the other three colors, but did draw a red marble for one of those three possibilities, how many possibilities would that be --- 3.  So, that's 6 in all.

 

Good answer.  Now, PLZ don't just get the right answer; PLZ learn the process used to solve the problem !

 

Answer 2

There are two different draws here, where one of the marbles needs to be red. I think it would help if I told you that the order of the draw matters, where drawing the red first or the red second are different ways to get to the same thing. Does this help?