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Comment on Same Color
Can we solve the following
A bag holds 4 red marbles, 5 blue marbles, and 2 green marbles. If 5 marbles are selected one after another without replacement, what is the probability of drawing 2 red marbles, 2 blue marbles, and 1 green marble?
No, you'd have to use a
No, you'd have to use a different approach, since the required outcomes for each draw aren't identical. In the question above, we must draw a red ball each time.
In your question, there are several possible orderings (e.g., red - red - blue - blue - green, OR blue - red - green - blue - red, OR ...
In my opinion, such a question would be out of scope (too difficult) for the GMAT.
Now I get the point.
Thanks a lot.
what if we use the
That approach works also.
That approach works also.
I followed this approach and
P (same color)= P(1st draw any color)*P(2nd draw same color)
Case 1: 1st draw was green. P(any color)* P(same color) = 1*3/8= 3/8
Case 2: 1st draw was red. P (any color)* P(same color)= 1*2/8= 2/8
Case 3: 1st draw was yellow. P (any color)* P(same color)= 1*1/8= 1/8
Total= case1+case2+case3
3/8 + 2/8 + 1/8 = 6/8 = 3/4
What is wrong with this approach?
Your cases don't match your
Your cases don't match your probabilities.,
For example, "Case 1: 1st draw was green. P(any color)* P(same color) = 1*3/8= 3/8"
If the 1st draw is green, then the first draw can't be "any color." The first draw must be green.
Here's what you need to do....
Case 1: 1st draw is green: P(1st is green) x P(2nd is green) = (4/9) x (3/8) = 12/72
Case 2: 1st draw is red. P(1st is red) x P(2nd is red) = (3/9) x (2/8) = 6/72
Case 3: 1st draw is yellow. P(1st is yellow) x P(2nd is yellow) = (2/9) x (1/8) = 2/72
etc..
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