On December 20, 2023, Brent will stop offering office hours.
- Video Course
- Video Course Overview
- General GMAT Strategies - 7 videos (free)
- Data Sufficiency - 16 videos (free)
- Arithmetic - 38 videos
- Powers and Roots - 36 videos
- Algebra and Equation Solving - 73 videos
- Word Problems - 48 videos
- Geometry - 42 videos
- Integer Properties - 38 videos
- Statistics - 20 videos
- Counting - 27 videos
- Probability - 23 videos
- Analytical Writing Assessment - 5 videos (free)
- Reading Comprehension - 10 videos (free)
- Critical Reasoning - 38 videos
- Sentence Correction - 70 videos
- Integrated Reasoning - 17 videos
- Study Guide
- Blog
- Philosophy
- Office Hours
- Extras
- Prices
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..
Is this approach valid?
(4C2 + 3C2 + 2C2)/9C2 = 10/36 = 5/18
Perfect!!
Perfect!!
If I were to list the
The main issue is that those
The main issue is that those three outcomes (GG, RR, YY) are not all equally likely.
To illustrate this, let's say there are 1,000,000 green balls and 2 red balls.
As you can imagine, the likelihood of getting 2 green balls is much much much greater than the likelihood of selecting 2 red balls.
Don't forget that, if we're going to use counting techniques to solve a probability question, each outcome must be equally likely.
For more see: https://www.gmatprepnow.com/module/gmat-probability/video/742
Here's another example: A barrel contains 999 red marbles and 1 blue marble. If one marble is selected at random from the barrel, what is the probability that the selected marble is blue?
If we list the possible outcomes the way you did we get two possibilities:
- Red
- Blue
Since 1 of the 2 possible outcomes is blue, one might incorrectly conclude that P(the selected ball is blue) = 1/2