pharmxanthan wrote:Q) A bag contains 3 red balls, 5 yellow balls, and 4 green balls. Two balls are drawn randomly. What is the probability that the balls drawn contain balls of different colors?
A) Number of ways to select 2 balls out of 12 balls = 12C2 = 66.
Number of ways to select two balls of different colors (red & yellow OR red & green OR yellow & green)= (3C1*5C1)+(3C1*4C1)+(5C1*4C1)= 47
Probability = 47/66 (Ans)
Is the approach correct?
Thanks!
A very similar way to do it would be to look at the 3 cases and use the probability formula.
First case: choose red first, non-red second.
Prob = 3/12 * 9/11 = 27/132
Second case: choose yellow first, non-yellow second.
Prob = 5/12 * 7/11 = 35/132
Third case: choose green first, non-green second.
Prob = 4/12 * 8/11 = 32/132
Prob of case 1 OR case 2 OR case 3:
27/132 + 35/132 + 32/132 = 94/132 = 47/66
Of course, we could also solve using the "1 minus" formula:
Probability (what you want) = 1 - Probability (what you don't want)
In this question, we don't want two of the same colour.
Prob red/red = 3/12 * 2/11 = 6/132
Prob yellow/yellow = 5/12 * 4/11 = 20/132
Prob green/green = 4/12 * 3/11 = 12/132
Prob double = (6+20+12)/132 = 38/132 = 19/66
Prob not double = 66/66 - 19/66 = 47/66
Or, using combinations as you did:
Number of ways to select 2 balls: 12C2 = 12*11/2 = 66
Number of ways to select 2 of the same colour: 3C2 + 4C2 + 5C2 = 3 + 6 + 10 = 19
Prob of selecting 2 of the same colour = 19/66
Prob of NOT selecting 2 of the same colour = 66/66 - 19/66 = 47/66
* * *
When practicing, it's good to think of all the different ways you can attack a question; that way, when it comes to Test Day, you'll know which approach works best for you.
