A bag contains 15 balls. 9 white , 4 black and 2 green. a)We pick 2 balls. Whats the probability of the balls to be different colors? b)We pick 3 balls. Whats the probability of the balls to be different colors?
1 Answer
a.
Explanation:
There are 15 balls in total.
The problem we run up against is that depending on what the first marble drawn is, getting a marble of a different colour on the subsequent draws changes. And so this will make for a long-ish answer.
We first look at the first draw and see that the probability of drawing the various colours is:
- White
#=9/15=3/5# - Black
#=4/15# - Green
#=2/15#
a
When we pick White first, we want the probability of not drawing another white. There are now 14 balls and 8 of them are white, so the probability of first drawing White then Not White is:
When we pick Black first, we want the probability of not drawing another black. There are now 14 balls and 3 of them are black, so the probability of first drawing Black then Not Black is:
When we pick Green first, we want the probability of not drawing another green. There are now 14 balls and 1 of them is green, so the probability of first drawing Green then Not Green is:
Therefore the probability of drawing 2 balls of different colours is:
b
I think the easiest way to work this is to simply show the calculation for each possible series of picks:
WBG
WGB
BWG
and we can stop here. Notice that the denominator isn't changing at all and the terms of the numerator are simply moving around. And so the answer can be found by looking at any one instance and multiplying by 6:
