If a six sided die is rolled three times, what is the probability of getting at least one even number and at least one odd number?
1/8 1/4 1/2 3/4 7/8
1/8 1/4 1/2 3/4 7/8
This question is much more complicated than it needs to be, so let's simplify it.CITI29 wrote:If a six sided die is rolled three times, what is the probability of getting at least one even number and at least one odd number?
1/8 1/4 1/2 3/4 7/8
50% of the time it will be even, 50% of the time it will be odd. So, we can pretend that the die has two sides (or even pretend that this is a coin flip question).
We could do this by brute force. If we roll 3 times, the diff events are:
EEE
EEO
EOE
EOO
OOO
OOE
OEO
OEE
6 of those events have at least one E and at least one O, so the answer is 6/8 = 3/4.
We could also do this using probability.
Prob = # desired events/total # of possibilities
Also, we know that Prob (what you want) = 1 - prob (what you don't want)
The "1 - what you don't want" approach is often the quickest way to attack tough probability questions on the GMAT.
Since this is a binary scenario (i.e. each option happens 50% of the time), the total # of possibilities is going to be 2^n, where n is the number of events. In this case, that's 3.
So, there are 2^3 = 8 total possibilities.
The only 2 events that we do NOT want are OOO and EEE. So, we don't want 2/8 = 1/4 of the things to happen.
So, Prob(do want) = 1 - 1/4 = 3/4
The second solution (using the formula) is more elegant, but brute force is probably quicker. Remember, on test day you don't get any bonus points for the beauty of your solution, we want fast and furious!
