Given that a fair dice is rolled twice and We need to find what is the probability that it will land on an even number at least once?As we are rolling two dice => Number of cases = \(6^2\) = 36
Let's solve the problem using two methods:
Method 1:Now there are 4 outcomes possible
(Odd, Odd), (Odd, Even), (Even Odd), (Even, Even) and there is an equal chance of each of them happening
=>
P(At least one number is Even) = P((Odd, Even) or (Even Odd) or (Even, Even)) = \(\frac{3}{4}\)
Method 2:Out of the 36 comes lets eliminate all options in which both the outcomes are odd
Both can be odd in 3*3 (=9 ways), as in the first roll we can get any number out of 1, 3, and 5. And in the second case also we have these 3 choices.
=>
P(At least one number is Even) = \(\frac{36 - 9}{36}\) = \(\frac{27}{36}\) = \(\frac{3}{4}\)
So,
Answer will be CHope it helps!
Playlist on Solved Problems on Probability hereWatch the following video to MASTER Dice Rolling Probability Problems