When two fair dice are rolled, what is the probability that at least

getting odd ; 3/6*3/6 ; 1/4
atleast 1 even ; 1-1/4; 3/4
IMO E

Prob of getting at least One even[P(A1E)] = 1- Prob of getting all odd [P(O)]

Total number of cases for 2 fair dice roll: 6*6 = 36
odd numbers: 1,3,5
So 3 cases each > 3*3 = 9 for both dice
P(O) = 9/36 => 1/4

Now P(A1E) = 1 - 1/4 => 3/4

Answer: Option E

↧↧↧ Detailed Video Solution to the Problem ↧↧↧



Given that Two dice are rolled and we need to find what is the probability that at least one of the numbers will be even?

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 E
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

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