One fair die has faces 1, 1, 2, 2, 3, 3 and another has faces 4, 4, 5

Question

One fair die has faces 1, 1, 2, 2, 3, 3 and another has faces 4, 4, 5, 5, 6, 6. The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?

(A) 1/3
(B) 4/9
(C) 1/2
(D) 5/9
(E) 2/3

itsanujsingh · Answer

Even+odd=odd

Probability, 1st die is odd and 2nd die is even = (2/3)*(2/3) = 4/9

probability 1st die is even and 2nd die is odd = (1/3)*(1/3) = 1/9

Thus, probability that sum is odd = 4/9 + 1/9 = 5/9

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Archit3110 · Answer

odd sum = even + odd integer
so 
case 1 : die 1 even integer and die 2 odd integer = 2/6 * 2/6; 1/9
case 2 ; die 1 odd integer and die 2 even integer = 4/6 *4/6 ; 4/9
sum 1/9 + 4/9 ; 5/9
IMO D

BrushMyQuant · Answer

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

https://www.youtube.com/watch?v=vkma0-VOQ7A

We need to find what is the probability that the sum will be odd?

We know that when we have two numbers then to get an Odd number we have following cases
 Sum equal Odd when we have Even + Odd or Odd + Even

P(Sum = Odd) = P(Even in 1st Dice) * P(Odd in 2nd Dice) + P(Odd in 1st Dice) * P(Even in 2nd Dice)

First Dice
  Odd numbers = 4 (1,1, 3, 3)
Even Number = 2 (2, 2)
P(O) = 4/6 = 2/3
P(E) = 2/6 = 1/3

Second Dice
  Odd numbers = 2 (5, 5)
Even Number = 4 (4, 4, 6, 6)
 P(O) = 2/6 = 1/3
P(E) = 4/6 = 2/3

P(Sum = Odd) = P(Even in 1st Dice) * P(Odd in 2nd Dice) + P(Odd in 1st Dice) * P(Even in 2nd Dice)
 =  \frac{1}{3}  *  \frac{1}{3}  +  \frac{2}{3}  *  \frac{2}{3} 
 =  \frac{1}{9}  +  \frac{4}{9}  =  \frac{5}{9}

So,  Answer will be D 
Hope it helps!

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

https://www.youtube.com/watch?v=5PVZYQXet18

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