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Question

A die is rolled twice. What is the probability of getting a sum equal to 9?
Options:
A .  $MF#%\dfrac{2}{3}$MF#%
B .  $MF#%\dfrac{2}{9}$MF#%
C .  $MF#%\dfrac{1}{3}$MF#%
D .  $MF#%\dfrac{1}{9}$MF#%
Answer: Option D

Answer : Option D

Explanation :

Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces)
Hence, total number of outcomes possible when a die is rolled twice, n(S) = 6 × 6 = 36
E = Getting a sum of 9 when the two dice fall = {(3, 6), {4, 5}, {5, 4}, (6, 3)}
Hence, n(E) = 4

$MF#%\text{P(E) = }\dfrac{\text{n(E)}}{\text{n(S)}} = \dfrac{4}{36} = \dfrac{1}{9}$MF#%



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