Question

Two dice of different colours are thrown simultaneously. The probability that the sum of the faces appeared is either 7 or 11 is?
up vote
0
down vote
favorite
The options are: (A) 7/36 (B) 1/9 (C)2/3 (D) 5/9

Answer 1

Dice are usually marked from 1 to 6

The possible sums are from 2 to 12

The total numbers of sums = 36

In the sums, 7 appears 6 times, while 11 appears 2 times

Probability = No. of desired outcomes/Total No. of possible outcome


Probability of getting a sum of 7 or 11 = 8/36

                                                             = 2/9

ALTERNATIVELY

Probability of getting sum of 7 = 6/36 = 1/6

Probability of getting sum of 11 = 2/36 = 1/18


Probability of getting sum of 7 or 11 = 1/6  + 1/18

                                                         = 2/9


NOTE:

As can be seen from the calculations above, none of the choices given is correct.

Answer 2

Two dice of different colors are thrown simultaneously.
∴ number of possible outcomes = 6 × 6 = 36

dice includes 1 to 6 number e.g., {1 , 2, 3, 4, 5, 6 }
Let two dice of different colors is red and blue
in red { 1, 2, 3, 4 , 5, 6}
In blue { 1, 2 , 3, 4, 5, 6}
Favourable outcome = sum of faces appeared is either 7 or 11
There are 6 ways , sum of faces appeared is 7
e.g., {1,6},{6,1},{4,3},{3,4},{5,2} and {2,5}
Similarly there are 2 ways , sum of faces appeared is 11
e.g., {5,6} ,{6,5)
Hence, favourable outcomes = 6 ways + 2 ways = 8 ways

Now, probability = favourable outcomes/total possible outcomes {sample space}
= 8/36 = 2/9