if a dice is thrown twice what probability is for  a)6 each b)8 total 

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gsarora17 | (Level 2) Associate Educator

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The probability of the happening of an event E , denoted by P(E) is defined as P(E) = (Number of favorable outcomes)/ Total number of possible outcomes

Since the dice is thrown twice , the total number of possible outcomes = 36

a) Favorable outcome for getting six each = 1

Therefore P(E) = 1/36

b) Favorable outcomes to get a sum of 8 will be (2,6) , (3,5) , (4,4) , (5,3) , (6,2)

Therefore number of favorable outcomes to get sum of 8 = 5 

Therefore P(E) = 5/36

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hkj1385 | (Level 1) Assistant Educator

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The possible outcomes when the dice is thrown twice are:-

(1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6)

(2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6)

(3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6)

(4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6)

(5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6)

(6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6)

Thus, total number of possible outcomes = 36

Now, a) 6 each

Total number of favourable outcomes = 1

Hence required probability = favourable outcomnes/total outcomes = 1/36

b) 8 total

Favourable outcomes = (2,6) , (6,2) , (3,5) , (5,3) , (4,4) 

Number of Favourable outcomes = 5

Hence required probability = favourable outcomnes/total outcomes = 5/36

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nisarg | Student, Grade 11 | (Level 1) Valedictorian

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on a 6 sided dice the possiblity of getting 6 on each 1/36th

this is because it is 1/6 on one dice and 1/6 on the other so 1/6*1/6=1/36

Now for the possiblity of getting 8 is different you have to find the numbers that add up to get an eight 

so (2,6) (3,5) (4,4) (5,3) (6,2)

that is 5 possiblities out of 36 because 2 dice are rolled each with 6 possiblies. so it is 5/36

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