If 2 dice are thrown find the probability that the sum of the numbers on their upper face is at least 9
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Answered by
1
total cases = 36
favourable cases = (3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6)
so probability is 10/36 = 5/18
favourable cases = (3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6)
so probability is 10/36 = 5/18
Answered by
3
Hi
Total cases -36
atleast 9 means sum can be 9,10,11,12
for 9 total cases = 4 I.e. 3,6 ; 4,5: 5,4:6,3
For 10 = 3 I.e. 4,5; 5,5; 6,4
For 11 = 2 I.e. 5,6:6,5
For 12 = 1 6,6
So total fav. = 4+3+2+1 = 10
Prob = 10/36
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Total cases -36
atleast 9 means sum can be 9,10,11,12
for 9 total cases = 4 I.e. 3,6 ; 4,5: 5,4:6,3
For 10 = 3 I.e. 4,5; 5,5; 6,4
For 11 = 2 I.e. 5,6:6,5
For 12 = 1 6,6
So total fav. = 4+3+2+1 = 10
Prob = 10/36
Mark as brainliest if helpful
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