Two unbiased dice marked from 1 to 6 are tossed together. The probability of the sum of the outcomes to be 7 in a single throw is
1/6
2/3
4/13
7/13
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UPSC CISF-AC-EXE – 2019
Number of outcomes for a single die = 6 (1, 2, 3, 4, 5, 6).
Total number of outcomes for two dice = $6 \times 6 = 36$.
We want to find the probability that the sum of the outcomes is 7.
Let (d1, d2) represent the outcome of the first and second die, respectively. The possible pairs (d1, d2) that sum up to 7 are:
(1, 6)
(2, 5)
(3, 4)
(4, 3)
(5, 2)
(6, 1)
There are 6 favorable outcomes.
The probability of an event is calculated as:
Probability = (Number of favorable outcomes) / (Total number of possible outcomes)
Probability of the sum being 7 = 6 / 36.
Simplifying the fraction:
Probability = 1 / 6.
– List all possible pairs of outcomes that result in the desired sum.
– Calculate probability as the ratio of favorable outcomes to total outcomes.
Sum 2: 1 way (1,1)
Sum 3: 2 ways (1,2), (2,1)
Sum 4: 3 ways (1,3), (2,2), (3,1)
Sum 5: 4 ways (1,4), (2,3), (3,2), (4,1)
Sum 6: 5 ways (1,5), (2,4), (3,3), (4,2), (5,1)
Sum 7: 6 ways (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
Sum 8: 5 ways (2,6), (3,5), (4,4), (5,3), (6,2)
Sum 9: 4 ways (3,6), (4,5), (5,4), (6,3)
Sum 10: 3 ways (4,6), (5,5), (6,4)
Sum 11: 2 ways (5,6), (6,5)
Sum 12: 1 way (6,6)
Total ways = 1+2+3+4+5+6+5+4+3+2+1 = 36.