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<\/p>\n
[amp_mcq option1=”1\/6″ option2=”2\/3″ option3=”4\/13″ option4=”7\/13″ correct=”option1″]<\/p>\n
We want to find the probability that the sum of the outcomes is 7.
\nLet (d1, d2) represent the outcome of the first and second die, respectively. The possible pairs (d1, d2) that sum up to 7 are:
\n(1, 6)
\n(2, 5)
\n(3, 4)
\n(4, 3)
\n(5, 2)
\n(6, 1)<\/p>\n
There are 6 favorable outcomes.<\/p>\n
The probability of an event is calculated as:
\nProbability = (Number of favorable outcomes) \/ (Total number of possible outcomes)<\/p>\n
Probability of the sum being 7 = 6 \/ 36. 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 [amp_mcq option1=”1\/6″ option2=”2\/3″ option3=”4\/13″ option4=”7\/13″ correct=”option1″] This question was previously asked in UPSC CISF-AC-EXE – 2019 Download PDFAttempt Online When two unbiased dice, each marked from 1 to … <\/p>\n
\nSimplifying the fraction:
\nProbability = 1 \/ 6.
\n<\/section>\n
\n– List all possible pairs of outcomes that result in the desired sum.
\n– Calculate probability as the ratio of favorable outcomes to total outcomes.
\n<\/section>\n
\nSum 2: 1 way (1,1)
\nSum 3: 2 ways (1,2), (2,1)
\nSum 4: 3 ways (1,3), (2,2), (3,1)
\nSum 5: 4 ways (1,4), (2,3), (3,2), (4,1)
\nSum 6: 5 ways (1,5), (2,4), (3,3), (4,2), (5,1)
\nSum 7: 6 ways (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
\nSum 8: 5 ways (2,6), (3,5), (4,4), (5,3), (6,2)
\nSum 9: 4 ways (3,6), (4,5), (5,4), (6,3)
\nSum 10: 3 ways (4,6), (5,5), (6,4)
\nSum 11: 2 ways (5,6), (6,5)
\nSum 12: 1 way (6,6)
\nTotal ways = 1+2+3+4+5+6+5+4+3+2+1 = 36.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"