You roll a dice twice and add the numbers.
Find the probability that you get a result less than 6 given that the tosses resulted in an odd number.
Baues' Theorem: P(A | B) = P(A & B) / P(B)
Number of cases that result in less than 6 and toss is odd = 6
[(1,2), (2,1), (1,4), (3,2), (2,3). (4,1)]
Number of cases in which result is odd = Number of cases with one number is odd and other number is even
= 6 x 3
= 18
P(less than 6 | tosses resulted in odd number) = Number of cases that result in less than 6 and tosses are odd / Number of cases in which result is odd)
= 6/18
= 1/3
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