Using Conditional Probability,
the probability that the sum of the numbers appearing on the two dice is 8, if 3 appears on the first is 1/5.
We have given that,
Two dice are thrown. Let S be the sample space S={1,2,3,4,5,6×{1,2,3,4,5,6}
Total number of possible outcomes (n(S)) =36
Let A ≡ the event that the sum of the numbers coming up is 8
and B ≡ the event of occurrence of 3 on the first die.
A ≡{(3,5),(6,2),(4,4),(5,3),(2,6)}
favourable cases for A (n(A) )= 5
B≡{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)}
favourable cases for B (n(B)) = 6
Now, A int B = { (3,5 ) } = 1
favourable cases for AnB (n(An B) )= 1
Probability of event A occur (P(A)) = 5/36
Probability of event B occur (P(B)) = 6/36
Probability of event B occur (P(AnB)) = 1/36
we have to calculate probability the sum of the numbers appearing on the two dice is 8, if 3 appears on the first i.e., P(A/B) --> Conditional Probability
Required probability, P(A/B) = P(ANB)/ P(B)
=> P(A/B) = 1/36/5/36 = 1/5
Hence, the required probability is 1/5.
To learn more about Conditional probability, refer:
https://brainly.com/question/10739997
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