P( Red Disc) = Number of red Disc / total number of Disc = 4 / 9 P( Blue Disc) = Number of Blue Disc / total number of Disc = 3 / 9 P( Green Disc) = Number of Green Disc / total number of Disc = 2 / 9 P( Not Blue Disc) = 1 – P( Blue Disc) = 1- 3 / 9 = 6 / 9 P( Either red or Blue) = P( red U Blue) = P( red) + P( Blue) – P( Red ∩ Blue) = 4 / 9 + 3 / 9 – 0 = 7 / 9
Numerical: Two students A & B appeared in an examination. The probability that A will qualify examination is 0.05 & that B will qualify examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that( a) Both A and B will not qualify the examination.( b) At least one of them will not qualify the examination( c) Only one of them will qualify the examination.
Solution: Given P( A) =. 05, P( B) =. 10 & P( A ∩ B) =. 02
Probability that Both A and B will not qualify the examination = P( AUB)’ = 1- P( AUB)
Probability that Both A and B will qualify the examination = P( AUB) P( A∪B) = P( A) + P( B) − P( A∩ B) = 0.05 + 0.1-0.02 = 0.13 P( AUB)’ = 1- P( AUB) = 1- 0.13 = 0.87
Probability that at least one of them will not qualify the examination = 1- P( A ∩ B) = 1-0.02 = 0.98
Probability that only one of them will qualify the examination = P( AUB)- P( A ∩ B) = 0.13 – 0.02 = 0.11