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