If the probabilities of a student succeeding in the entrance tests for institutes A, B and C are 0.6 , 0.5 and 0.4 respectively, while the probability of succeeding in both A and B is 0.3 , in both B and C is 0.2 , in both A and C is 0.2 , and in all three is 0.1 , then the probability that the student succeeds in exactly one of these tests is......

Step-by-step solution

Let A , B , and C be the events that the student succeeds in the entrance tests for institutes A, B, and C respectively. Given: P(A) = 0.6 P(B) = 0.5 P(C) = 0.4 P(A B) = 0.3 P(B C) = 0.2 P(A C) = 0.2 P(A B C) = 0.1 The probability that the student succeeds in exactly one of the tests is given by the formula: P( exactly one ) = P(A) + P(B) + P(C) - 2[P(A B) + P(B C) + P(A C)] + 3P(A B C) Substituting the given values into the formula: P( exactly one ) = (0.6 + 0.5 + 0.4) - 2(0.3 + 0.2 + 0.2) + 3(0.1) P( exactly one