If the probability of A to fail in an examination is 0.2 and that for B is 0.3, then probability that either A or B is fail, is

Question

If the probability of A to fail in an examination is 0.2 and that for B is 0.3, then probability that either A or B is fail, is (A) 0.5 (B) 0.44 (C) 0.8 (D) 0.25

Tardigrade Admin · Accepted Answer

Given, Probability of A to fail exam = 0.2 Probability of B to fail exam = 0.3 Since, A and B are independent events ∴ P(A∩ B)=P(A).P(B)=0.06 Now, P(A∪ B)=P(A)+P(B)-P(A∩ B) =0.2+0.3-0.06=0.44

Q. If the probability of A to fail in an examination is 0.2 and that for B is 0.3, then probability that either A or B is fail, is

0.5

0.44

0.8

0.25

Given, Probability of A to fail exam = 0.2 Probability of B to fail exam = 0.3 Since, A and B are independent events $∴$ $P (A \cap B) = P (A) . P (B) = 0.06$ Now, $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ $= 0.2 + 0.3 - 0.06 = 0.44$

Q. If the probability of A to fail in an examination is 0.2 and that for B is 0.3, then probability that either A or B is fail, is

0.5

0.44

0.8

0.25

Given, Probability of A to fail exam = 0.2 Probability of B to fail exam = 0.3 Since, A and B are independent events ∴ P(A∩B)=P(A).P(B)=0.06 Now, P(A∪B)=P(A)+P(B)−P(A∩B) =0.2+0.3−0.06=0.44

Given, Probability of A to fail exam = 0.2 Probability of B to fail exam = 0.3 Since, A and B are independent events $∴$ $P (A \cap B) = P (A) . P (B) = 0.06$ Now, $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ $= 0.2 + 0.3 - 0.06 = 0.44$