  Question

# Two buses A and B are scheduled to arrive at a town central bus station at noon. The probability that bus A and bus B will be late are 15 and 725 respectively. The probability that the bus B is late given that bus A is late is 910. Thenthe probability that neither of the two bus is late on a particular day is 710the probability that bus A is late given that bus B is late is 914  the probability that at least one bus is late is 310the probability that at least one bus is in time is 4150

Solution

## The correct options are A the probability that neither of the two bus is late on a particular day is 710 B the probability that bus A is late given that bus B is late is 914   C the probability that at least one bus is late is 310 D the probability that at least one bus is in time is 4150 P(A): Probability that bus A will be late. P(B): Probability that bus B will be late. P(A)=15, P(B)=725, P(B|A)=910​  P(¯¯¯¯A∩¯¯¯¯B)=1−P(A∪B) =1−[P(A)+P(B)−P(A∩B)] =1−[P(A)+P(B)−P(A)P(B|A)] =1−[15+725−15×910] =710  Therefore, the probability that neither of the two bus is late on a particular day is 710. P(A|B)=P(A∩B)P(B)               =P(A)P(B|A)P(B)               =15×910725               =914 Therefore, the probability that bus A is late given that bus B is late is 914.  Probability that at least one bus is late =1−P(none of the bus is late) =1−710 =310  P(A′∪B′)=1−P(A∩B)                    =1−P(A)P(B|A)                    =1−15×910                    =4150 Therefore, the probability that at least one bus is in time is 4150.  Suggest corrections   