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. Then

- the probability that neither of the two bus is late on a particular day is 710
- the probability that bus A is late given that bus B is late is 914
- the probability that at least one bus is late is 310
- the 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.

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.

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