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Question

Two buses A and B are scheduled to arrive at a town central bus station at noon. The probability that bus A will late is 15. The probability that bus B will be late is 725. The probability that the bus B is late given that bus A is late is 910. Then probability that

A
neither bus will be late on a particular day is 710
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B
the bus A is late given than bus B is late is 914
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C
at least one bus is late is 310
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D
at least one bus is in time is 45
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Solution

The correct option is C at least one bus is late is 310
Let A,B be the events that bus A, bus B are late respectively.
Given P(A)=15,P(B)=725 and P(BA)=910P(AB)=910×P(A)=950
a) P(¯A¯B)=P¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(AB)=1P(AB)=1[P(A)+P(B)P(AB)]=1[15+725950]=11550=1310=710
b) P(A/B)=P(AB)P(B)=(950)(725)=914
c) P(AB)=(15+725950)=310
d) P(¯A¯B)=P(¯¯¯¯¯¯¯¯¯¯¯¯¯¯AB)=1P(AB)=1950=4150

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