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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 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
  1. the probability that neither of the two bus is late on a particular day is 710
  2. the probability that bus A is late given that bus B is late is 914  
  3. the probability that at least one bus is late is 310
  4. 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)=1P(AB)
=1[P(A)+P(B)P(AB)]
=1[P(A)+P(B)P(A)P(B|A)]
=1[15+72515×910]
=710 
Therefore, the probability that neither of the two bus is late on a particular day is 710.

P(A|B)=P(AB)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
=1P(none of the bus is late)
=1710
=310 

P(AB)=1P(AB)
                   =1P(A)P(B|A)
                   =115×910
                   =4150
Therefore, the probability that at least one bus is in time is 4150.   

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