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Question

Let A, B, C be any three events in a sample space of a random experiment. Let the events E1= exactly one of A, B occurs, E2=  exactly one of B, C occurs, E3= exactly one of C, A occurs, E4 = all of A, B, C occurs, E5= atleast one of A, B, C occurs. P(E1)=P(E2)=P(E3)=13, P(E4)=19 then P(E5)=


  1. 518

  2. 1118

  3. 19

  4. 79


Solution

The correct option is B

1118


E1=(A¯B)(¯AB)=13E2=(B¯C)(¯BC)=13E3=(C¯A)(¯CA)=13E4=ABC=19E5=ABC

From figure,

E1=a+d+e+f=13

E2=b+e+d+g=13

E3=a+b+f+g=13

Adding all three, we get

2(a+b+d+e+f+g)=1a+b+d+e+f+g=12

E4=c=19

E5=a+b+c+d+e+f+g=12+19=1118

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