a=161
c+d+e+f=41...(i)
b+d+e+g=41.....(ii)
b+f+c+g=41....(iii)
Adding (i),(ii) and (iii)
⇒2(b+c+d+e+f+g)=43
⇒b+c+d+e+f+g=83
Now ⇒P(A∪B∪C)=a+(b+c+d+e+f+g)
⇒P(A∪B∪C)=161+83
⇒P(A∪B∪C)=167
For three events, A,B and C, P(Exactly one of A or B occurs)
=P(Exactly one of B or C occurs)
=P(Exactly one of C or A occurs) =41 and P(All the three events occur simultaneously) =161.
Then the probability that at least one of the events occurs, is:
Held on 2 Apr 2017 · Verified 6 Jul 2026.
327
167
647
163
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