## Question

For the three events *A*, *B* and *C*, *P*(exactly one of the events *A* or *B*occurs) = *P*(exactly one of the events *B* or *C* occurs) = *P*(exactly one of the events *C* or *A* occurs) = *p* and *P* (all the three events occur simultaneously) =** ***p*^{2}, where 0 < *p* < 1/2. Then the probability of at least one of the three events *A*, *B* and *C* occurring is

### Solution

we know that

*P*(exactly one of *A* or *B* occurs)

= *P*(*A*) + *P*(*B*) – 2*P*(*A* ∩ *B*)

Therefore, *P*(*A*) + *P*(*B*) – 2*P*(*A* ∩ *B*) = *p* ... (1)

Similarly, *P*(*B*) + *P*(*C*) – 2*P*(*B* ∩ *C*) = *p* ...(2)

and, *P*(*C*) + *P*(*A*) – 2*P*(*C* ∩ *A*) = *p* ...(3)

Adding (1), (2) and (3) we get

2[P(A) + P (B) + P (C) – P (*A* ∩ *B*) – P(*B* ∩ *C*) – P (*C* ∩ *A*)] = 3*p*

⇒ P(A) + P (B) + P (C) – P (*A* ∩ *B*) – P(*B* ∩ *C*) – P (*C* ∩ *A*)] = 3*p*/2 (4)

We are also given that P (*A* ∩ *B *∩ *C*) = *p*^{5} (5)

Now, *P* (at least one of *A*, *B* and *C*)

= P(A) + P (B) + P (C) – P (*A* ∩ *B*) – P(*B* ∩ *C*) – P (*C* ∩ *A*) + P (*A* ∩ *B *∩ *C*)

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