A, B, C are events such that

P (A) = 0.3, P (B) = 0.4, P (C) = 0.8

P (AB) = 0.008, P (AC) = 0.28, P (ABC) = 0.09

If P (A ∪ B ∪ C) ≥ 0.75, then show that P (BC) lies in the interval 0.23 ≤ x ≤ 0.48

Hello Student,

Please find the answer to your question

Given that
P (A) = 0.3, P (B) = 0.4, P (C) = 0.8
P (AB) = 0.08, P (AC) = 0.28, P (ABC) = 0.09
P (A ∪ B ∪ C) ≥ 0.75
To find P (BC) = x (say)
Now we know,
P (A ∪ B ∪ C) = P (A) + P (B) + P (C) – P (AB) – P (BC) – P (CA) + P (ABC)
⇒ P (A ∪ B ∪ C) = 0.3 + 0.4 + 0.8 – 0.08 – x – 0.28 + 0.09 = 123 – x
Also we have,
P (A ∪ B ∪ C) ≥ 0.75 and P (A ∪ B ∪ C) ≤ 1
∴ 0.75 ≤ P (A ∪ B ∪ C) ≤ 1
⇒ 0.75 ≤ 1.23 – x ≤ 1
⇒ 0.23 ≤ x ≤ 0.48

Thanks
Jitender Pal
askIITians Faculty