Which of the following is false?

Question

Which of the following is false? (A) P (φ) = 0  (B) For any three M.E. events A, B and C, one can have P(A) = (2/3) , P(B) = (1/4), P(C) = (1/6) (C) P (A+ B) = P (A) + P (B) - P (AB) (D) P (Ac)= 1 - P (A).

Tardigrade Admin · Accepted Answer

In alternative we must have P (A) + P (B) + P (C) le 1 as A, B and C are mutually exclusive events. Here, P (A) + P (B) + P (C) = (2/3) + (1/4) + (1/6). is greater than 1, which is not possible

Q. Which of the following is false?

$P (ϕ) = 0$

For any three M.E. events A, B and C, one can have $P (A) = \frac{2}{3}, P (B) = \frac{1}{4}, P (C) = \frac{1}{6}$

P (A+ B) = P (A) + P (B) - P (AB)

$P (A^{c}) = 1 - P (A) .$

In alternative we must have
P (A) + P (B) + P (C) $\leq$ 1 as A, B and C are mutually exclusive events. Here,
P (A) + P (B) + P (C) = $\frac{2}{3} + \frac{1}{4} + \frac{1}{6}$ . is greater than 1, which is not possible

Q. Which of the following is false?

P(ϕ)=0

For any three M.E. events A, B and C, one can have P(A)=32​,P(B)=41​,P(C)=61​

P (A+ B) = P (A) + P (B) - P (AB)

P(Ac)=1−P(A).

In alternative we must have P (A) + P (B) + P (C) ≤ 1 as A, B and C are mutually exclusive events. Here, P (A) + P (B) + P (C) = 32​+41​+61​. is greater than 1, which is not possible

$P (ϕ) = 0$

For any three M.E. events A, B and C, one can have $P (A) = \frac{2}{3}, P (B) = \frac{1}{4}, P (C) = \frac{1}{6}$

$P (A^{c}) = 1 - P (A) .$

In alternative we must have
P (A) + P (B) + P (C) $\leq$ 1 as A, B and C are mutually exclusive events. Here,
P (A) + P (B) + P (C) = $\frac{2}{3} + \frac{1}{4} + \frac{1}{6}$ . is greater than 1, which is not possible