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+16 votes

Let $A$ and $B$ be any two arbitrary events, then, which one of the following is TRUE?

  1. $P (A \cap B) = P(A)P(B)$
  2. $P (A \cup B) = P(A)+P(B)$
  3. $P (A \mid B) = P(A \cap B)P(B)$
  4. $P (A \cup B) \leq P(A) + P(B)$
asked in Probability by Veteran (52.1k points) | 5.5k views
Why should it be? The operator is multiplication not division.

5 Answers

+27 votes
Best answer
$(a)$ is true only if events are independent.

$(b)$ is true only if events are mutually exclusive i.e. $P (A \cap B) = 0$

$(c)$ is false everywhere

$(d)$ is always true as $P (A \cup B) = P(A)+P(B)-P (A \cap B)$

Since, $P (A \cap B) >=0$, $P (A \cup B) \leq P(A)+P(B)$

Correct Answer: $D$
answered by Boss (11.3k points)
edited by

in the case of  option 1 what is P(A ∩ B) if a and b are not independent?

In general, P(A $\cap$ B) = P(A|B) * P(B).

If A and B are independent, then knowing B doesn't affect probability of A, hence P(A|B) = P(A)

So P(A $\cap$ B) = P(A) * P(B) when A and B are independent.
Arbitrary means anything you can take it may be mutually exclusive or mutually independent.
Why is C false??
There must be a division sign after P(A∩B) that is why c is incorrect

correct formula is

+8 votes
If A and B are mutually exclusive events then:

A$\cap$B = $\Phi$ and A$\cup$B = A+B

If A and B are non mutually exclusive events then:

A$\cap$B != $\Phi$ and A$\cup$B = A+B-A$\cap$B (principle of inclusion-exclusion)

therefore A$\cup$B <= A + B

so D is the answer...
answered by Active (1.8k points)
edited by
+6 votes
Option D is Ans

If A and B be two arbitrary events, then

P (A ∪ B) = P (A) +P (B) - P(A $\cap$ B)

Let say 10=7+8-5

So now  10 <= 7+8

So option D ....P (A ∪ B) <= P (A) +P (B) is correct Ans.
answered by Boss (23.4k points)
0 votes
What are arbitrary events?If they are independent then Option A and if disjoint then Option C.
answered by Active (2.8k points)
Option A is correct
Option D should be correct.
0 votes
answered by (205 points)

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