GATE CSE 1994 | Question: 1.4, ISRO2017-2

Question

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

• A. $P (A \cap B) = P(A)P(B)$
• B. $P (A \cup B) = P(A)+P(B)$
• C. $P (A \mid B) = P(A \cap B)P(B)$
• D. $P (A \cup B) \leq P(A) + P(B)$

Comments

we are given with 2 arbitary events A and B

2 events A and B may or may not be independant, without knowing P(A U B) , P(A) and P(B) , we can't comment on that

if P(A ^ B) = 0 , then only P(A ^ B) = P(A) + P(B) , we don't have any info about their intersection as well

it is not given that B has already happened or something like that, applying conditional probability rule does not makes sense

 

last option is a general property

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Mutually Exclusive (Disjoint) Events

• Meaning: Two events cannot occur at the same time. If one event happens, the other is impossible.
• Probability Rule:The probability of both occurring is zero P(A,B)=0

Independent Events

• Meaning: The occurrence of one event has no effect on the probability of the other happening. Both events can potentially happen.
• Probability Rule: The probability of both occurring is the product of their individual probabilities P(A,B)=P(A)*P(B)

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@Omkar_Shelke your explanation of C is wrong, it doesn't make any sense. There is only a single reason of C being invalid which is we don't know P(b). ONLY in the case of P(b) = 0 the formula becomes invalid so C is false. If something to rule out this possibility was given then C would've been TRUE too like if in question it was given that P(b) != 0 or in option C it was given when/if P(b) != 0. 

Answer 1

• 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.

Comments

Just to add, option D is a famous inequality called as the Union bound aka the Boole’s Inequality.

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Just want to add, it seems option C isn’t false $\textbf{everywhere}$ if $B = \Omega$(sample space) then the equality holds!

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That is smart

Answer 2

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...

Answer 3

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.

Answer 4

What are arbitrary events?If they are independent then Option A and if disjoint then Option C.

Comments

Option A is correct

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Option D should be correct.

Answer 5

A and B are two arbitrary events.

When A and B are independent events,then P(A∩B)=P(A)P(B) but in the question it is not mentioned that A and B are independent events.

When (A∩B)=Φ then P(A∪B)=P(A)+P(B) but here (A∩B)=Φ is not specified.

P(A|B)=P(A∩B)/P(B)

P(A∪B)=P(A)+P(B)-P(A∩B) ,P(A∩B)>=0 so P(A∪B)<=P(A)+P(B)

So option D is the right option.

Answer 6

(a) P(A ∩B) = P(A) P(B) is false since this true if and only if A and B are independent events.

(b) P(AUB) = P(A) + P(B) is false since P(A∩ B) is zero if and only if A and B are mutually

exclusive.

(c) P(A|B) = P(A ∩ B)/P(B) is true.

(d) P(AUB) <P(A) + P(B) is false. Since P(AUB) ≤ P(A) + P(B)