edited by
2,905 views
7 votes
7 votes
If P and Q are two random events, then the following is TRUE:

(A)  Independence of P and Q implies that probability (P ∩ Q) = 0
(B)  Probability (P ∪ Q) ≥ Probability (P) + Probability (Q)
(C)  If P and Q are mutually exclusive, then they must be independent
(D)  Probability (P ∩ Q) ≤ Probability (P)
edited by

4 Answers

Best answer
6 votes
6 votes

here i check option---> i let P = probability,  P and Q is replaced by A and B.

A) it  is saying independent means P(A $\cap$ B) =  P(A)* P(B)  so option A is false

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

hence P(A $\cup$ B) >= P(A) + P(B)  its false option B is false

C) if A and B is mutually exclusive means P(A $\cap$ B) =  0 then its independent ........its false  there is no relation independent and mutually exclusive

D)  P(A $\cap$ B) <= P(A)

   its true

hence option D is correct

edited by
6 votes
6 votes

Let's analyse options one by one.

(a)  Independence of P and Q implies that probability (P ∩ Q) = 0

   No. This is the case for mutually exclusive events that if one occurs then the chance of another one occuring is ruled out.

  Infact Independence of P and Q means if P happens, then outcome of Q won't be affected by that.

  Means  P(Q|P) =Q because it has not effect whether P happens or not

which makes P(P $\cap$ Q) =P(P).P(Q) because again P(P|Q)= P.

(b) Probability (P ∪ Q) ≥ Probability (P) + Probability (Q)

This can hold true for mutually exclusive events P and Q but if P and Q can happen together(implies independence)

 then P(P $\cap$ Q) $\neq$ 0.

So P( P U Q )=  P(P) + P(Q) - P(P $\cap$ Q)

So, in this case P(PUQ) $\leq$ P(P) +P(Q)

(c) If P and Q are mutually exclusive, then they must be independent.

Absolutely false. When P and Q are mutually exclusive, then either P occurs or Q occurs but not both simultaneously. So if P happens, chance of Q happening gets ruled out and vice-versa. Hence, mutually exclusive events are dependent.

(d) Probability (P ∩ Q) ≤ Probability (P)

Consider 2 cases

Case 1 : P and Q are mutually exclusive.

Let S be sample space of outcomes of a single toss of a coin.

Let P : Outcome is head

   Q : Outcome is tails.

 Obviously, P(P $\cap$ Q) =0 and P(P)= $\frac{1}{2}$

Equality satisfied.

Case 2: P and Q are independent.

P : Outcome of getting head on a single toss of a coin

Q : Outcome of getting 4 on dice.

P(P $\cap$ Q) = $\frac{1}{12}$

P(P)=$\frac{1}{2}$

Equality satisfied.

Option (d) TRUE.

0 votes
0 votes
A) Let random experiment be tossing 2 coins ... Let E1 be probability of getting a head in first coin and E2 be probability of getting       tails in 2 nd coin ... Here both E1 and E2 are independent but P(E1 AND E2) = 1/4 != 0 ..so A) is eliminated ...

B) P(P OR Q) = P(P) + P(Q) WHEN P and Q are mutually exclusive . Else P(P OR Q) < P(P) + P(Q) .So P(P OR Q) <= P(P) +P(Q) ... So B) is FALSE ..

C) There is no relation between mutually exclusive and independent events ...

D) is TRUE always ... Ex : Probability of getting good rank in gate >= probability of getting good rank in gate AND getting into iits ...
Answer:

Related questions

72 votes
72 votes
8 answers
3
Ishrat Jahan asked Nov 3, 2014
30,037 views
An unbiased coin is tossed repeatedly until the outcome of two successive tosses is the same. Assuming that the trials are independent, the expected number of tosses is$3...
37 votes
37 votes
3 answers
4