BMS College of Engineering, Bangalore -19 I-Internals, PSQ

Scheme

a) Define Sample Space, Event with an example each. (1+1 = 2 M)

Sample Space is set of all possible outcomes of an experiment.

Example: While throwing a single die Sample Space S = {1,2,3,4,5,6}.

An event is a set of possible outcomes of an experiment (a subset of a sample space) to which

probabilities is assigned.

Example : “ A face card drawn out of a deck of cards”

1. (b) Using Axioms of the algebra of events prove the relations: (4 x 2 = 8M)

i) A U A = A A = A U Φ (Identity Law) = A U (A ∩ Ac) (complementation Law) = (A U A) ∩ ( A U Ac) (Distributive Law) = ( A U A ) ∩ S (Complementation Law) = A U A (Identify Law)

ii) A U S = S S = A U Ac (Complementation Law) = A U (Ac ∩ S) (Identify Law) = (A U Ac) ∩ ( A U S) (Distributive law) = S ∩ ( A U S) (Complementation law) = (A U S) ∩ S (Commutative law) = A U S (Identity law)

iii) A ∩ Φ = Φ

Φ = A ∩ Ac (Complementation law)

= A ∩ (Ac U Φ) (Identity Law)

= (A ∩ Ac) U ( A ∩ Φ ) (Distributive Law)

= Φ U ( A ∩ Φ) (Complementation Law)

= (A ∩ Φ) U Φ (Commutative Law)

= A ∩ Φ (Identity Law)

iv) A ∩ ( A U B) = A

A = A U Φ (Identify Law)

= A U (B ∩ Φ) (By (iii))

= (A U B) ∩ ( A U Φ) (Distributive Law)

= (A U B) ∩ A (Identity Law)

= A ∩ ( A U B) (Commutative Law)

2. (a) ( 3 x2 =6M)

2. b) If A and B are independent events prove that �̅� and �̅� are also Independent. (4M)

3. Consider a pool of six I/O (input/output) buffers. Assume that any buffer is just a likely to be

available (or occupied) as any other. Compute the probabilities associated with the following events.

A = “ At least 2 but no more than 5 buffers occupied” 5M

B = “ At least 3 but no more than 5 occupied” 5M

Ans: Problem can be solved by using sample space or using combinatorial or using Bernoulli trails

The total number of buffers can be arranged in 26 possible ways. i.e, |S| = 64.

A = “ At least 2 but not more than 5 buffers occupied”

P(A) = (P(A=2) + P( A=3) + P(A=4) + P(A=5))/ |S|

= (6 C 2 + 6 C 3 + 6 C 4 + 6 C 5) / 64 = 56/64 = 7/8

B = “ At least 3 but no more than 5 occupied”

P(B) = (P(A=3) + P(A=4) + P(A=5) / |S|

= ( 6 C 3 + 6 C 4 + 6 C 5) / 64 = 41/64 = 7/8

4. (6 M (for pairwise)+3 M (mutual independence) + 1 M (Sample space representation))

A ∩ B = {(1, 4), (1, 5), (1, 6), (2,4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

And A ∩ C = B ∩ C

= A ∩ B ∩ C

= { (1,6), (2, 5), (3,4)}

Therefore

P ( A ∩ B ) = 1/4 = P(A). P(B)

P ( A ∩ C ) = 1/12 = P(A). P(C)

P ( B ∩ C ) = 1/12 = P(B). P(C)

But P (A ∩ B ∩ C ) = 1/12 = P(A). P(B).P(C) = 1/24

Therefore Events A, B and C are pairwise independent but not mutually independent.

5. (a) 5M

5. (b) 5M

6. (a) 5M

6. (b) 5x1 =5M

P(A) = 0.6, P(B) = 0.3, P( A ∩ B) = 0.2

P(AC) = 1- P(A) = 0.4

P(BC) = 1- P(B) = 0.7

i) P( A | B) = 𝑃 ( 𝐴 ∩𝐵)

𝑃(𝐵) = 0.66

ii) P (A U B) = P(A) + P(B) – P(A ∩B) = 0.7

iii) P (AC) = 1-P(A) = 0.4

iv) P(BC) = 1- P(B) = 0.7

v) P ( BC | AC) = 0.75