Q1.
An urn contains 3 pink and 6 white balls . A, B, C and D draw one ball in turn without replacement, one who gets pink ball first wins .
find the probability of winning of C.
Solution.
If A and B both will lose then C will get chance to win
probability of winning of C = P(A' ∩ B' ∩ C)
= P(A') . P(B') . P(C) =
6/9
.
5/8
.
3/7
=
5/28
Solution
Q2.
If E(x2) = 3 and E(x) = 1.5 , find E(x - 3)2
E(x - 3)2 = E(x2 - 6x + 9)
= E(x2) - 6E(x) + 9 = 3 - 9 + 9 = 3
Solution
Q3.
P(A/B') is defined only when
(a) A is a sure event
(b) B is a sure event
(c) B is not a sure event
(d) None
Solution.
(c) B is not a sure event
P(
A/B'
) =
P(A ∩ B')/P(B')
=
P(A ∩ B')/1 - P(B)
if P(B) = 1
1 - P(B) = 0
Solution
Q4.
If A and B are mutually exclusive events then P(A ∪ B) is equal to _________________ .
Solution.
P(A ∪ B) = P(A) + P(B)
∵ P(A ∩ B) = 0
Solution
Q5.
If P(A - B) = P(B - A), which statement is correct
(a) P(A ∪ B) = 1 (b)P(A ∩ B) = 0
(c) P(A) + P(B) = 1 (d) P(A) = P(B)
Solution.
(d) P(A) = P(B)
P(A - B) = P(B - A)
P(A) - P(B - A)
P(A) - P(A ∩ B) = P(B) - P(A ∩ B)
P(A) = P(B)
Solution
Q6.
A, B and C are three mutually exclusive and exhaustive events such that P(A) = 2P(B) = 4P(C) find the value of P(C) .
Solution.
A, B and C are exhaustive events
so P(A) + P(B) + P(C) = 1
let P(A) = 2P(B) = 4P(C) = x
P(A) = x , P(B) =
x/2
, P(C)
=
x/4
x +
x/2
+
x/4
= 1 ⇒
3x/2
+
x/4
= 1
14x/8
= 1 ⇒
7x/4
= 1 ⇒ x =
4/7
P(C) =
x/4
=
4/7.4
=
1/7
Solution
Q7.
A and B be events of a sample space S of an experiment , then find the value of
P(
S/A
)
Solution.
P(
S/A
) =
P(S ∩ A)/P(A)
=
P(A)/P(A)
= 1
Solution
Q8.
A boy contains 5 Red balls, 4 blue balls and x balls . If the random probability of picking two green ballsis
1/7
, find x .
Solution.
1/7
=
xC2/9 + xC2
=
x!/2!(x - 2)!
/
(9 + x)!/(7 + x)! 2!
1/7
=
x!/(x - 2)!
x
(7 + x)!/(9 + x)!
1/7
=
x (x - 1)(x - 2)!/(x - 2)!
x
(7 + x)!/(9 + x)(8 + x)(7 + x)!
1/7
=
x(x - 1)/(9 + x)(8 + x)
solving this x = 6
Solution
Q9.
If P(A ∪ B) = P(A), find P(A ∩ B).
Solution.
P(A ∩ B) = P(B)
Solution
Q10.
P(A) =
1/2
, P(B') =
5/8
, , P(A ∪ B) =
3/4
. find
P(A' ∩ B')
Solution.
P(A' ∩ B') = P(A ∪ B)' = 1 - P(A ∪ B) =
1/4
Solution
