Q1.
Three unbased coins are tossed together . what is the probability of getting at least 2 heads .
Solution.
Let A is the event to get at least 2 heads
S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
A = {HHH, HHT, HTH, THH}
n(S) = 8, n(A) = 4
P(A)=
n(A)/n(S)
=
4/8
=
1/2
Solution
Q2.
A card is drawn at random from a pack of 52 cards . find the probability of getting a club or a king .
In this card n(S) = 52
P(A or B) = P(A ∪ B)
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
A = shows king , B = shows club
P(A ∪ B) =
4/52
+
13/52
-
1/52
=
4/13
Direct one can do this , there are 13 club cards and 4 king but one king is already exist in 13 club cards so favourable outcome are 16 ,
probability is
16/52
=
4/13
Solution
Q3.
3 Boys and 2 girls are to be seated in a row . find the probability that both the girls are together .
Solution.
There are 5! to seat 3 boys and 2 girls in a row .
favourable outcomes when both the girls are seating together = 4! . 2!
So required probability is =
4! . 2!/5!
=
4! x 2/5 x 4!
=
2/5
Solution
Q4.
Five boys and 3 girls are to be seated in a row . find the probability that no two girls are together .
Solution.
Total number of ways to arrange 8
person in a row is 8! = 40320
The number of ways in which no two girls are
together 5! x
6P
3 = 14400
So required probability is =
14400/40320
=
5/14
Solution
Q5.
3 books of maths , 2 books of physics and 4 books of statistics are randomly arranged on a shelf in a vertical row . find the
probability that the book of the same language are together .
Solution.
Total number of ways to arrange the book are 9! and
the number of ways to arrange the books of same language together = 3! . 2! . 4! . 3!
the required probability is
3! . 2! . 4! . 3!/9!
Solution
Q6.
A dice thrown 2 time find the probability of getting sum 14 on upper faces .
Solution.
getting sum 14 is impossible event , so probability will be zero .
Solution
Q7.
In a probability distribution function the sum of all probabilities will be equal to _________ .
Solution.
1
Solution
Q8.
Which statement is/are correct .
(a) probability value must be non-negative .
(b) E(a) = a (Expectation of a constant is a
constant)
(c) E(x + y) = E(x) + E(y)
(d) All
Solution.
(d) All
Solution
Q9.
If v(x) = 10 , find the value of v(10x + b) where b is an constant .
Solution.
= v(10x + b)
= 100 v(x)
= 1000
∴ v(b) = 0 (variance of constant is zero .)
Solution
Q10.
Conditional probability P(A/B) is defined only when
(a) A is an sure event
(b) B is an certain event .
(c) B is not an
impossible event .
(d) None of these .
Solution.
(c) B is not an impossible event .
P(A/B) =
P(A ∩ B)/P(B)
∴ P(B) ≠ 0
Solution