Question-1
Solution:
(i) P(A)+ P(B)+ P(C)+ P(D) = 0.37+ 0.17+ 0.14 + 0.32 = 1.00
Total Probability = 1
âˆ´ Permissible.
(ii) P(A)+ P(B)+ P(C)+ P(D) = 0.30 + 0.28 + 0.26 + 0.18 = 1.02 > 1
âˆ´ not permissible.
(iii) Not possible since P (C) is negative.
(iv) P (A) + P (B) + P (C) + P (D) =
= = 1
Since the total probability â‰ 1. It is not permissible.
Question-2
Solution:
(i) Sample space = {(1,1) (1,2), (1,3), (1,4), (1,5), (1,6), (2,1) (2,2), (2,3), (2,4), (2,5), (2,6), (3,1) (3,2), (3,3), (3,4), (3,5), (3,6), (4,1) (4,2), (4,3), (4,4), (4,5), (4,6),(5,1) (5,2), (5,3), (5,4), (5,5), (5,6), (6,1) (6,2), (6,3), (6,4), (6,5), (6,6)}
n(S) = 36
A = obtaining sum less than 5
= {(1,1) (1,2), (2,1), (3,1), (2,2), (1,3)}
n(A) = 6
âˆ´ P(A) = = =
(ii) B = obtaining sum greater than 10 = {(5,6) (6,5) (6,6)}
âˆ´ n(B) = 3
âˆ´ P(B) = = =
(iii) C = a sum of 9 or 11
C = {(4,5) (5,4) (3,6) (6,3) (6,5) (5,6)}
âˆ´ n (C) = 6
âˆ´ P(C) = = =
Question-3
(i) exactly two heads
(ii) atleast two heads
(iii) almost two heads
Solution:
n (S) = 8
Probability of getting head = and Probability of getting tail =
(i) Probability of getting exactly two heads = 3C_{2} =
(ii) Probability of getting atleast two heads
= P(2) + P(3)
= 3C_{2}
=
(iii) Probability of getting almost two heads = P(0)+ P(1)+ P(2)
=
= =
Question-4
Solution:
Probability of the card is jack or king
= Probability (jack) + Probability (king)
= =
Probability of getting the card that will be 5 or smaller = = =
Probability of getting queen or 7 = = =
Question-5
(i) all are white
(ii) one white and 2 black.
Solution:
(i) n (S) = 10C_{3 }
A = event of getting all the three are white balls
âˆ´ n (A) = 5C_{3 }
\ P(A) = = =
(ii) B = event of getting one white and 2 black balls
n (B) = 5C_{1 }Ã—_{ }7C_{2 }
âˆ´ P(B) = = 5 Ã— Ã— =
Question-6
Solution:
Probability of getting a defective bulb = = âˆ´ Probability of getting a non defective bulb =
While 5 bulbs are chosen at random, the probability of getting none of the defective bulb is = 10C_{5 }
= = 10C_{5}
Question-7
(ii) both are of the same variety.
Solution:
n(S) = 7C_{2 }
A = getting 1 mango and 1 apple
n(A) = 4C_{1} Ã— 3C_{1} = 4 Ã— 3
âˆ´ P(A) = =
B = getting both as same variety
n(B) = 4C_{2} + 3C_{2 }= = 6 + 3 = 9
âˆ´ P(B) = =
Question-8
Solution:
Possibilities |
Girls 6 |
Boys 4 |
Combinations |
1 |
2 |
2 |
6C_{2} x 4C_{2} = 90 |
2 |
3 |
1 |
6C_{3} x 4C_{1} = 80 |
3 |
4 |
0 |
6C_{4} x 4C_{0} = 15 |
Total = 185 |
n(S) = 10C_{4}
âˆ´ Probability = = =
Question-9
Solution:
An ordinary year consists of 365 days.
365 days = 52 weeks + 1 day. 52 weeks will have 52 Sundays.
Let one day be any one of the following Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday.
The probability that an ordinary year may contain 53 Sundays =
A leap year contains 366 days
366 days = 52 weeks + 2 days
These two days may be any one of the following combinations
1) Monday + Tuesday
2) Tuesday + Wednesday
3) Wednesday + Thursday
4) Thursday + Friday
5) Friday + Saturday
6) Saturday + Sunday
7) Sunday + Monday
In the above seven, Sunday appears only on two.
Required Probability = =
Question-10
Solution:
n (S) = 50
A = {2,3,5,7, 11, 13,17, 19,23,29,31,37,41,43,47, 4,8,12,16,20,24,28,32,36,40,44,48}
n(A) = 27
âˆ´ P(A) =
Question-11
(ii) P()
Solution:
(i) P(A) = 0.36, P(A âˆª B) = 0.9, P(A âˆ© B) = 0.25
P(A âˆª B) = P(A)+ P(B) - P(A âˆ© B)
0.9 = 0.36 + P (B) - 0.25
P(B) = 0.9 + 0.25- 0.36= 0.79
(ii) P() =1 - P(A âˆª B)= 1-0.9=0.1
Question-12
(ii) P(A âˆª B)
(iii) (
(iv) P(
Solution:
(i) P(A) = 0.28, P (B) = 0.44. A and B are mutually exclusive.
P() = 1 - P (A) = 1 - 0.28 = 0.72
(ii) P(A âˆª B) = P (A) + P (B) = 0.28 + 0.44 = 0.72
(iii) ( = P (A) - P (A âˆ© B) = 0.28 - 0 = 0.28
(iv) P( = 1- P(A âˆª B)= 1- 0.72= 0.28
Question-13
(i) P(A âˆª B)
(ii) P(
(iii) P(
(iv)
(v) P(
Solution:
(i) P (A âˆª B) = P (A) + P (B) - P (A âˆ© B) = 0.5 + 0.6 - 0.24 = 0.86
(ii) P( = P (B) - P (A âˆ© B) = 0.6 - 0.24 = 0.36
(iii) P(= P(A)- P(A âˆ© B) = 0.5- 0.24 = 0.26
(iv) P = 1 - P(A âˆ© B) = 1 - 0.24 = 0.76
(v) P= 1 - P(A âˆª B) = 1 - 0.86 = 0.14
Question-14
Solution:
n (S) = 36;
A = event of "first die shows 4"
n(A) = 6 ;
âˆ´ P(A) =
B = event of "second die shows 4"
n(B) = 6 ;
âˆ´ P(A) =
A âˆ© B = event of first die showing 4 and second die showing 4
n(A âˆ© B) = 1 ;
âˆ´ P(A âˆ© B) =
âˆ´ P (A âˆª B) = P (A) + P (B) - P (A âˆ© B) = =
Question-15
Solution:
P(A) = 0.5 ; P (B) = 0.3 ; A âˆ© B = Ï† (given)
P = 1- P(Aâˆª B)
= 1-{P(A)+P(B)}
= 1- {0.5 + 0.3} = 0.2
Question-16
(i) a queen or club card?
(ii) a queen or a black card?
Solution:
(i) n(S) = 52
A = getting a queen
n(A) = 4 ; P (A) =
B = getting a club
n(B) = 13 ; P(B) =
A âˆ© B = getting a club queen
n(A âˆ© B) = 1 ; P(A âˆ© B) =
âˆ´ P(A âˆª B) = = =
(ii) A = getting a queen
n (A) = 4; P(A) =
B = getting a black card;
n(B) = 26 ; P (B) =
Aâˆ© B = getting a black queen
n(Aâˆ© B) =2; P(Aâˆ© B) =
âˆ´ P(Aâˆª B) = P (A) + P (B) - P (A âˆ© B) = = =
Question-17
(i) it will get atleast one of the two awards?
(ii) it will get only one of the awards?
Solution:
(i) Probability for design award = 0.25 = P (A)
Probability for efficient award = 0.35 = P(B)
Probability for both the awards = 0.15 = P(Aâˆ© B)
Probability that it will get atleast one award
P(Aâˆª B) = P(A) + P(B) - P(A âˆ© B)= 0.25 + 0.35 - 0.15 = 0.45
(ii) += P(A)- P(Aâˆ© B) + P(B) - P(Aâˆ© B)
= 0.25-0.15+0.35-0.15
= 0.60 -0.30
= 0.30
Question-18
Solution:
Independent events
Two events A and B are said to be independent events if happening of one does not depend on happening of the other.
Mutually exclusive events
Two events A and B are said to be mutually exclusive if happening one prevents the happening of the other.
These two events cannot be mutually exclusive and independent simultaneously for non-empty events. Of course possible for any one being null event that is impossible event
Question-19
Solution:
Given A and B are independents.
P(A âˆ© B) = P(A) . P(B)
To prove that and and also independents.
That is P() = P().P()
LHS P() = 1 - P()
= 1 â€“ P(A) + P(B) â€“ P()
= 1 â€“ P(A) + P(B) â€“ P(A). P(B)
=[1 â€“ P(A)] [1 â€“ P(B)]
= P().P() = R.H.S
Question-20
Solution:
P (A) = 0.4, P(B) = 0.4 and P(B/A) = 0.5
P (B/A) =
0.5 =
P (Aâˆ© B) = 0.2
(i) P (A/B) = = =
(ii) P (Aâˆª B) = P(A) + P(B) â€“ P(Aâˆ© B) = =