3. In tossing a coin, events
are:
P ( Head) = 1/
2
P ( Tail ) = 1/
2
The Resource Center
4. Probability (Head or Tail)
1/ + 1/ =1
2 2
Probability ( A or B)
= P(A) + P(B)
Probability 1
The Resource Center
5. Example of 2 coins tossed
Case 1: 2 H
Case 2: 2 T
Case 3: H & T
The Resource Center
6. What is the probability of
“a head and tail”?
Favorable outcome
Total outcome
1/
= 3
The Resource Center
7. Wrong
Case 1: HH
Case 2 : TT
Case 3 : HT
Case 4 : TH
2 / = 1/
So P = 4 2
The Resource Center
8. With 3 coins, what is the
probability of only 1 head?
P (HTT) = 2
(1/ )*(1/ 1/ ) 1/
= 8 2)*( 2
Joint P = P(A) * P(B) * P(C)
P (THT) and P (TTH) are also
1/ each.
8
1/ + 1/ + 1/ = 3/
So P = 8 8 8 8
The Resource Center
9. 2 dice are thrown
Find the probability that the
sum of faces is 1
P=0
Probability
The Resource Center
0
10. Find the probability that
the sum of faces is 2
Probability = 1/
36
The Resource Center
11. Find the probability that
the sum of faces is 12
Probability = 1/
36
The Resource Center
12. Find the probability that
the sum of faces is 6
Total Sum Set = { 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 12}
So Probability = 1/
11
The Resource Center
13. Wrong
Total outcomes = 6*6 = 36
Favorable outcomes
{ (3,3), (2,4), (1,5), (4,2), (5,1) }
5/
P = 36
Probability of all outcomes
should be equal
The Resource Center
14. 2 dice are thrown
Find the probability that the
sum is single digit
The Resource Center
15. Prob ( Sum = 2) = 1/
36
Prob ( Sum = 3) = 2/
36
Prob ( Sum = 4) = 3/
36
Prob ( Sum = 5) = 4/
36
The Resource Center
16. Prob ( Sum = 6) = 36 5/
Prob ( Sum = 7) = 366/
Prob ( Sum = 8) = 365/
Prob ( Sum = 9) = 364/
Total =
(1 + 2 + 3 + 4 + 5 + 6 + 5 + 4)
36
The Resource Center
17. Alternatively
Prob ( Sum = 10) = 3/
36
Prob ( Sum = 11) = 2/
36
Prob ( Sum = 12) = 1/
36
The Resource Center
18. P ( Sum 10,11 or 12) = 36 6/
P (Sum is not 10, 11 or 12)
6/ 30/
= 1 - 36 = 36
P(A c) = 1 - P(A)
Its easier to work out the
complement sometimes...
The Resource Center
19. 4 dice are thrown -
probability that there are 3
sixes and another number
Favorable outcome = 1*1*1*5*4
Total outcomes = 6*6*6*6
P = 20 / 6 4
The Resource Center
20. Probability that a number
is divisible by 2 is
P= 1/
2
The Resource Center
21. Probability that a number
is divisible by 3 is
P= 1/
3
The Resource Center
22. Probability that a number
is divisible by 2 or 3 is
P= 1/ + 1/ = 5/
2 3 6
The Resource Center
23. Is it?
Probability that number is
divisible by 2 or 3 is
1/ + 1/ - 1/ = 4/ = 2 /
2 3 6 6 3
P (A B) =
P(A) + P(B) - P(A B)
The Resource Center
24. 10 coins are tossed, what
is the probability that only
one is a tail or head?
The Resource Center
25. Prob of 1 H, 9 T is = 1/ but 210,
there are 10 such cases. So
prob = 10/2 10
Prob of 9 H, 1 T is = 10/2 10
Prob of one tail or head is =
10/2 10 + 10/210 = 10/29
The Resource Center
26. In a single toss of 4 coins
what is the probability of
getting at least one head?
P (no heads) = 1/
16.
P (at least 1 head) =
1- 1/ = 15/
16 16
The Resource Center
27. In a single toss of 4 coins
what is the probability of
getting at most 1 head?
P (no heads) = 16. 1/
1/
P (1 H & 3 T) = 16 * 4.
P (at most 1 H)
5/
= P (0 H) + P (1 H) = 16
The Resource Center
28. 2 cards are successively
drawn without
replacement. Probability
that both are greater than
2 & less than 9 is
The Resource Center
29. There are 6 * 4 = 24 cards
P (1st card) = 24/
52
P (2nd card) = 23/
51
Joint P = 24/ * 23/
52 51
The Resource Center
30. Probability of 1 st
ball blue
& 2 nd white without
replacement?
The Resource Center
31. first ball blue = 4/ 2/
10 = 5.
second ball white = 9 6/ = 2/
3
Joint P = 2/ x 2/ = 4/
5 3 15
The Resource Center
32. Probability of all 4 being
blue - without
replacement?
The Resource Center
33. = (7/ 6/ )*(5/ )*(4/ )
13)*( 12 11 10
7/
= 143
The Resource Center
34. Probability of all four not
blue - without
replacement?
The Resource Center
37. All white =
6/ x 5/ x 4/ x 3/ = 3/
13 12 11 10 143
Probability all same color =
P (all B) + P (all W) = 10/
143
The Resource Center
38. The probability that a card
drawn from a pack of 52
cards is a diamond/king?
There are 13 diamonds and 4
kings, but one of them is a king
of diamond. Therefore,
P= (13 + 4 - 1) / 16/
52 = 52
The Resource Center
39. In a lottery of tickets
numbered 1 to 50, 2
tickets are drawn. The
probability that both are
prime numbers is:
The Resource Center
40. (15 prime numbers from 1 - 50)
15/
P (1st being prime) = 50
P (2nd being prime) = 4914/
P (both prime) = 15*14 / 50*49
The Resource Center
41. An AA gun can take max
4 shots. P of hitting the
plane at the 1 st, 2nd, 3rd
and 4 th shot are 0.4, 0.3,
0.2, 0.1 resp. P that the
gun hits the plane is:
The Resource Center
42. Probability of hitting the plane =
P (the plane is hit at least once)
= 1 - (1-p1)(1-p2)(1-p3)(1-p4) =
1 - (0.6)*(0.7)*(0.8)*(0.9) =
(1 - 0.3024) = 0.6976
The Resource Center
43. A contains 4 red and 5
white balls and B contains
3 red and 7 white balls. 1
ball is drawn from A and 2
from B. Find the
probability that out of 3, 2
are white and 1 is red.
The Resource Center
44. P(2 W & 1 R) =
P[(1 R from 1 st and 2 W from
2 nd or
(1 W from 1 st,1 W and 1 R from
2 nd)]
4/ )*(7/ )*(6/ )+ (5/ )*(7/ )*(3/ )
( 9 10 9 9 9 10
7/
= 15
The Resource Center
45. A speaks the truth in 75%
and B in 80% of the
cases. In how many cases
will they contradict each
other narrating the same
incident.
The Resource Center
46. P(A and B contradict each
other)
= P[(A tells truth and B tells a
lie)
or
P(A tells a lie and B tells truth)
= 3 /4 * 1 /5 + 1 /4 * 4 / 5
= 207/
The Resource Center
47. P of two events E and F
are 0.25 and 0.50 resp. P
of their simultaneous
occurrence is 0.14. P that
neither E occurs nor F is:
The Resource Center
48. P (E U F) =
0.25 + 0.50 - 0.14
= 0.61.
Now, P (neither E nor F)
= 1 - P(E U F)
= 1 - 0.61 = 0.39
The Resource Center
49. The probability that a leap
year selected at random
will contain 53 Sundays,
is:
The Resource Center
50. A leap year contains 366 days
and therefore, 52 weeks and 2
days. Now, there can be 7
different combinations of the
remaining two days, where 2
favour the required event.
Hence, P = 2/7.
The Resource Center
51. The probability of solving
a problem by A, B, and C
1/ , 1/ and 1/
are 2 3 4
respectively. The
probability that the
problem is solved is:
The Resource Center
52. P = 1 - P(that the problem
remains unsolved)
= 1 - (1-p1)(1- p2)(1- p3)
=1-( 2 1/ )*(2/ )*(3/ )
3 4
= 1 - 1 /4
= 43/
The Resource Center
53. If P that A and B will die
within a year are a and b
resp, P that only one of
them will be alive at the
end of the year is:
The Resource Center
54. Required probability
= P(A dies & B alive) or
P(A is alive and B dies.) =
a (1 - b) + (1 - a) b
= a + b - 2ab.
The Resource Center
55. In a row of 6 Mathematics
books and 4 Physics
books, the probability that
3 particular Maths books
will be together is:
The Resource Center
56. Taking 3 books as one book,
we have 8 books which can be
arranged in 8! ways.
These 3 books can be arranged
in 3! ways. Hence n(F) = 8!*3!
Also n(T) = 10!.
1/
Therefore P = (8!*3!) / 10! = 15
The Resource Center
57. N cadets stand in a row. If
all possible permutations
are equally likely, P of 2
particular cadets standing
side by side is:
P = [(N-1)! * 2!] / N! = 2/
N
The Resource Center
58. A can solve 90% of the
problems given in a book
and B can solve 70%.
What is P that at least 1 of
them will solve a problem,
selected at random?
The Resource Center
59. P(E) = 0.9, P(F) = 0.7
Also P(E F) = P(E)*P(F) =
0.63.
Hence required probability is
0.9 + 0.7 - 0.63
= 0.97
The Resource Center
60. Bag X contains 3 B and 4
W balls. Bag Y contains 4
B and 2 W balls. A bag is
selected at random. From
the selected bag, 1 ball is
drawn. Find P that it is W.
The Resource Center
61. P (drawing W from X) = 7. 4/
P (drawing W from Y) = 6 2/ .
Choosing either bag gives a
1/ .
probability of 2
Hence, total probability
1/ (4/ + 2/ ) = 19/
= 2 7 6 42
The Resource Center
62. 3 urns contain 2 W & 3 B
balls, 3 W & 2 B and 4 W
& 1 B resp. A ball is drawn
from an urn at random
and it is W. What it P that
it was drawn from 1 st urn?
The Resource Center
63. P=
2/ )*(1/ )
( 5 3
------------------------------------------
(2/5)*(1/3) + (3/5)*(1/3) + (4/5)*(1/3)
= 92/
The Resource Center
64. X contains 5 W and 7 B
balls. Y contains 7 W and 8
B. 1 ball is drawn from X
and put into Y without
noticing its colour. If now a
ball is drawn from Y, what is
P that it is W?
The Resource Center
65. P(transfer W ball from X & then
5/ * 8/
select W ball from Y) = 12 16
5/ .
= 24
P( transfer B ball from X and
then draw B ball from Y) =
7/ )*(7/ ) = 49/
( 12 16 192.
Total probability is =
5/ ) + (49/
( 24 89/
192) = 192
The Resource Center
