Probability

A and B are said to be independent, if the occurrence of A does not depend on B.
)()()( BPAPBAP 
If A and B are not independent,
)/()()( ABPAPBAP 
)/()()( BAPBPABP 
P(AB)=
P(BA)=

Addition Theorem
)()()()( BAPBPAPBAP 
A and B are said to be mutually exclusive, if BA 
0)(  BAP

A and B are said to be mutually exclusive and exhaustive, if BA 
and SBA 
Question 1
In a certain school, 20 % of the students failed in English, 15 % failed in Mathematics,
and 10 % failed in both English and Mathematics, A student is selected at random. If he
passed in English, find the probability that he also passed in Mathematics.

100
20
)( EP
100
80
)( C
EP
100
15
)( MP
100
85
)( C
MP
100
10
)(  MEP
)()()()( EMPEPMPEMP 
100
25
100
10
100
15
100
20

Let E – failed in English
M-failed in Math

)(
)(
)/( C
CC
CC
EP
EMP
EMP


100
75
)()(  CCC
EMPEMP
16
15
80
75
)/( CC
EMP

R
8
W
5
2) A bag contains 8 red and 5 white balls. 2 draws of 3 balls are made without replacement.
Find the probability that the 1st draw gives 3 white balls and the 2nd draw gives 3 red balls
429
7
)()(
3
10
3
8
3
13
3
5

C
C
C
C
RRRPWWWP
3) A and B throw a pair of dice alternately. In order to win, they have to get a sum of 8. Find their
respective probabilities of winning if A starts the game.

{(2,6),(3,5),(4,4),(5,3),(6,2)}
36
31
,
36
5
 qp
P(A wins) = p +qqp + qqqqp+--------
)1( 42
 qqp
2
2
)
36
31
(1
36
5
1


 q
p
=0.537
Given that A starts the game
P(A wins) = p(A wins in 1st trial)+P(Awins in the 3rd trial)
+P(A wins in the 5th trial)+---------------

P(A wins) = 0.537
P( B wins) = 1- 0.537=0.463
Question 4
A bag contains 2n+1 coins. n of these coins have a head on both sides while the rest of
the coins are fair. A coin is picked up at random from the bag and tossed. If the probability
that the toss results in a head is 31/42, find n

P(head) = P(fair coin)P(head/fair coin appears)+P(biased coin)P(head/biased coin appears)
)1(
12
)
2
1
(
12
1
42
31





n
n
n
n
2)12(
21
42
31



n
nn
12
13
21
31



n
n
31(2n+1)=21(3n+1)
n=10
n+1 coins are fair
2n+1-(n+1)=n coins are biased
**********************************

Probability

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