1.Algebra Booster

Probability 8.71
[P(A) + P(B) + P(C) – P(A « B)
3p
- PB ( « C) - PC ( « A)
] =
2
…(iv)
Also, P(A « B « C)
…(v)
Adding Eqs (iii) and (iv), we get
[P(A) + P(B) + P(C) – P(A « B)
– P(B « C) – P(C « A) + P(A « B « C)]
3p
2
= + p
2
Thus,
3p
2
PA ( » B» C)
= + p
2
53. The roots of x 2 + px + q = 0 are imaginary if and only if
p 2 – 4q < 0, i.e. p 2 < 4q
The possible pairs of p and q can happen is as follows,
q p No. of pairs of p and q
1 1 1
2 1,2 2
3 1,2,3 3
4 1,2,3 3
5 1,2,3,4 4
6 1,2,3,4 4
q p No. of pairs of p and q
7 1,2,3,4,5 5
8 1,2,3,4,5 5
9 1,2,3,4,5 5
10 1,2,3,4,5,6 6
-----------------------------------------------------
Thus, the total possible pairs of p and q = 38
Required probability,
P(roots are real)
= 1 – P(roots are imaginary)
Ê 38 ˆ
= Á1 -
Ë
˜ 100 ¯
62
= = 0.62
100
54. (a) The number of ways of choosing 8 winners out of
16 is 16 C 8
.
The number of ways of choosing S 1
and 7 other
winners out of 15 is 15 C 7
.
Required probability,
15
C7
P(S 1
will win) =
16
C8
=
( 15 )! ( 8 )! ¥ ( 8 )! 8 1
¥ = =
7! ¥ 8! 16! 16 2
( ) ( )
( )
(b) Let E 1
denotes the event S 1
and S 2
are paired and
E 2
denotes the event S 1
and S 2
are not paired and A
denotes the event that one of the two players S 1
and
S 2
is among the winners.
1 14
So, PE ( 1) = and PE ( 2)
=
15 15
In case E 2
occurs, the probability that exactly one
of S 1
and S 2
is among the winners., i.e.
PS ( 1« S2 or S1«
S2)
= PS ( 1« S2) + PS ( 1«
S2)
= PS ( 1) ◊ PS ( 2) + PS ( 1) ◊PS
( 2)
1Ê 1ˆ Ê 1ˆ
1
= Á1- + 1-
2Ë ˜
2¯ Á
Ë
˜
2¯
2
1 1 1
= + =
4 4 2
1
Now, PAE ( / 1) = 1and PAE ( / 2)
=
2
Hence, the required probability,
Ê Aˆ Ê A ˆ
P(A) = PE ( 1) PÁ + PE ( 2)
P
ËE
˜
1¯ Á
ËE
˜
2¯
Ê 1 ˆ Ê14ˆ
1
= Á ◊ 1+ ◊
Ë
˜
15¯ Á
Ë
˜
15¯
2
16 8
= =
30 15
55. Let A the event, minimum of the chosen nos is 3 and B
the event, maximum of the chosen nos is 7.
Now,
P(A) = P(Choosing 3 and two other numbers
from 4 to 10)
7
C2
7.6 6 7
= = ¥ =
10
C 2 10.9.8 40
3
P(B) = P(Choosing 7 and two others numbers
from 1 to 6)
6
C2
6.5 6 1
= = ¥ =
10
C 2 10.9.8 8
3
and P(A « B) = P(Choosing 3 and 7 and one number
from 4 to 6)
3 3.6 1
= = =
10
C 10.9.8 40
3
Hence, the required probability,
P(A » B) = P(A) + P(B) – P(A « B)
7 1 1
= + -
40 8 40
11
=
40
56. Here, we fill 3 black balls in between 7 white balls.
8
C3 8.7.6 7
Required probability = = =
( 10 )! 10.9.8 15
( 7! ) ¥ ( 3! )
57. Required probability
P(2 white ball and 1 black ball)
= P(W 1
W 2
B 3
or W 1
B 2
W 3
or B 1
W 2
W 3
)
= P(W 1
W 2
B 3
) + P(W 1
B 2
W 3
) + P(B 1
W 2
W 3
)