1.Algebra Booster

Probability 8.63
= Co-efficients of x 8 in (1 – 4x 6 ) ¥ (1 – x) –4
= Co-efficients of x 8 in (1 – 4x 6 ) ¥ (1 + 4 C 1
x
+ 5 C 2
x 2 + … + 11 C 8
x 8 + …)
= ( 11 C 8
– 4 ◊ 5 C 2
)
= 125
Hence, the required probability
3
125 Ê5ˆ
1
= = Á ¥
1296 Ë
˜
6¯
6
Thus, m = 5 and n = 6.
Hence, the value of (n – m + 2) = 6 – 5 + 2 = 3.
7. Two numbers a and b can be chosen as
(3, 2), (6, 4), (9, 16), (12, 8), (15, 10).
Hence, the required probability,
5 5¥
2 1
= =
15
C2
15 ¥ 14 21
1
Thus, p =
21
Hence, the value of (84p + 2) = 4 + 2 = 6.
8. First we arrange the given set of numbers as
1, 4, 7, ..., 97
2, 5, 8, ..., 98
3, 6, 9, ..., 99
Now, (x 3 + y 3 ) is divisible by 3 only when one number
from the first row and 1 number from the second row
or 2 numbers from the third row.
Thus, the number of favourable ways,
33.32
33
C 1
¥ 33 C 1
+ 33 C 2
= 33.33 +
2
= 33(33 + 16) = 33 ¥ 49
Hence, the required probability,
33 ¥ 49 33 ¥ 49
=
99
C2
99 ¥ 49
1
=
3
Thus, p = 1 and q = 3
Hence, the value of p + q + pq = 1 + 3 + 7 = 7.
9. Probability of drawing a white ball 4th time at 7th
draw.
= Probability of drawing 3 white balls in first 6 draws
¥ probability of drawing a white ball at the 7th draw.
6 3 3 16
= ( Cpq 3 ) ¥
24
Ê
3 3
Ê2ˆ Ê1ˆ
ˆ 2
= Á20
¥ Á ˜ Á ˜ ˜ ¥
Ë Ë3¯ Ë3¯
¯ 3
3
Ê2ˆ
40
= Á ¥
Ë
˜
3¯
81
Clearly a = 2, b = 3.
Hence, the value of a + b + 1 = is 2 + 3 + 1 = 6.
10. Let the number of elements in set X is m.
\ Number of selections of two subsets is 2m C 2
ways.
Let E be the event for which two subsets A and B are
selected in such a way that
A » B = j, A » B = X
1 m m m m
Thus, nE ( ) = ( C0 + C1+ C2
+ ... + Cm)
2
m
2 m-1
= = 2
2
Hence, the required probability,
Thus,
m-1 m - 1
¥
=
m
2 m m
C2
m
2 2 2
2 (2 -1)
2 1
= =
m m m
2 (2 -1) (2 -1)
1 1
=
m
(2 -1)
127
fi (2 m – 1) = 127
fi 2 m = 127 + 1 = 128 = 2 7
fi m = 7
Hence, the number of elements in set X is 7.
3 1
11. Here, P (white ball) = =
9 3
and
6 2
P (non-white ball) = =
9 3
Let X be the number of white balls drawn.
Clearly, n = 4
Thus, the required probability,
P(X = 1) = 4 C 1
¥ p 1 ¥ q 3
3 4
Ê1ˆ Ê2ˆ Ê2ˆ
= 4¥ Á ¥ = 2¥
Ë
˜
3¯ Á
Ë
˜ Á ˜
3¯ Ë3¯
Clearly, a = 2 and b = 3
Hence, the value of a + b + 1 = 2 + 3 + 1 = 6.
1
12. Here, P (leap year) = and
4
3
P (non-leap year) =
4
Let A and B be the events that a non-leap year and a
leap year, respectively have 53 Sundays.
1 2 3 1
Thus, PA ( ) = ¥ and PB ( ) = ¥
4 7 4 7
Hence, the required probability,
P(A or B) = P(A » B)
= P(A) + P(B)
1 2 3 1
= ¥ + ¥
4 7 4 7
5 5 1
= = ¥
28 4 7