1.Algebra Booster

Probability 8.77
g
Similarly, z =
g + p
a g + p p
1+
PE ( 1)
a + p g g
= = =
PE ( 2)
g a + p p
1+
g + p p a
ab 2bg
Also, given = p =
a -2b b -3g
5ag
fi b =
a + 4g
Ê Ê 5ag ˆˆ
a ◊5ag
Thus,
Áa
- 2 p =
Ë
Á
Ë a + 4g ˜˜
¯ ¯ a + 4g
fi ap – 6pg = 5ag
Ê p ˆ Ê p ˆ
fi Á + 1 = 6Á
+ 1˜
Ëg
˜
¯ Ëa
¯
Ê p ˆ
Á + 1
Ëg
˜
¯
fi
= 6
Ê p ˆ
Á + 1
Ë
˜
a ¯
96. We have
1 + 2 + 3 + … + n – 2 £ 1224 £ 3 + 4 + … + n
( 1)( 2) ( 2)( 3)
fi n - n -
1224
n - n +
£ £
2 2
fi n 2 – 3n – 2446 £ 0 and n 2 + n – 2454 ≥ 0
fi 49 < n < 51
fi n = 50
nn ( + 1)
fi - (2k
+ 1) = 1224
2
fi k = 25
Thus, k – 20 = 25 – 20 = 5
97. (i) P(WWW) + P(RRR) + P(BBB)
1 2 3 3 3 4 2 4 5
= ¥ ¥ + ¥ ¥ + ¥ ¥
6 9 12 6 9 12 6 9 12
6+ 36+
40
=
6¥ 9¥
12
82
=
648
(ii) P(Ball drawn from box 2/one is W one is R)
PA ( « B)
=
PB ( )
1 2.3
¥
9
3 C2
=
1 Ê 1.3 2.3 3.4 ˆ
6 9 12
3
Á + + ˜
Ë C2 C2 C2¯
1
6 55
= =
1 1 2 181
+ +
5 6 12
98. Either a girl will start the sequence or will be at second
position and will not acquire the last position as well.
Required probability =
3 3
C1+
C1
5
C2
99. (i) Case I: One odd, 2 even
Total number of ways = 2.2.3 + 1.3.3 + 1.2.4 = 29
=
1
2
Case II: All 3 odd
Number of ways = 2.3.4 = 24
Favourable ways = 29 + 24 = 53
53 53
Required probability = =
3 ¥ 5 ¥ 7 105
(ii) Here, 2x 2
= x 1
+ x 3
fi x 1
+ x 3
= Even
Hence number of favourable ways
= 2 C 1
¥ 4 C 2
+ 1 C 1
¥ 3 C 1
= 8 + 3
= 11