1.Algebra Booster

8.36 Algebra Booster
Hence, the required probability = P(A or B)
= P(A » B)
= P(A) + P(B)
1 2 3 1
= ¥ + ¥
4 7 4 7
5
=
28
36. We have,
( x -10)( x - 20)
< 0
( x - 30)
– + – +
10 20 30
Thus, x Œ (0, 10) » (20, 30)
Hence, the required probability
37. We have,
x 2 – 6x 2 + 11x – 6 = 0
fi (x – 1) (x – 2) (x – 3) = 0
fi x = 1, 2, 3
3
Hence, the required probability =
10
38. (i) Hence, the required probability
=
50 33
C1¥
C1
100
C2
50 ¥ 33 ¥ 2
=
100 ¥ 99
1
=
3
(ii) Hence, the required probability
=
50 33 16
C2 + C2 - C2
100
C 2
18 9
= =
100 50
C2
(iii) Hence, the required probability =
100
C2
4
=
165
16
C2
(iv) Hence, the required probability = 1-
100
C2
4
= 1-
165
161
=
165
(v) Hence, the required probability =
16
25 33
C1¥
C1
100
C2
25 ¥ 33 ¥ 2
=
100 ¥ 99
1
=
6
7¥ 8+ 7¥
8
39. Hence, the required probability =
64
C2
2 ¥ 112
=
64 ¥ 63
2¥
7 7 1
= = =
4 ¥ 63 126 18
40. Hence, the required probability = P(S ≥ 5)
= 1 – P(S < 5)
= 1 – {(P(S = 3) + P(S = 4)}
Ï 1 3 ¸
=1- Ì + ˝
Ó216 216˛
Ê 4 ˆ Ê 1 ˆ 53
= Á1- = 1- =
Ë
˜
216¯ Á
Ë
˜
54¯
54
41. Hence, the required probability = P(S ≥ 15)
= P(S = 15) + P(S = 16) + P(S = 17) + P(S = 18)
10 + 6 + 3 + 1 20 5
= = =
216 216 54
42. Let S be the sample space.
Then n(S) = 2 ¥ 2 ¥ 2 ¥ 2 = 16
Let E be the event that the determinant is non-negative
Then E¢ be the event that the determinant is negative
Ï11 01 01 ¸
= Ì ,, ,, , ˝
ÔÓ10 11 10 Ô˛
nE ( ¢ ) 3
Now, PE ( ¢ ) = =
nS ( ) 16
Hence, the required probability, P(E)
3 13
= 1 - PE ( ¢ ) = 1- = 16 16
43. When 4 Æ (1, 1, 2)
3!
PS ( = 4) = = 3
2!
16 Æ (4,6,6),(5,5,6)
3! 3!
PS ( = 16) = + = 6
2! 2!
9 Æ (1, 2, 6), (1, 3, 5), (2, 3, 4), (1, 4, 4)
(2,2,5),(3,3,3)
3!
PS= ( 9) = 3 ¥ 3! + 2 ¥ + 1 = 25
2!
Hence, the required probability
= P(S = 4) + P(S = 9) + P(S = 16)
3 + 6 + 25 34 17
= = =
216 216 108
44. Let E be the event, where the product of two numbers
is equal to the third number.
= {(2, 5, 10), (2, 4, 8), (2, 3, 6)}