1.Algebra Booster

8.68 Algebra Booster
1 Ê13ˆÊ1ˆ Ê13ˆ Ê1ˆ
= + Á + + ...
9 Ë
˜Á
18¯Ë ˜
9¯ Á
Ë
˜
18¯ Á
Ë
˜
9¯
2
1Ê
Ê13ˆ Ê13ˆ
ˆ
= Á1 + Á ˜ + Á ˜ + ... ˜
9Ë
Ë18¯ Ë18¯
¯
Ê ˆ
1Á
1 ˜
= Á ˜
9 Ê13ˆ
1 Á - Á ˜
Ë Ë18¯˜
¯
1Ê
18 ˆ
= Á
9Ë
˜
18-13¯
2
=
5
35. Let A = {a 1
, a 2
, …, a n
}
For each a i
Œ A)1 £ i £ n)
We have the following four choices:
(i) a i
ΠP, a i
ΠQ (ii) a i
œ P, a i
ΠQ
(iii) a i
ΠP, a i
œ Q (iv) a i
œ P, a i
œ Q
Thus, the total number of ways of choosing P and Q is
4 n .
Out of these four choices (i) is not favourable for
P « Q = j
n
Ê3ˆ
Hence, the required probability is = Á
Ë
˜
4¯ .
36. Let P(B) = x
Given P(A) = 0.3 and P(A » B) = 0.8
Now, P(A » B) = 0.8
fi P(A) + P(B) – P(A « B) = 0.8
fi P(A) + P(B) – P(A) ◊ P(B) = 0.8
fi 0.3 + x – (0.3)x = 0.8
fi (0.7)x = 0.8 – 0.3
fi (0.7)x = 0.5
0.5
fi x =
0.7
5
fi x =
7
5
fi PB ( ) =
7
Ê Aˆ P( A«
B)
37. PÁ
=
Ë
˜ B ¯ P ( B )
PA ( ) + PB ( )- PA ( » B)
PB ( )
PA ( ) + PB ( )-1
≥
PB ( )
Thus, (a) is correct.
Also, P(A « B¢) = P(A) – P(A « B) holds.
So (b) is incorrect.
Also, P(A » B) = 1 – P(A¢ «B¢)
= 1 – P(A¢)P(B¢), since A and B are independent events.
Hence (c) is correct.
Again, P(A » B) = P(A) + P(B)
2
π P(A¢) P(B¢)
So (d) is incorrect.
38. Let E 1
, E 2
and E 3
be the events that the answer is
guessed, copied and knows the answer and E be the
event that the examinee answers correctly.
1 1
Given PE ( 1) = , PE ( 2)
=
3 6
Here, E 1
, E 2
and E 3
are exhaustive events.
Thus P(E 1
) + P(E 2
) + P(E 3
) = 1
fi P(E 3
) = 1 – P(E 1
) – P(E 2
)
1 1 6-2-1 3 1
= 1- - = = =
3 6 6 6 2
Now,
P(E/E 1
) = Probability of getting correct answer
by guessing = 1/4
P(E/E 2
) = Probability of getting correct answer
by copying = 1/8.
P(E/E 3
) = Probability of getting correct answer
by knowing = 1.
Hence, the required probability,
= PE ( 3 / E)
PE ( 3) ◊PEE
( / 3)
=
PE ( 1) ◊ PEE ( / 1) + PE ( 2) ◊ PEE ( / 2) + PE ( 3) ◊PEE
( / 3)
1
¥ 1
2
24
= =
1 1 1 1 1 29
¥ + ¥ + ¥ 1
3 4 6 8 2
39. Given np =2 and npq = 1
1 1 1
Thus, q = , p = 1- = and n = 4
2 2 2
Now,
P(X > 1) = 1 – P(X £ 1)
= 1 – {P(X = 0) + P(X = 1)}
= 1 – { 4 C 0
◊ (p) 0 q 4 + 4 C 1
◊ (p) 1 q 3 }
= 1 – {q 4 + 4 ◊ pq 3 }
Ï
4 4
ÔÊ1ˆ Ê1ˆ
¸Ô
=1- ÌÁ ˜ + 4◊Á ˜ ˝
ÔË2¯ Ë2¯
Ó
Ô˛
Ï 1 4 ¸
= 1 - Ì + ˝
Ó16 16˛
5 11
= 1 - = 16 16
40. Required probability
P(X ≥ 7) = P(X = 7 or X = 8)
= P(X = 7) + P(X = 8)
=P(India wins 3 matches and draws one)
+ P(India wins all 4 matches)
= 4 C 3
(0.5) 3 ◊ (0.05) + 4 C 4
(0.5) 4
= (0.5) 3 (0.2 + 0.5)
= (0.125) ◊ (0.7) = 0.0875