1.Algebra Booster

8.70 Algebra Booster
46. We have,
P(A « (B « C) = P(A)P(B « C)
= P(A)P(B)P(C)
Thus, the statement S 2
is true.
Also,
P(A « (B » C))
= P((A « B) » (A « C))
= P(A « B) + P(A « C) – P((A « B) « (A « C))
= P(A « B) + P(A « C) – P((A « B « C)
= P(A)P(B) + P(A)P(C) – P(A)P(B)P(C)
= P(A)(P(B) + P(C) – P(B)P(C))
= P(A)(P(B » C))
= P(A)P(B » C)
Thus, the statement S 1
is true.
47. Let E 1
, E 2
denote the events that the coin shows a head,
tail and A be the event that the noted number is either 7
or 8.
1 1
We have PE ( 1) = and PE ( 2)
=
2 2
Now, 7 Æ {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}
and 8 Æ {(2, 6), (6, 2), (3, 5), (5, 3), (4, 4)}
11 2
Thus, PAE ( / 1) = , PAE ( / 2)
=
36 11
Hence, the required probability,
P(A) = P(E 1
)P(A/E 1
) + P(E 2
)P(A/E 2
)
Ê1ˆÊ11ˆ Ê1ˆÊ 2 ˆ 193
= Á + = .
Ë
˜Á
2¯Ë ˜
36¯ Á
Ë
˜Á ˜
2¯Ë11¯
792
48. Given PA ( ) = 0.3
PA ( ) = 1 - PA ( ) = 1- 0.3 = 0.7
Also, PA ( « B) = 0.5
Now, P{B « (A » B¢)} = P((B « A) » (B « B))
= P((B « A) » j)
= P(B « A)
= P(A) – P(A « B)
= 0.7 – 0.5
= 0.2
PA ( » B)
= PA ( ) + PB ( )- PA ( « B)
= PA ( ) + PB ( )- PA ( « B)
= 0.7 + 0.6 – 0.5
= 0.8
We have,
Ê B ˆ
P(B/(A » B c )) = PÁ
Ë
˜ A » B ¯
PB ( «( A»
B)
=
PA ( » B)
0.2
=
0.8
1
=
4
49. Required probability
= P(2nd win at the third test)
= P(Exactly one win in first two matches
¥ P(winning the third test)
= P(WL or LW) ¥ P(W)
= (P(WL) + P(LW)) ¥ P(W)
= (P(W)P(L) + P(L)P(W)) ¥ P(W)
Ê1 1 1 1ˆ
1
= Á ◊ + ◊ ¥
Ë
˜
2 2 2 2¯
2
Ê1 1ˆ
1
= Á + ¥
Ë
˜
4 4¯
2
1 1
= ¥
2 2
1
=
4
50. Given P(A » B) = P(A) + P(B) – P(A)P(B)
fi P(A) + P(B) – P(A « B)
= P(A) + P(B) – P(A)P(B)
fi P(A « B) = P(A)P(B)
A and B are independent events.
Now,
P(A » B)¢ = P(A¢ «B¢)
= P(A¢) « P(B¢)
A
Also,
Ê ˆ PA ( « B)
PÁ
Ë
˜ = B ¯ PB ( )
PA ( ) ◊PB
( )
=
PB ( )
= PA ( )
51. Required probability,
2 2
6
C = (6◊5◊4)/6
3
2
=
20
1
=
10
52. P(Exactly one of A or B occurs),
PA ( « Bor A«
B)
= PA ( « B) + PA ( « B)
= P(B) – P(A « B) + P(A) – P(A « B)
= P(A) + P(B) – 2P(A « B) = p …(i)
Similarly, P(Exactly one of B or C occurs),
P(B) + P(C) – 2P(B « C) = p
…(ii)
and P(Exactly one of C or A occurs),
P(C) + P(A) – 2P(A « C) .= p …(iii)
Adding all these results, we get,
2[(A) + P(B) + P(C) – P(A « B)
–P(B « C) – P(C « A)] = 3p