1.Algebra Booster

8.54 Algebra Booster
We are interested in the event x + y < a/3.
Here, first we make a square of side a/2 and its subregion
triangle with base a/3 and height is also a/3.
Y
y =1
y = 1/2
O
X
x = 1/2 x =1
1 ÊaˆÊaˆ
Á
2 Ë
˜Á
2
Thus, the required probability
3¯Ë ˜
3¯
= =
ÊaˆÊaˆ
9
Á
Ë
˜Á ˜
2¯Ë2¯
180. Let the two numbers be x and y such that
0 £ x, y £ 1.
1/3
½
1
A B C
Now, the probability of
1 1 1 1 1 1
¥ ¥ + ¥ ¥
Ê 1ˆ
1
PÁ
x- y >
2 2 2 2 2 2
Ë
˜ = =
2¯
1¥
1 4
181. Let us consider two numbers be x and y where
0 £ x, y £ 1, so that the geometric configuration is a
square.
Y
M
L 1/3 X
O 1/3
Let M = max. {x, y} and m = min. {x, y}
O
Y
B
X = 1/3
A
X = 3/4 X =1
y =1
y = 3/4
y = 1/3
X
Hence the required probability,
ÊM ≥3/4 ˆ P( M ≥ 3/4, m£
1/3)
PÁ
=
Ë
˜ m £ 1/3 ¯ P ( m £ 1/3)
1 1 1
2◊
◊
3 4 6 3
= = =
1 1 1 5 10
+ -
3 3 9 9
182. Let AB be the straight line divided at P and Q such that
AB = a.
Let AP = x and BQ = y
X
A P Q
a
Now in favourable cases,
we must have,
Y
x < a/2, y < a/2
and x + y > a/2
L
M
Hence, the required probability
1 a a
O N X
◊ ◊
2 2 2 1
= =
1 4
◊aa
◊
2
183 Let two numbers be x and y such that
x + y = 24.
For extremum, it is clear that, the greatest value of xy is
(12) 2
Hence the required probability,
P(xy ≥ 108) = P(x(24 – x) ≥ 108)
= P((x – 18)(x – 6) £ 0)
= P(6 £ x £ 18)
= 12 =
1
24 2
184. Let the probability of a shot
falling in the region C of
radius 1/3 is
p ◊(1/3) 1
p1 = =
2
p ◊(1)
9
the probability of a shot
falling in the region B of
radius 1/2 is
2
2 2
p◊(1/2) -p(1/3) 5
p2 = =
2
p ◊(1)
36
and the probability of a shot falling in the region A of
radius 1 is
p
2 2
p◊(1) -p(1/2) 3
= =
p ◊(1)
4
3 2
Y
½
C
B
B
1/3 O
1
A