1.Algebra Booster

Probability 8.69
41. (i) Let D i
and N i
respectively denote the occurrence of
a defective and non-defective bulb at the ith draw
where i = 1, 2
50 1
We have, PD ( i ) = = , i=
1,2
100 2
50 1
PN ( i ) = = , i=
1,2
100 2
Here, D 1
N 2
, D 2
N 2
, D 1
D 2
and N 1
N 2
all are independents.
Given A = {the first bulb is defective}
= {D 1
D 2
, D 1
N 2
}
B = {the 2nd bulb is non-defective)
= {D 1
N 2
, N 1
N 2
}
and C = {both bulbs are defective or non
defective}
= {D 1
D 2
, N 1
N 2
}
(i) P(A) = P(D 1
D 2
) + P(N 1
N 2
)
= P(D 1
)P(D 2
) + P(N 1
)P(N 2
)
1 1 1 1 1
= ◊ + ◊ =
2 2 2 2 2
1
Similarly, PB ( ) = = PC ( )
2
Also, P(A « B) = P(D 1
N 2
) = P(D 1
)P(N 2
)
1 1 1
= ◊ =
2 2 4
1
Similarly, PB ( « C) = = PC ( « A)
4
Thus, P(A « B) = P(A)P(B)
P(B « B) = P(B)P(C)
and P(C « A) = P(C)P(A)
Hence, A, B and C are pairwise independents.
(ii) Now, (A « B « C) = j
Thus, P(A « B « C) π P(A)P(B)P(C)
Hence, A, B and C are not independent.
3 1
42. P(yellow at the first toss) = =
6 2
2 1
P(red at the second toss) = =
6 3
1
P(blue at the third toss) =
6
1 1 1
Hence, the required probability = ¥ ¥
2 3 6
1
=
36
43. Let E be the event whose face value not less than 2 and
not more than 5 in a single throw of a die, i.e.
E = {2, 3, 4, 5}
4 2
So, PE ( ) = =
6 3
Required probability,
P(EEEE) = P(E) ◊ P(E) ◊ P(E) ◊ P(E)
4
Ê2ˆ
16
= Á =
Ë
˜
3¯
81
44. Given
1 1
PE ( « F) = and PE ( « F)
=
12 2
1
Thus, PE ( ) ◊ PF ( ) =
12
…(i)
1
Also, PE ( « F)
=
2
fi
1
PE ( » F)
=
2
fi
1
1 - PE ( » F)
=
2
fi
1 1
PE ( » F) = 1- = 2 2
fi
1
PE ( ) + PF ( )- PE ( « F)
=
2
fi
1 1
PE ( ) + PF ( )- =
12 2
fi
1 1 7
PE ( ) + PF ( ) = + =
2 12 12
…(ii)
From Eqs (i) and (ii), we can say that, P(E) and P(F)
are the roots of
2 7 1
x - x+ = 0
12 12
fi 12x 2 – 7x + 1 = 0
fi (4x – 1)(3x – 1) = 0
fi
1 1
x = ,
4 3
Thus,
1 1 1 1
PE ( ) = , PF ( ) = or PE ( ) = , PF ( ) =
4 3 3 4
45. Let E be the event, where the product of two-digit selected
number is 18, i.e.
E = {(2, 9), (9, 2), (3, 6), (6, 3)}
4 1 24
Thus, PE ( ) = = and PE ( ) =
100 25 25
Required probability,
P(X ≥ 3) = P(X = 3) + P(X = 4)
= 4 C 3
p 3 q + 4 C 4
p 4 q 0
= 4p 3 q + p 4
= p 3 (4q + p)
= p 3 (1 + 3q)
3
Ê 1 ˆ Ê 24ˆ
= Á 1+ 3◊
Ë
˜
25¯ Á
Ë
˜
25¯
3
Ê 1 ˆ Ê 25+
72ˆ
= Á .
Ë ˜
25¯ Á
Ë ˜
25 ¯
Ê 97 ˆ
= Á .
Ë 25
4 ˜
¯