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JEE Main Math Class 12 Probability Part 2 Questions
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© examsnet.com
Question : 95
Total: 95
Let there be three independent events
E
1
,
E
2
and
E
3
. The probability that only
E
1
occurs is
α
, only
E
2
occurs is
β
and only
E
3
occurs is
γ
. Let
p
denote the probability of none of events occur that satisfies the equations
(
α
−
2
β
)
p
=
α
β
and
(
β
−
3
γ
)
p
=
2
β
γ
.
All the given probabilities are assumed to lie in the interval
(
0
,
1
)
. Then, probability of occurrence of
E
1
probability of occurrence of
E
3
is equal to ..........
[2021, 17 March Shift-1]
Your Answer:
Validate
Solution:
Let
P
(
E
1
)
=
x
,
P
(
E
2
)
=
y
,
P
(
E
3
)
=
z
Then,
(
1
−
x
)
(
1
−
y
)
(
1
−
z
)
=
P
. . . (A)
Probability of only
E
1
=
P
(
E
1
∩
E
2
∩
E
3
)
α
=
x
(
1
−
y
)
(
1
−
z
)
. . . (B)
Similarly,
β
=
(
1
−
x
)
y
(
1
−
z
)
. . . (C)
γ
=
(
1
−
x
)
(
1
−
y
)
z
. . . (D)
On dividing Eq. (A)
÷
(
B
)
,
P
α
=
1
−
x
x
⇒
x
P
=
α
−
α
x
⇒
x
(
P
+
α
)
=
α
⇒
x
=
α
α
+
P
Similarly,
y
=
β
β
+
P
z
=
γ
γ
+
P
P
(
E
1
)
P
(
E
3
)
=
x
z
=
(
α
α
+
P
)
×
(
γ
+
P
γ
)
x
z
=
(
α
γ
)
(
γ
+
P
α
+
P
)
=
(
1
+
P
∕
γ
1
+
P
∕
α
)
Also given,
P
=
α
β
α
−
2
β
=
2
β
γ
β
−
3
γ
α
P
−
2
β
P
=
α
β
⇒
α
P
α
+
2
P
=
β
β
P
−
3
P
γ
=
2
β
γ
β
(
P
−
2
Y
)
=
3
P
Y
⇒
β
=
3
P
γ
P
−
2
γ
From Eqs. (i)and(ii),
⇒
α
P
α
+
2
P
=
3
P
γ
P
−
2
γ
⇒
1
1
+
2
(
P
α
)
=
3
(
P
γ
)
−
2
⇒
(
P
∕
γ
)
−
2
=
3
+
6
(
P
∕
α
)
P
γ
=
5
+
6
(
P
α
)
x
z
=
1
+
P
∕
γ
1
+
P
∕
α
=
1
+
5
+
6
(
P
∕
α
)
1
+
P
∕
α
=
6
(
1
+
P
∕
α
1
+
P
∕
α
)
=
6
© examsnet.com
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