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JEE Mains 26-July-2022 Shift 2 Solved Paper
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© examsnet.com
Question : 18
Total: 90
Let
X
be a binomially distributed random variable with mean 4 and variance
4
3
. Then,
54
P
(
X
≤
2
)
is equal to
[26-Jul-2022-Shift-2]
73
27
146
27
146
81
126
81
Validate
Solution:
👈: Video Solution
Mean
=
4
=
µ
=
n
p
Variance
=
σ
2
=
n
p
(
1
−
P
)
=
4
3
4
(
1
−
P
)
=
4
3
P
=
2
3
n
×
2
3
=
4
n
=
6
P
(
X
=
k
)
=
n
C
k
P
k
(
1
−
P
)
n
−
k
P
(
X
≤
2
)
=
P
(
X
=
0
)
+
P
(
X
=
1
)
+
P
(
X
=
2
)
=
6
C
0
P
0
(
1
−
P
)
6
+
6
C
1
P
1
(
1
−
P
)
5
+
6
C
2
P
2
(
1
−
P
)
4
=
6
C
0
(
1
3
)
6
+
6
C
1
(
2
3
)
(
1
3
)
5
+
6
C
2
(
2
3
)
2
(
1
3
)
4
=
(
1
3
)
6
[
1
+
12
+
60
]
=
73
3
6
54
P
(
X
≤
2
)
=
73
3
6
×
54
=
146
27
© examsnet.com
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