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Byju's Answer
Standard XII
Mathematics
Probability Distribution
X 1 2 3 4 5 6...
Question
X
1
2
3
4
5
6
P(X=x)
k
2k
3k
4k
5k
6k
The probability distribution of discrete r.v.
X
is
Then
P
(
X
≤
4
)
=
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Solution
Sum of the probabilities is equal to 1
∴
k
+
2
k
+
3
k
+
4
k
+
5
k
+
6
k
=
1
21
k
=
1
k
=
1
21
P
(
X
≤
4
)
=
P
(
1
)
+
P
(
2
)
+
P
(
3
)
+
P
(
4
)
=
k
+
2
k
+
3
k
+
4
k
=
1
21
+
2
21
+
3
21
+
4
21
=
1
+
2
+
3
+
4
21
=
10
21
∴
P
(
X
≤
4
)
=
0.476
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0
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