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Question

The probability distribution of X is

X0123
P(x)0.3k2k3k
The value of k is

A
0.116
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B
0.7
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C
1
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D
0.3
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Solution

The correct option is A 0.116
x=0P(x)=1

3x=0P(x)=1
P(0)+P(1)+P(2)+P(3)=1
0.3+k+2k+3k=1
6k=10.3
6k=0.7
k=0.76; k=0.116

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