Which of the following values can serve as the values of a probability distribution?
A. P(1) = 0.42, P(2) = 0.31, P(3) = 0.37
B. P(1) = 9/14, P(2) = 4/14, P(3) = 1/14
C. P(1) = 0.08, P(2) = 0.12, P(3) = 1.03
D. P(1) = 10/33, P(2) = 12/33, P(3) = 10/33
Answers
Answer:
SOLUTION
TO CHOOSE THE CORRECT OPTION
Which of the following values can serve as the values of a probability distribution
A. P(1) = 0.42, P(2) = 0.31, P(3)= 0.37
B. P(1) = 9/14, P(2) = 4/14, P(3)= 1/14
C. P(1) = 0.08, P(2) = 0.12, P(3)= 1.03
D. P(1) = 10/33, P(2) = 12/33, P(3)= 10/33
CONCEPT TO BE IMPLEMENTED
The given values can serve as the values of a probability distribution if
\sf{ \sum \: P(x) = 1}∑P(x)=1
EVALUATION
CHECKING FOR OPTION A
Here the given values are
P(1) = 0.42, P(2) = 0.31, P(3)= 0.37
Now
P(1) + P(2) + P(3)
= 0.42 + 0.31 + 0.37
= 1.1 ≠ 1
Since P(1) + P(2) + P(3) ≠ 1
So the given values can not serve as the values of a probability distribution
So this option is not correct
CHECKING FOR OPTION B
Here the given values are
P(1) = 9/14, P(2) = 4/14, P(3)= 1/14
Now
P(1) + P(2) + P(3)
= 9/14 + 4/14 + 1/14
= 14/14
= 1
Since P(1) + P(2) + P(3) = 1
So the given values can serve as the values of a probability distribution
So this option is correct
CHECKING FOR OPTION C
Here the given values are
Now
P(1) + P(2) + P(3)
= 0.08 + 0.12 + 1.03
= 1.23 ≠ 1
Since P(1) + P(2) + P(3) ≠ 1
So the given values can not serve as the values of a probability distribution
So this option is not correct
CHECKING FOR OPTION D
Here the given values are
Now
P(1) + P(2) + P(3)
= 10/33 + 12/33 + 10/33
= 32/33 ≠ 1
Since P(1) + P(2) + P(3) ≠ 1
So the given values can not serve as the values of a probability distribution
So this option is not correct
FINAL ANSWER
Hence the correct option is
B. P(1) = 9/14, P(2) = 4/14, P(3)= 1/14
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