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Question

Mathematics Question on Probability Distribution

A random variable XX has the following probability distribution:

X01234567
P(X)0m2m2m3m2m²7m² + m

The value of m is:

A

10

B

110\frac{1}{10}

C

-1 and 110\frac{1}{10}

D

120\frac{1}{20}

Answer

110\frac{1}{10}

Explanation

Solution

The sum of probabilities in a probability distribution must equal 1:

P(X)=1\sum P(X) = 1.

Substitute the given probabilities:

0m + 2m + 2m + 3m + 3m + 2m2+2m2+(7m2+m)2m^2 + 2m^2 + (7m^2 + m) = 1.

Simplify:

12m+11m2=112m + 11m^2 = 1.

Factorize:

m(12+11m)=1m(12 + 11m) = 1.

m=110m = \frac{1}{10} (since m>0m > 0).

Thus, m=110m = \frac{1}{10}.