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The random variable X has the following probability distribution:

X12345
P(X)110\frac{1}{10}110\frac{1}{10}13\frac{1}{3}k13\frac{1}{3}

The correct value of k is:

Solution

✅ Correct Option: 1

Probabilities must sum to 1: 110+110+13+k+13=1\frac{1}{10}+\frac{1}{10}+\frac{1}{3}+k+\frac{1}{3} = 1. So k=1−15−23=215k = 1 - \frac{1}{5} - \frac{2}{3} = \frac{2}{15}.

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