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Let X be the random variable with probability distribution given by the following table.

X0123
P(X = x)18\frac{1}{8}k38\frac{3}{8}18\frac{1}{8}

The value of P(X≤1)P(X \leq 1) is:

Solution

✅ Correct Option: 4

Sum of probabilities equals 1: 18+k+38+18=1\frac{1}{8} + k + \frac{3}{8} + \frac{1}{8} = 1, so k=38k = \frac{3}{8}. Then P(X≤1)=P(X=0)+P(X=1)=18+38=12P(X \leq 1) = P(X=0) + P(X=1) = \frac{1}{8} + \frac{3}{8} = \frac{1}{2}.

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