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A random variable X has the following probability distribution:

X=xiX=x_i1234
P(X=xi)P(X=x_i)k2k3k4k

The value of P(X≥3)P(X\ge 3) is

Solution

✅ Correct Option: 2

The probabilities must sum to 11: k+2k+3k+4k=10k=1k+2k+3k+4k=10k=1, so k=110k=\frac{1}{10}.

P(X≥3)=3k+4k=7k=710P(X\ge3)=3k+4k=7k=\frac{7}{10}.

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