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A random variable X has the following probability distribution:

X0123456
P(X)K3K7K5K4KK2K

The value of K is

Solution

✅ Correct Option: 4

The probabilities must sum to 1: K+3K+7K+5K+4K+K+2K=23K=1K+3K+7K+5K+4K+K+2K=23K=1, so K=123K=\frac{1}{23}.

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