Question

If a random variable X has the following probability distribution values:

X	0	1	2	3	4	5	6	7
P(X)	1/12	1/12	1/12	1/12	1/12	1/12	1/12	1/12

Then P(X ≥ 6) has the value:

• A. \( \frac{16}{100} \)
• B. \( \frac{81}{100} \)
• C. \( \frac{1}{100} \)
• D. \( \frac{91}{100} \)

Hint:

When working with discrete probability distributions, remember to sum the individual probabilities for the range of values you're interested in.