1. State the problem. We are given the probability distribution of the daily demand X for a movie magazine with values 0 through 9 and their probabilities as follows. $$P(0)=0.08,\;P(1)=0.14,\;P(2)=0.16,\;P(3)=0.14,\;P(4)=0.12,\;P(5)=0.10,\;P(6)=0.08,\;P(7)=0.07,\;P(8)=0.07,\;P(9)=0.04$$ 2. Formula and important rules. For any event A in a discrete distribution we sum the probabilities: $P(A)=\sum_{x\in A} P(x)$. Important rules: probabilities are nonnegative and the total sum is 1: $\sum_x P(x)=1$. 3. Part (a): Find $P(X\ge 3)$. Use A={3,4,5,6,7,8,9} and sum the probabilities. $$P(X\ge 3)=P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)$$ $$=0.14+0.12+0.10+0.08+0.07+0.07+0.04$$ $$=0.62$$ Therefore the probability that three or more copies will be demanded is $0.62$. 4. Part (b): Find $P(2\le X\le 6)$. Sum P(2) through P(6). $$P(2\le X\le 6)=P(2)+P(3)+P(4)+P(5)+P(6)$$ $$=0.16+0.14+0.12+0.10+0.08$$ $$=0.60$$ Therefore the probability that the demand is at least two but not more than six is $0.60$. 5. Check: the probabilities sum to 1. $$0.08+0.14+0.16+0.14+0.12+0.10+0.08+0.07+0.07+0.04=1.00$$ This confirms a valid distribution.