1-Suppose the time it takes a nine-year old to eat a donutis between 0.5 and 4 minutes, inclusive. Let X = the time,in minutes, it takes a nine-year old child to eat a donut.Then X^ U (0.5, 4).a. The probability that a randomly selected nine-year oldchild eats a donut in at least two minutes is ?b. find the probability that a different nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes.​

Answer:0.8571,0.6667Step-by-step explanation:Given that the time it takes a nine-year old to eat a donut is between 0.5 and 4 minutes,inclusive. Let X = the time, in minutes, it takes a nine-year old child to eat a donut. Then X:   U (0.5, 4).The probability distribution function of X would be[tex]P(X) = \frac{1}{4-0.5} =0.2857[/tex]Cumulative function CDF would be[tex]F(x) = \frac{x-0.5}{3.5}[/tex]a) The probability that a randomly selected nine-year old child eats a donut in at least two minutes is=[tex]P(X\geq 2) = 1-F(2)\\= 1-\frac{2-0.5}{3.5} \\=0.8571[/tex]b) the probability that a different nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes.=[tex]P(X>2/x>1.5) = \frac{P(X>2)}{P(X>1,5)} \\= \frac{1-\frac{2-0.5}{3.5} }{1-\frac{1.5-0.5}{3.5} } \\=\frac{0.571429}{0.857143} \\=0.6667[/tex]