Question 1183249
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Part (a)


I'm assuming we're dealing with a uniform distribution.


*[Tex \Large P(t_0 \le T \le t_1) = \frac{t_1 - t_0}{t - 0}]


*[Tex \Large P(t_0 \le T \le t_1) = \frac{t_1 - t_0}{t}]


There's not much more I can say since we don't know the values of {{{t[0]}}}, {{{t[1]}}}, and {{{t}}}


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Part (b)


Let *[Tex \Large A = \frac{t_1 - t_0}{t}] which represents the probability T is between {{{t[0]}}} and {{{t[1]}}}. We found this back in part (a)


Then we can say that
*[Tex \Large P(T \ge t_0) = 1 - P(T \le t_0)]


*[Tex \Large P(T \ge t_0) = 1 - \frac{t_0}{t}]


*[Tex \Large P(T \ge t_0) = \frac{t}{t} - \frac{t_0}{t}]


*[Tex \Large P(T \ge t_0) = \frac{t-t_0}{t}]


We'll assign this to the variable B. 


So,
*[Tex \Large P(t_0 \le T \le t_1 \ \text{ given } \ T \ge t_0) = \frac{A}{B}]


*[Tex \Large P(t_0 \le T \le t_1 \ \text{ given } \ T \ge t_0) = A \div B]


*[Tex \Large P(t_0 \le T \le t_1 \ \text{ given } \ T \ge t_0) = \frac{t_1 - t_0}{t} \div \frac{t-t_0}{t}]


*[Tex \Large P(t_0 \le T \le t_1 \ \text{ given } \ T \ge t_0) = \frac{t_1 - t_0}{t} \times \frac{t}{t-t_0}]


*[Tex \Large P(t_0 \le T \le t_1 \ \text{ given } \ T \ge t_0) = \frac{(t_1 - t_0)*t}{t*(t-t_0)}]


*[Tex \Large P(t_0 \le T \le t_1 \ \text{ given } \ T \ge t_0) = \frac{t_1 - t_0}{t-t_0}]


I'm using the idea of conditional probability
P(A given B) = P(A and B)/P(B)
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