Let X1, X2 and X3 be three independent random variables that are uniformly distributed between 50 and 100. A) Find the probability that the minimum of the three is between 75 and 90.
B) Find the probability that the second smallest of the three is between 75 and 90.

a) the probability that the minimum of the three is between 75 and 90 is 0.00072

b) the probability that the second smallest of the three is between 75 and 90 is 0.396

Step-by-step explanation:

Given that;

fx(x) = { 1/5 ; 50 < x < 100

              0, otherwise}

Fx(x) = { x-50 / 50 ; 50 < x < 100

                          1 ;   x > 100

a)

n = 3

F(1) (x) = nf(x) ( 1-F(x)^n-1

= 3 × 1/50 ( 1 - ((x-50)/50)²

= 3/50 (( 100 - x)/50)²

=3/50³ ( 100 - x)²

Therefore P ( 75 < (x) < 90) =  ⁹⁰∫₇₅ 3/50³ ( 100 - x)² dx

= 3/50³ [ -2 (100 - x ]₇₅⁹⁰

= (3 ( -20 + 50)) / 50₃

= 9 / 12500 = 0.00072

b)

f(k) (x) = nf(x)  ( ⁿ⁻¹_k₋ ₁) ( F(x) )^k-1 ; ( 1 - F(x) )^n-k

Now for n = 3, k = 2

f(2) (x) = 3f(x) × 2 × (x-50 / 50) ( 1 - (x-50 / 50))

= 6 × 1/50 × ( x-50 / 50) ( 100-x / 50)

= 6/50³ ( 150x - x² - 5000 )

therefore

P( 75 < x2 < 90 ) = 6/50³ ⁹⁰∫₇₅ ( 150x - x² - 5000 ) dx

= 99 / 250 = 0.396

dude thats to long figure it

step-by-step explanation: