Question

(a) A parachutist lands at a point on the line between the points A and B, and the target is an operation at A. The operation fails, if parachutist's distance to A is more than five times as much as her distance to B. What is the probability of success?

(b) Let X be a Gamma(2, 1) random variable. Find P(X > 1).

Answer 1

a)
`(B-(A+5B)/(6))/(B-A)= (6B -A-5B)/(6(B-A))=(B-A)/(6(B-A))=(1)/(6)`

b)
`P(X>1) = 1-P(X\leq 1)=1-\int_(0)^1 (1^(2) x^(2-1) e^(- x))/(\gamma(2))=0.736`

Explanation:

Part a

We assume that the parachutist lands at random point in the interval (x=A,y=B) we have a continuous random variable X. And the distribution of X would be uniform
`Y\sim Unif(A,B)`. And the density function would be given by:

`f(x) =(1)/(B-A) , A <x<B`

And 0 for other case.

The operation fails, if parachutist's distance to A is more than five times as much as her distance to B.

So the point P in the interval (A,B) at which the distance to A is exactly 5 times the distance to B is given by:

`P= A + (5)/(6) (B-A)= (6A +5B -5A)/(6)=(A+5B)/(6)`

And we can find the probability desired like this:

`P(d(P,A) \geq 5 d(P,B))= P((A+5B)/(6) < X< B)`

And from the cumulative distribution function of X ficen by
`F(X)(X-A)/(B-A)` we got:

`(B-(A+5B)/(6))/(B-A)= (6B -A-5B)/(6(B-A))=(B-A)/(6(B-A))=(1)/(6)`

Part b

For this case we assume that
`X\sim Gamma (2,1)`

On this case we assume that
`\alpha=2, \beta= 1`

The density function for the Gamma distribution is given by:

`P(X)= (\beta^(\alpha) x^(\alpha-1) e^(-\beta x))/(\gamma(\alpha))`

And on this case we can find the probability using the complement rule like this:

`P(X>1) = 1-P(X\leq 1)=0.736`

We can solve this problem with the following excel code:

"=1-GAMMA.DIST(1;2;1;TRUE)"

And if we do it by hand we need to do this:

`P(X>1) = 1-P(X\leq 1)=1-\int_(0)^1 (1^(2) x^(2-1) e^(- x))/(\gamma(2))=0.736`