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

Answer:

a) $\frac{B-\frac{A+5B}{6}}{B-A}= \frac{6B -A-5B}{6(B-A)}=\frac{B-A}{6(B-A)}=\frac{1}{6}$

b) $P(X>1) = 1-P(X\leq 1)=1-\int_{0}^1 \frac{1^{2} x^{2-1} e^{- x}}{\gamma(2)}=0.736$

Step-by-step 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:

$f(x) =\frac{1}{B-A} , A$

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:

$P= A + \frac{5}{6} (B-A)= \frac{6A +5B -5A}{6}=\frac{A+5B}{6}$

And we can find the probability desired like this:

$P(d(P,A) \geq 5 d(P,B))= P(\frac{A+5B}{6} < X< B)$

And from the cumulative distribution function of X ficen by $F(X)\frac{X-A}{B-A}$ we got:

$\frac{B-\frac{A+5B}{6}}{B-A}= \frac{6B -A-5B}{6(B-A)}=\frac{B-A}{6(B-A)}=\frac{1}{6}$

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:

$P(X)= \frac{\beta^{\alpha} x^{\alpha-1} e^{-\beta x}}{\gamma(\alpha)}$

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:

$P(X>1) = 1-P(X\leq 1)=1-\int_{0}^1 \frac{1^{2} x^{2-1} e^{- x}}{\gamma(2)}=0.736$