# Thread: Probability problem

1. ## Probability problem

Could anyone suggest the solution for this problem?

Consider Omega=[0,1] with probability defined on intervals by P([a,b])=b-a, 0<= a <= b <=1,

1. Define a random variable by: X(omega)=min{omega,1-omega}.

Find P({omega:X(omega)<= 4/5}), P({omega:X(omega)<=1/4}), P({omega:X(omega)<=0}), P({omega:X(omega)<=-1})

2. ## Re: Probability problem

The first probability is 1, as the maximum value for X is 0.5. The second probability will be 0.5. The third probability is 0. The last probability is 0.

The first probability is $P([0,1]) = 1-0 = 1$.
The second probability is $P\left( \left[ 0, \dfrac{1}{4} \right] \cup \left[ \dfrac{3}{4}, 1 \right] \right) = P\left( \left[ 0, \dfrac{1}{4} \right] \right) + P\left( \left[ \dfrac{3}{4}, 1 \right] \right) = \dfrac{1}{4} - 0 + 1 - \dfrac{3}{4} = \dfrac{1}{2}$
The third probability is $P(\{0,1\}) = P([0,0]) + P([1,1]) = 0-0+1-1 = 0$
The last probability is $P(\emptyset) = 0$.

3. ## Re: Probability problem Originally Posted by SlipEternal The first probability is 1, as the maximum value for X is 0.5. The second probability will be 0.5. The third probability is 0. The last probability is 0.
The first probability is $P([0,1]) = 1-0 = 1$.
The second probability is $P\left( \left[ 0, \dfrac{1}{4} \right] \cup \left[ \dfrac{3}{4}, 1 \right] \right) = P\left( \left[ 0, \dfrac{1}{4} \right] \right) + P\left( \left[ \dfrac{3}{4}, 1 \right] \right) = \dfrac{1}{4} - 0 + 1 - \dfrac{3}{4} = \dfrac{1}{2}$
The third probability is $P(\{0,1\}) = P([0,0]) + P([1,1]) = 0-0+1-1 = 0$
The last probability is $P(\emptyset) = 0$.
I am totally confused. What in the definition gives this: $P(\{0,1\}) = P([0,0]) + P([1,1])$ ?
It seems to me in the O.P. $\omega$ is used in two different ways.

Surely, $\mathcal{P}([0.2,0.7])=0.5$ But what does X(omega)=min{omega,1-omega} mean?
"with probability defined on intervals by P([a,b])=b-a, 0<= a <= b <=1" WHAT?

4. ## Re: Probability problem Originally Posted by Plato I am totally confused. What in the definition gives this: $P(\{0,1\}) = P([0,0]) + P([1,1])$ ?
It seems to me in the O.P. $\omega$ is used in two different ways.

Surely, $\mathcal{P}([0.2,0.7])=0.5$ But what does X(omega)=min{omega,1-omega} mean?
"with probability defined on intervals by P([a,b])=b-a, 0<= a <= b <=1" WHAT?
I took omega = $\omega$ while Omega = $\Omega$. So, $\omega \in [0,1] = \Omega$.

Then, to evaluate a probability on a set containing only two points, I turned them into two intervals, as $[0,0] \cup [1,1] = \{0,1\}$