Cumulative Distribution function

[boneill3]

Homework Statement

I have been given a CDF of T value probabilites for t >= 0
I have been given P(t=3)=0.59 P(t=5)=0.85

Homework Equations

The Attempt at a Solution

I have been asked to find P(T > 5 | T>3 )

I was wondering how to work this out.

As this is a conditional probabilty I was heading towards

P(T > 5 | T>3 ) =
$\frac{P(T > 5 \and T>3 )}{P(t>5)}$
so the probabilty of P(T > 5 \and T>3 )} = .41
and P(t>5) = .15

But you can't use that calculation.

But wouldn't the probability of of being > 5 still just be .15

Any help would be appreciated

 

[HallsofIvy]

Homework Statement

I have been given a CDF of T value probabilites for t >= 0
I have been given P(t=3)=0.59 P(t=5)=0.85

Homework Equations

The Attempt at a Solution

I have been asked to find P(T > 5 | T>3 )

I was wondering how to work this out.

As this is a conditional probabilty I was heading towards

P(T > 5 | T>3 ) =
$\frac{P(T > 5 \and T>3 )}{P(t>5)}$
so the probabilty of P(T > 5 \and T>3 )} = .41
and P(t>5) = .15

But you can't use that calculation.

But wouldn't the probability of of being > 5 still just be .15

Any help would be appreciated

Yes, P(t> 5)= 1- .85= .15. But that is NOT "P(t>v5|P>3)"

Remember the basic formula P(A and B)= P(A|B)P(B).

Now if x> 5 then it MUST be >3 so P((x>.5) and (x>3))= P(x> 5).

P(t>5)= P(t>5|P>3)P(t> 3).

You know P(T>5) and you know P(t> 3). Put those into that formula and solve for P(t> 5|t> 3).

 

[boneill3]

Thanks for your help.

So we need to find


P(A|B) = P(A and B)/P(B)

we have:

P(t>5)= P(t>5|P>3)P(t> 3)

and need to find

P(t>5|P>3)= P(t>5)/P(t> 3)

P(t>5|P>3)= P(0.15)/P(0.41)

therefore

P(t>5|P>3)= 0.365

regards