Exponential Distribution

[Thomas74]

Good evening; I want to make sure I did the right thing; if you guys have time, please kindly do check my work.

Here's the problem:

-Random variable X comes with a probability of 0.3 from an exponential distribution with lamda=0.5 and with the remaining probability from an exponential distribution with lamda=0.25

-What is the probability that this random variable has a value larger than 5?

If this random variable has a value larger than 5, what is the probability that it comes from the first distribution (E(0.5)) and the second one(E(0.25))?

Here's what I did as a picture.

I appreciate your help!

 

[BigBeachBanana]

Can you post the original question? I find the wording ambiguous. Just want to make sure what you posted is correct.

 

[Thomas74]

Can you post the original question? I find the wording ambiguous. Just want to make sure what you posted is correct.

Ok, Here it is.

 

[Thomas74]

Ok, Here it is.

If you have free time, I kinda have another one too, I did half way, thank you https://www.freemathhelp.com/forum/threads/power-of-test-of-the-hypothesis.132595/

 

[BigBeachBanana]

Part 1 should be a single value. In fact, you calculated as in part 2. The P_all(X>5) should be the answer for part 1. It's good practice to start setting up your problem using probabilities. To get you started, can you continue?
$$Pr(X>5) = Pr(X>5 \cap E(0.5)) + Pr(X>5 \cap E(0.25))$$Part 2,the question is asking for Pr[E(0.5) | X>5] and Pr[E(0.25) | X>5]. Can you set this up using probability rules?
Your answers are correct.

 

[Thomas74]

Part 1 should be a single value. In fact, you calculated as in part 2. The P_all(X>5) should be the answer for part 1. It's good practice to start setting up your problem using probabilities. To get you started, can you continue?
$$Pr(X>5) = Pr(X>5 \cap E(0.5)) + Pr(X>5 \cap E(0.25))$$

$$\Pr(X>5)=Pr(X>5) X Pr(E(0.5))+Pr(X>5) X Pr(E(0.25))$$

Part 2,the question is asking for Pr[E(0.5) | X>5] and Pr[E(0.25) | X>5]. Can you set this up using probability rules?
Your answers are correct.

$$\Pr[E(0.5) | X>5] =Pr(X>5∩E(0.5)/(P(X>5))$$$$\Pr[E(0.25) | X>5] =Pr(X>5∩E(0.25)/(P(X>5))$$

 

[BigBeachBanana]

$$\Pr(X>5)=Pr(X>5) X Pr(E(0.5))+Pr(X>5) X Pr(E(0.25))$$
$$\Pr[E(0.5) | X>5] =Pr(X>5∩0.5)/(P(X>5))$$$$\Pr[E(0.25) | X>5] =Pr(X>5∩0.25)/(P(X>5))$$

Part 1) should be $$\Pr(X>5)=Pr(X>5|E(.5)) *Pr(E(0.5))+Pr(X>5|E(0.25)) * Pr(E(0.25))$$Notice it should be conditional
Part2) looks fine.
Good practice to write your solution in this manner, then plug in the values.

 

[Thomas74]

Part 1) should be $$\Pr(X>5)=Pr(X>5|E(.5)) *Pr(E(0.5))+Pr(X>5|E(0.25)) * Pr(E(0.25))$$Notice it should be conditional
Part2) looks fine.
Good practice to write your solution in this manner, then plug in the values.

ok, thank you, I'll try that from now on, I have another one if you don't mind https://www.freemathhelp.com/forum/threads/power-of-test-of-the-hypothesis.132595/