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# s2 continuous random variable question help watch

1. Hi, can I have help with part d and e of the questions?

for part d I know that the probaibilty that one cartridge has been used for 600 hours is a 1/3 from part c. so I'm assuming that you would need to do 1/3*1/3=1/9 to find the probabilty of two-but I'm not sure if this is the correct method even though the answer is 1/9.
For part d I worked out that the one cartridge lasts for more than 600 hours is 2/3. but im not usre what tod do next the final answers should be 8/27.
2. part d) your logic is right since the cartridges are independent, and therefore the probability of one being less than 600 doesn't affect the other one so it would just be as you have said. In general stats speak its P(A and B)= P(A)*P(B) for independent events.
part e) Try a binomial distribution, you have the probability of being more than 600 as 2/3, the number of events as 4, and are trying to find P(X=2).
3. (Original post by IhatePickingName)
part d) your logic is right since the cartridges are independent, and therefore the probability of one being less than 600 doesn't affect the other one so it would just be as you have said. In general stats speak its P(A and B)= P(A)*P(B) for independent events.
part e) Try a binomial distribution, you have the probability of being more than 600 as 2/3, the number of events as 4, and are trying to find P(X=2).
Using the binomial distrubtion formula I did get the right answer but I am a bit confused with why it works. The question says to find the probabiltity that two last more than 600 and 2 two last less than 600. Wouldn't you need to use the 1/9 probability. Thanks for your help.
4. Okay, so they key here is to recognise that the order doesn't matter, the events can occur in any order. So the first two failing before 600 and the second to failing above 600 is exactly the same probability as the first one lasting for less than 600 then the next two lasting for more and then the final one failing at below 600 hours and more importantly, they are both examples of two failing before 600 and two failing afterwards. This means when we calculate the total probability we have to take into account every possible combination of two failing before 600 and two after.

The way I like to do it involves dividing my probability into two parts:
The first part involves the probabilities themselves so for two failing before 600 and two failing after we have:
(1/3)^2*(2/3)^2=4/81
The second part takes into account all possible combinations so this is were we use our binomial expansion/pascals triangle.
Since we want X=2 with 4 events we want 4C2 which is 6.
Finally, we just multiply these together to get the total probability of 8/27.
5. (Original post by IhatePickingName)
Okay, so they key here is to recognise that the order doesn't matter, the events can occur in any order. So the first two failing before 600 and the second to failing above 600 is exactly the same probability as the first one lasting for less than 600 then the next two lasting for more and then the final one failing at below 600 hours and more importantly, they are both examples of two failing before 600 and two failing afterwards. This means when we calculate the total probability we have to take into account every possible combination of two failing before 600 and two after.

The way I like to do it involves dividing my probability into two parts:
The first part involves the probabilities themselves so for two failing before 600 and two failing after we have:
(1/3)^2*(2/3)^2=4/81
The second part takes into account all possible combinations so this is were we use our binomial expansion/pascals triangle.
Since we want X=2 with 4 events we want 4C2 which is 6.
Finally, we just multiply these together to get the total probability of 8/27.
Thank you so much for taking the time to explain it to me, I understand it now

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