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Probability help needed AGAIN :(

Urgh I just keep getting so confused over these probability questions they're really hard.

Okay so this question is in two parts.

Every Sunday afternoon two children, Andy and Betty, try to fly their kites. Suppose that Andy is successful with a probability 0.2 and Betty is successful with a probability 0.25. Find the probability that it will take a least three Sundays until a kite flies.

Continuing on from this question, how many Sundays should the Children expect to wait until at least one kite flies?

So for the first part I tried to do the Geometric Distribution but I am really worried my notes have been written wrong. It says for At least 3 we take the Pr(X>=2)??? Which is confusing to me. I'm really stumped.

Anyone who can take the time to help me with this will be greatly appreciated.

Reply 1

DFranklin

Original post by Mr XcX

So for the first part I tried to do the Geometric Distribution but I am really worried my notes have been written wrong. It says for At least 3 we take the Pr(X>=2)??? Which is confusing to me. I'm really stumped. For this to make sense I would guess that your definition of X here(*) is supposed to be the number of trials before success. If you don't have success for at least 2 trials, then the number of trials until success must be at least 3.

(*) A difficulty for people trying to answer these questions over the internet is that there are two different definitions of Geometric Distribution. It can either be number of trials before success, or number of trials until success, and without seeing your notes it's hard to know which definition you are using.

Reply 2

Mr XcX

Original post by DFranklin

For this to make sense I would guess that your definition of X here(*) is supposed to be the number of trials before success. If you don't have success for at least 2 trials, then the number of trials until success must be at least 3.

(*) A difficulty for people trying to answer these questions over the internet is that there are two different definitions of Geometric Distribution. It can either be number of trials before success, or number of trials until success, and without seeing your notes it's hard to know which definition you are using.

For the first part could I not do

Pr(3)=(0.6)^2*(0.4).

Reply 3

DFranklin

Original post by Mr XcX

For the first part could I not do

Pr(3)=(0.6)^2*(0.4).

No. (That looks like the probabilty of it taking exactly 3 Sundays).

(edited 8 years ago)

Reply 4

Mr XcX

Original post by DFranklin

No. (That looks like the probabilty of it taking exactly 3 Sundays).

I got 0.360 for the first part.

If that's wrong then I have no clue what to do.

I did

Pr(X>=3) = 1 - 0.4 - 0.6(0.4)

Reply 5

Mr XcX

Can anyone give me a hint for how I do the 2nd part??

Reply 6

DFranklin

Original post by Mr XcX

I got 0.360 for the first part.

If that's wrong then I have no clue what to do.

I did

Pr(X>=3) = 1 - 0.4 - 0.6(0.4)

This is correct, but you are making it much harder than it needs to be. You have calculated 1 - p(success on first Sunday) - p(no success on first Sunday, success on 2nd Sunday).

But X>=3 is the same as saying "X <= 2 is not true". So you could simply calculate p(no success on 1st and 2nd Sundays), whicih is just 0.6^2 = 0.36.

Original post by Mr XcX

Can anyone give me a hint for how I do the 2nd part??

What you have is a geometric series. You need to look at your definitions and results for geometric series and the result should be immediate.

Reply 7

HadiLadi

As far as I understand this can be just solved with a simple tree diagram right?

Reply 8

DFranklin

Original post by HadiLadi

As far as I understand this can be just solved with a simple tree diagram right?

The last part of the problem (the expectation bit) can't be solved with a tree diagram (well, there's a very non-obvious way of doing it via tree diagram but it's not what you're supposed to do).

Reply 9

Stevenson Dude

Original post by DFranklin

The last part of the problem (the expectation bit) can't be solved with a tree diagram (well, there's a very non-obvious way of doing it via tree diagram but it's not what you're supposed to do).