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# using binomial distribution watch

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1. i do not get this q.
it is do to with which is most likely rolling one 6 in 6 throws or rolling two sixes in 12 throws,
On a single throw of a die the probability of getting a 6 is a 1/6 and that of not getting a 6 is 5/6.
So the probability distributions for the 2 situations in the problem are B(6,1/6) and B(12,1/6) giving probabilities of:

1-6C0 (5/6)^6 = 1-0.335=0.665 (at least one 6 in 6 throws)

and

1-[12C0(5/6)^12+12C1(5/6)^11(1/6)]=1-(0.112+0.269)
=0.619 (at least 2 sixes in 12 throws)

So at least one 6 in 6 throws is more likely.

what i dont get is why in 6 throws^^ the r value is 0 if they need one 6 as usually P(X=R) in this case i would of thought that P(X=1) instead of P(X=0)
2. can ne 1 help plz
3. (Original post by kam130)
i do not get this q.
it is do to with which is most likely rolling one 6 in 6 throws or rolling two sixes in 12 throws,
On a single throw of a die the probability of getting a 6 is a 1/6 and that of not getting a 6 is 5/6.
So the probability distributions for the 2 situations in the problem are B(6,1/6) and B(12,1/6) giving probabilities of:

1-6C0 (5/6)^6 = 1-0.335=0.665 (at least one 6 in 6 throws)

and

1-[12C0(5/6)^12+12C1(5/6)^11(1/6)]=1-(0.112+0.269)
=0.619 (at least 2 sixes in 12 throws)

So at least one 6 in 6 throws is more likely.

what i dont get is why in 6 throws^^ the r value is 0 if they need one 6 as usually P(X=R) in this case i would of thought that P(X=1) instead of P(X=0)
are you sure its P(X=0) cuz that doesn't make sense! Maybe its because the prob of at least 1 is the prob of 1 - P(X=0)... is that it?
4. na i jus dont get it wot exam board u doin
5. (Original post by kam130)
na i jus dont get it wot exam board u doin
OCR/MEI... is this S2 for you? if so do you have your exam this friday?
6. na s1, if ya got dat book chek out top of page 149
7. by the way if i was taclking this question i wouldn't have gone about it that way... as the question doesn' ask you to find "at least" i would do it like this:

for the prob of 1 in 6 throws:

6C1.(1/6)^1.(5/6)^5 = 0.402

for the prob of 2 in 12 throws:

12C2.(1/6)^2.(5/6)^10 = 0.296

therefore there is a higher probability of 1 in 6 throws...
8. (Original post by kam130)
na s1, if ya got dat book chek out top of page 149
which board?
9. same as u, chek out page 149 if ya got it, (mei, ocr)
10. iv had a look in the s1 book i have and the pages only go up to 140 so i must have the old version... what section is it in? probability or binomial dis?
11. binomial distribution
12. ok no i don't have that question in my book... is the answer i did above not the correct one then? if not try typing the exact question out and il have a look
13. the exact q is at d start of dis thread, but i dont understand, y P(X=0) when it should be P(X=1) i think newayz, btw wot dya get in S1
14. but where in the question does it specify that P(X=0)??
I got an A in s1
15. it doesnt 6C0 tells me that r=0 and P(X=0) where R=0
16. but theres no need to use 6C0 in the question... you would need to if the question was to find the probability of "at least" 1 six in 6 throws in which case the prob is 1 - P(X=0) . If this is what the question asks then it would be:

for at least 1 six in 6 throws:

6C0.(1/6)^0.(5/6)^6 = 0.335
1 - 0.335 = 0.665
17. which is what you'v written so i don't see where the problem is...
18. The question is asking for the probability of AT LEAST 1 Six.

P(X=1) is the probability of EXACTLY 1 Six. The question DOES NOT ASK THIS.

You are asked to calculate p(X>0) = 1 - P(X=0)

The probability of P(X=0) is just (5/6)^6 - you can multiply by 6c0 if you want to, but that just equals 1 anyway, so it's not required (because there is only one combination of not getting a six 6 times)

Likewise in question 2, you are asked for the p(X>1) = 1 - P(X=0) - P(X=1)

P(X=0) is easy....just (5/6)^12 again (the 12c0 is not required)

P(X=1) is just (5/6)^11 * (1/6), because that is the probability of getting one 6 in 12 throws. BUT THEN you must multiply by 12C1 (which equals 12), because there are 12 combinations of getting just one six (you could throw the 6 in your first throw, or in your second, or third, or fourth...and so on - which equals 12 different possiblities)

Hope that's clear

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Updated: June 8, 2004
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