Question

Kazuki is playing a Scratch-and-Win competition in which the chance of winning a pet albatross is 1/15.

Show probability that he wins an albatros before his tenth attempt.

Answer 1

@amistre64 Do you know how to go about this?

Answer 2

\[20C1 (0.666)^n (.933)^(n-1)\]

Do I simply take the 9th value? Or is there some other way to do this?

Answer 3

P(first) + P(second) + P(third)+...+P(ninth)

Answer 4

we want 9 choose 1 attempts

1/15^1 success, 14/15^8 fail

Answer 5

So I got 34% is that correct?

Answer 6

9*14/15^9 is very small

Answer 7

0.57

Answer 8

s+fs+ffs+fffs+ffffs+fffffs+ffffffs+fffffffs+ffffffffs

or maybe we just say his chances of winning are 9/15 with 9 tickets? then we get something like 60%

otherwise the sum of the probs of the binomial setup is like .0000003%

Answer 9

So is the odd 60% he'll win before his 10th? That seems really high.

Answer 10

i know, but .0000003% seems very low

Answer 11

So what would I do for Still hasn't won an albatross after his 20th attempt

Answer 12

where does 20th attempt come from?

Answer 13

Its another question.

Answer 14

the chance of winning on the first ticket:
1/15

the chance of losing on the first ticket, and winning on the second:
14/15 *1/15

the chance of losing on the first 2 tickets, and winning on the third:
14/15 * 14/15 * 1/15

the chance of losing on the first 3 tickets, and winning on the forth:
14/15 * 14/15 * 14/15 * 1/15

etc....

right? then adding up all the probabilities

Answer 15

if we run the sum of a geometric sequence ....

1/15 (1+14/15+(14/15)^2+...+(14/15)^(n-1))

http://www.wolframalpha.com/input/?i=1%2F15+%281-%2814%2F15%29%5E9%29%2F%281-14%2F15%29

Answer 16

same chances out of 20 we get ....

http://www.wolframalpha.com/input/?i=1%2F15+%281-%2814%2F15%29%5E20%29%2F%281-14%2F15%29

Answer 17

So why doesn't this apply to the first question?

Answer 18

i think it does .... which is what i was actually working some bugs out of

Answer 19

the binomial assumes that you only have one winning ticket out of so many tries. the geometric sum is similar but it doesnt count the tickets "after" the winning ticket as anything useful

Answer 20

would you agree with all these "AND" setups? and that we should "OR" all the probabilities?

the chance of winning on the first ticket:
1/15

the chance of losing on the first ticket, and winning on the second:
14/15 *1/15

the chance of losing on the first 2 tickets, and winning on the third:
14/15 * 14/15 * 1/15

the chance of losing on the first 3 tickets, and winning on the forth:
14/15 * 14/15 * 14/15 * 1/15

Answer 21

in other words: spose we have 9 winning tickets,


w w w w w w w w w

this conforms to scratching the first one and winning before the 10th try; it also conforms to all the positional results before the 10th try. If we try to sort out a binomial on this it can get messy:

9 choose 1, exactly 1 win in 1 position
or 9 choose 2, exactly 2 win in 2 position
or 9 choose 3, exactly 3 win in 3 position
or 9 choose 4, exactly 4 win in 4 position
or 9 choose 5, exactly 5 win in 5 position
or 9 choose 6, exactly 6 win in 6 position
or 9 choose 7, exactly 7 win in 7 position
or 9 choose 8, exactly 8 win in 8 position
or 9 choose 9, exactly 9 win in 9 position

maybe .....

Answer 22

here is that mess :)

http://www.wolframalpha.com/input/?i=9%2814%5E8%2F15%5E9%29%2B36%2814%5E7%2F15%5E9%29%2B84%2814%5E6%2F15%5E9%29%2B126%2814%5E5%2F15%5E9%29%2B126%2814%5E4%2F15%5E9%29%2B84%2814%5E3%2F15%5E9%29%2B36%2814%5E2%2F15%5E9%29%2B9%2814%2F15%5E9%29%2B1%2F15%5E9

which is equal to the geometric sum i presented before

Answer 23

as i laid there in the middle of the might, it finally hit me that the question is simply: 1 - (never win)

1 - (14/15)^9

which is the same answer, but in a more succinct form :)