Hint: this question cannot be solved explicitly with the infomation given above, contrary to the logic of many. (COUGH! jimbo abe lincoln cough cough.) I'm looking instead for an "algebraic formula"; if you solved it right you'll know what i'm talking about.
2006-09-14 04:57:11 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
And yes i know I took this problem from Justin L.
2006-09-14 07:13:59 · update #1
Also the number 5.59% seems to come up a lot; this number is arrived at though INCORRECT reasoning.
2006-09-16 01:14:57 · update #2
So that means 75% of the time you lose 5 times straight
let p = prob win
q = prob loss
then q^5 = 0.75
q = (0.75)^(1/5)
and p = 1-q = 1- (0.75)^1/5) = 0.055912 approximately
about 5.6%
EDIT
This does assume however each game is indipendant with equal probability.
2006-09-14 05:06:14 · answer #1 · answered by Anonymous · 1⤊ 0⤋
1:20
original probability is 1:5, but only 1:4 of those will be successful.
2006-09-14 12:06:58 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Okay, here is my answer, though i'm not sure its right.
I say its (2^4/(2^5-1)) * (1/4) = (16/31)*(1/4) = 12.9% approximately.
2006-09-14 13:31:50 · answer #3 · answered by yljacktt 5 · 0⤊ 0⤋
x(>or =)5%, where x is the number of times you will win.
2006-09-14 12:08:05 · answer #4 · answered by Akfek_Branford 4 · 0⤊ 0⤋
1in 100
2006-09-14 12:20:23 · answer #5 · answered by jack 7 · 0⤊ 0⤋
the 'at least' part of the equation makes a real answer impossible.
2006-09-14 11:59:11 · answer #6 · answered by Anonymous · 0⤊ 0⤋
20%.
2006-09-14 12:01:31 · answer #7 · answered by isaac a 3 · 0⤊ 1⤋
12.5% is your probability
2006-09-14 12:08:43 · answer #8 · answered by bostoncity_guy 2 · 0⤊ 0⤋
www.whocares.com
2006-09-14 12:06:36 · answer #9 · answered by triplesixkoe 2 · 0⤊ 1⤋