They do play = x
They don't = y
x+y = 1
x=9y
now replace x in the first equation
9y+y = 1
10y = 1
y = .1 = 10%
now since x+y = 1
x + .1 = 1
x = .9 = 90%
San Francisco supposedly has a 90% chance, which is definitely not true... GO BEARZ!
2007-04-07 16:09:33
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answer #1
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answered by Sean Walker 3
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let a be the probability that San Francisco plays in the next Super Bowl
let b be the probability that they do not play in the next Superbowl
=> a=9b
The probability that SF plays in the next Superbowl plus the probability that they do not play is 1
a+b =1
=> a= 10% or 1/10
b=90% or 9/10
hope it's correct
2007-04-07 16:14:41
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answer #2
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answered by manlygirl_tatice_4ever 1
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You are correct that you need two eqns
So first you want to set up the variables so lets make SF in the Super Bowl = S and not in the Super Bowl = N
the first statement should look like this
9N = S
and the second should be
S + N = 1
So substitute S into the second eqn and you get
9N + N = 1 this is the same as
10N =1 and therefore N = 1/10
now put this into one of the two eqns I choose eqn 1 so we have now
9(1/10) = S therefore S = 9/10
hope this helps
2007-04-07 16:17:31
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answer #3
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answered by Art Vandalay 2
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"The probability that SF plays in the next Superbowl plus the probability that they do not play is 1."
This is true for all probabilities. The probability something will + won't = 1...
1 = x+9x
1=10x
x=.1
x = won't play = .1 therefor probability will = .9 or 90%
2007-04-07 16:13:06
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answer #4
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answered by Paul T 2
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I will use W as the probability they play, and L as the probability they won't. Since the problem says the probability of playing is nine times as likely as not playing, the first equation will be:
W=9*L
It also says the two probabilities added up equals 1 so:
W+L=1
You should be able to find the exact numbers now by substitution.
2007-04-07 16:12:19
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answer #5
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answered by Supermatt100 4
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Well basically. SF was a 10% chance of getting to a superbowl the next ten years.
2007-04-07 16:12:55
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answer #6
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answered by H3rBz 1
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OK.
say prob. that SA wins = W
and say prob. that SA loses = L
then..
W = 9 * L
W + L = 1 ( or L = 1- W)
(these are the 2 equations)
there fore..
W = 9 * (1- W)
W = 9 - 9W
10 W = 9
and W = 9/10
W = 0.9
Which means L = 0.1
(W + L = 0.9 + 0.1 = 1)
Hope this helps....
Good Luck..
2007-04-07 16:14:36
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answer #7
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answered by cool_guy 2
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Theoretically 0. wearing in basic terms one thousand at a time and eating a million in line with mile for one thousand miles could effect in there being 0 bananas left whilst Corey arrived on the marketplace. besides, Corey could no longer stroll lower back because of the fact there are not any bananas left for the holiday.
2016-10-21 07:55:11
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answer #8
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answered by ? 4
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so they are nine times more likely to go on then not. Nine chances to go and 1 to not. They have a 90% chance of going on and a 10% of not, i think
2007-04-07 16:09:31
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answer #9
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answered by Anonymous
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Let P( do not play) = x
Then P(play) = 9*x
And P(play) + P(do not play) = 9x + x = 10x = 1
So x = 1/10 = 0.1 = P(do not play)
And 9 * 0.1 = 0.9 = P(play)
hope that helps!
2007-04-07 16:10:10
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answer #10
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answered by birdwoman1 4
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