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young_gun
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binomial distrib 23 Oct 2007, 17:59
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is there a way to solve the below problem without using the binomial distribution?

There is a 90% chance that a registered voter in Burghtwon voted in the last election. If five registered voters are chosen at random, what is the approximate likelihood that exactly four of them voted in the last election?

a) 26.2 %
b) 32.8%
c) 43.7 %
d) 59.0 %
e) 65.6%



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binomial distrib 23 Oct 2007, 18:03
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B

5C4 * (.9)^4* (.1)
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binomial distrib 23 Oct 2007, 19:20
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trivikram
B

5C4 * (.9)^4* (.1)
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isnt that just the binomial distrib? is there any other practical way to solve this? also, can someone post the equation for binomial, does it end with P(not happening)^(k-n) or just (P not happening)?
thx
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binomial distrib 23 Oct 2007, 20:11
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P(voted) = 0.9
P(!voted) = 0.1

The winning word is VVVVN where V = voted, N = not voted. This can be arranged 5!/4!1! = 5 ways.

Each word has a probability of (0.9)^4 (0.1). So 5 words will have a probability of 0.328 or 32.8%.

Ans B
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binomial distrib 23 Oct 2007, 21:43
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ywilfred
P(voted) = 0.9
P(!voted) = 0.1

The winning word is VVVVN where V = voted, N = not voted. This can be arranged 5!/4!1! = 5 ways.

Each word has a probability of (0.9)^4 (0.1). So 5 words will have a probability of 0.328 or 32.8%.

Ans B
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Very nicely explained. thanks.
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binomial distrib 23 Oct 2007, 21:45
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young_gun
trivikram
B

5C4 * (.9)^4* (.1)

isnt that just the binomial distrib? is there any other practical way to solve this? also, can someone post the equation for binomial, does it end with P(not happening)^(k-n) or just (P not happening)?
thx
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C(n, k) * p^k * (1-p)^n-k

n = number of draws, 5 in this case
k = exact number we want, 4 in our case



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