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There is a 90% chance that a registered voter in Burghtown 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%
A. 26.2%
B. 32.8%
C. 43.7%
D. 59.0%
E. 65.6%
25 Apr 2010, 04:08
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fruit
If the probability of a certain event is \(p\), then the probability of it occurring \(k\) times in \(n\)-time sequence is: \(P = C^k_n*p^k*(1-p)^{n-k}\)
In our case:
n=5
k=4
p=0.9
So, \(P = C^k_n*p^k*(1-p)^{n-k}=C^4_5*0.9^4*0.1\)
OR: probability of scenario V-V-V-V-N is \(0.9^4*0.1\), but V-V-V-V-N can occur in different ways:
V-V-V-V-N - first four voted and fifth didn't;
N-V-V-V-V - first didn't vote and last four did;
V-N-V-V-V first voted, second didn't and last three did;
...
Certain # of combinations. How many combinations are there? Basically we are looking at # of permutations of five letters V-V-V-V-N, which is 5!/4!.
Hence \(P=\frac{5!}{4!}*0.9^4*0.1\).
Check this links for similar problems:
combinatorics-at-least-none-56728.html
probability-qs-attention-88945.html#p671944
probability-q-91460.html#p666937
permutation-86687.html#p650790
Also you can check Probability chapter of Math Book for more (link in my signature).
22 Jan 2013, 20:55
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manimgoindowndown
The point is that the 5 voters are unique. Say they are A, B, C, D and E.
Now the case where A, B, C and D voted in the last election is different from the case where B, C, D and E voted in the last election. These are two different ways in which we can have 4 of the 5 who voted in the last election. There are 5 such different cases.
When you .9 * .9 * .9 * .9 * .1, you are finding the probability that A, B, C and D voted in the last election while E did not.
So you have to count 4 more cases e.g. .1 * .9 * .9 * .9 * .9 (Probability that A did not vote in the last election while B, C, D , E voted in the last election) etc.
The calculation is identical to the first case so you just need to count the first case 5 times.
