algebra homework:probability. can anyone help me?
Here’s the question….
A three judge panel is being used to settle a dispute. Both sides in the dispute have decide that a majority decision will be upheld. If each judge will render a favorable decision based on the evidence presented two-thirds of the time, what is the probability that the correct side will win the dispute?
The answer to this problem’s probability is 20 out of 27.
My homework for tonight is to figure out how to get that answer as well as being able to explain it.
I really appreciate anyone who is able to help. =]]
Thanks.
Favorite Answer
to have two judges you’ll have j1, j2; j2,j3; j1,j3: so three different ways to have two judges being right. the probability for one of these is 1/3 * 2/3 * 2/3, one wrong * two right
3 * 1/3 * 2/3 * 2/3 = 4/9 = 12/27
the sum of the two is 20/27 so WELL DONE!!!
———
the more interesting way to do this is Let X be the number of judges that are right.
X has the binomial distribution with n = 3 trials and success probability 2/3
In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, …, n
find P(X ≥ 2) = P(X=2) + P(X=3) = 20/27
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