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!!!

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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