Suppose you tossed a die six times. What is the probability of getting exactly three fours

Question

Probability 

Answer 1

Let X be the number of 4's among 6 tosses of a fair die.

 

Then letting a an outcome of 4 on a toss be a success, we have n=6 trials and probability of success per trial is p=1/6.

 

Thus we are looking for P(X=3) where X ~ Bin(6,1/6).

 

Recall that for X ~ Bin(n,p), P(X=k) = C(n,k) p^k (1-p)^n-k

 

Putting n=6, p=1/6, k=3 gives:

 

P(X=3) = C(6,3)(1/6)³(5/6)^6-3 = 20(1/216)(125/216) = 625/11664 = 0.05358…

Answer 2

my wrong answer was:  (1/6)³ = 1/216 = 0.00463  <------- If there were only three tosses.

 

I messed up on this answer, forgetting that it was 6 tosses, in which case it's a binomial distribution (Bernoulli trials).

 

My thanks to Roman & my apology to Cheyenne!