If there really were 8 vanilla and 11 peanut butter candies in the​ box, what is the probability that you would have picked 2 vanillas in a​ row?

Question

Someone hands you a box of 19 chocolate-covered candies, telling you that 8 are vanilla creams and the rest are peanut butter. You pick candies at random and discover the first 2 you eat are all vanilla. Consider a result to be surprising if there is less than a​ 5% chance of observing it given your expectations.

If there really were 8 vanilla and 11 peanut butter candies in the​ box, what is the probability that you would have picked 2 vanillas in a​ row?

Answer 1

8v, 11p

Pr(1st v) = 8/(8+11)= 8/19

Pr(2nd v) =(8-1)/(19-1) = 7/18

Pr(v1v2) =(8/19)(7/18) = (56)/342= about 0.164

Pr(2v's in a row) = 1 - Pr(no v's) - Pr(vp)-P(pv)

= 1- (11/19)(10/18) - (11/19)(8/18) - (8/19)(11/18)

= 1 - 110/342-88/342-88/342

= 1 - 286/342

= 56/342

= about 0.164

= about 16.4% chance of two vanillas in a row