Two people took turns tossing a fair die until one of them tossed a 6. PersonA tossed first, B second, A third, and so on. Given that person B threw the first 6, whatis the probability that B obtained the first 6 on her second toss (that is, on the fourth tossoverall) ?

Answer: 0.0965

Step-by-step explanation:

This would happen if:

First toss: Here we must have any number that is not 6.

the options are 1, 2, 3, 4, 5 so the probablity is p1 = 5/6

The same happens for the second toss, p2 = 5/6

and for the third one: p3 = 5/6

for the fourth toss, person B must roll a 6, so the probability here is p4 = 1/6

Now, the joint probability is equal to the product of the probabilities for each toss, this is:

P = p1*p2*p3*p4 = (5/6) ^3 * (1/6) = 0.0965