Three people toss a fair coin and the odd one pays for coffee. if the coins all turn up the same, they are tossed again. find the probability that fewer than 4 tosses are needed.

Since the problem doesn’t give a definite question, here are the possible questions and answer:

P (fewer than 4 tosses)

= P (one toss) + P (two toss) + P (3 toss)

= (3/4) + (3/4) (1/4) + (3/4) (1/4) ^2

= 0.984375

Expected value

= 1 / p

= 1 / (3/4)

= 4 / 3

Variance

= (1 - p) / p^2

= (1 - (3/4)) / (3/4) ^2

= (1/4) / (9/16)

= 4 / 9

Standard deviation

= sqrt (Variance)

= sqrt (4 / 9)

= 2 / 3