Question

In a certain game of chance, your chances of winning are 0.2. if you play the game five times and outcomes are independent, the probability that you win at most once is

Answer 1

Winning Probablity = 0.2, hence Losing Probability = 0.8 Probablity of winning atmost one time, that means win one and lose four times or lose all the times. So p(W1 or W0) = p (W1) + p(W0) Winning once W1 is equal to L4, winning zero times is losing 5 times. p(W1) = p(W1&L4) and this happens 5 times; p(W0) = p(L5); p (W1) + p(W0) = p(L4) + p(L5) p(L4) + p(L5) = (5 x 0.2 x 0.8^4) + (0.8^5) => 0.8^4 + 0.8^5 p(W1 or W0) = 0.4096 + 0.32768 = 0.7373