Bunuel wrote:
Messi and Ronaldo are playing chess till one of them wins a total of four games. If each of them is equally likely to win any game and no game ends in a draw, what is the probability that at least six games will be played?
(A) 3/8
(B) 7/16
(C) 1/2
(D) 9/16
(E) 5/8
Given information:
1) Messi and Ronaldo are playing chess till one of them wins a total of 4 games.
2) Each likely to win
3) No draw
To determine: Probability(At least 6 games will be played)
Inferences:
1) Total possible options for a single game = 2 (Messi win or Ronaldo win, because no draw)
2) P(Either to win a game) = 1/2
3) P(At least 6 games) = 1 - Probability(Less than 6 games)So, we need to actually determine P(Less than 6 games) {and subtract the result from 1} and minimum possible number of games for match to finish is 4 so we need to find the P(game finishing in 4 games) or P(game finishing in 5 games)
P(Less than 6 games) = P(4 games) + P(5 games)P(4 games)Case 1: Messi winning all
Total possible outcomes = 2*2*2*2 = 16
Favorable outcomes = 1
P = 1/16
Case 2: Ronaldo winning all
Total possible outcomes = 2*2*2*2 = 16
Favorable outcomes = 1
P = 1/16
Total = Messi winning all or Ronaldo winning all = 1/16 + 1/16 = 1/8
P(5 games)Case 1: Messi winning 4 and Ronaldo winning 1 {RMMMM}
Total possible outcomes = 2*2*2*2*2 = 32
Total favorable outcomes = {5!/4!} - 1 = 4 [The reason for -1 is that there is one combination which will be MMMMR which means Messi wins the first 4 games but game will finish at that point so cannot have 5 games at that point and already covered 4 games scenario in previous part]
P = 4/32 = 1/8
Case 2: Ronaldo winning 4 and Messi winning 1 {MRRRR}
Total possible outcomes = 2*2*2*2*2 = 32
Total favorable outcomes = {5!/4!} - 1 = 4 [The reason for -1 is same as above]
P = 4/32 = 1/8
Total P = 2/8 = 2/8
Now: P(Less than 6) = P(4) + P(5) = 1/8 + 2/8 = 3/8
P(At least 6) = 1 - P(Less than 6) = 1 - (3/8) = 5/8Answer - E