A and B play one another at chess. The probability of A winning a particular game is ⅓ and the 3 probability of a draw is ⅙. In a tournament, A and B play 4 games each other. Find the probability that 6
(i) none is drawn
(ii) A wins 3 games with the other drawn
(iii) B wins exactly 3 games
prob(A wins) = 1/3
prob(A draw) = 1/6
prob(A loses) = 1 - 1/2 - 1/6 = 1/2
Let's look at it as the outcomes of player A
a) no draws:
we could have WWWW = (1/3)^4 = 81
We could have WLLL, in its permutations = 4(1/3)(1/2)^3 = 4/24 = 1/6
We could have WWLL, in its permutations = 4!/(2!2!)(1/3)^2 (1/2)^2 = 1/6
We could have WWWL, in its permutations = 4((1/3)^3 (1/2) = 2/27
prob(no draws) = 1/6 + 1/6 + 2/27 = 11/27
b) must be a permutation of WWWL, which we did above and is 2/27
c) B wins exactly 3 games ----> A loses exactly 3 games
= permutation of WLLL = 4(1/3)(1/2)^3 = 1/6
Your 2 typos could change my interpretation of the question.
for a)
should say:
we could have WWWW = (1/3)^4 = 1/81
...
prob(no draws) = 1/81 + 1/6 + 1/6 + 2/27 = 34/81
To find the probability of different outcomes in this tournament, we can use the concept of binomial probability. The binomial probability formula is used when there are two possible outcomes (in this case, A winning or B winning) and you want to find the probability of a specific number of successes (games won) in a fixed number of trials (games played).
In this case, A and B play 4 games each. The probability of A winning a game is ⅓, and the probability of a draw is ⅙.
(i) Probability that none of the games is drawn:
To find the probability of A winning all 4 games and there being no draws, we multiply the probability of A winning a game (⅓) by itself four times (since A plays 4 games). So,
P(A winning a game) = ⅓
P(A winning 4 games and no draws) = (⅓)^4 = 1/81
(ii) Probability that A wins 3 games with the other drawn:
To find the probability of A winning 3 games and one game being a draw, we need to consider different combinations of wins and draws. We can use the binomial coefficient to calculate the number of ways to choose 3 games out of 4 to be won by A. The binomial coefficient can be calculated as C(4,3) = 4.
The probability of A winning a game is ⅓, and the probability of a draw is ⅙. So,
P(A winning a game) = ⅓
P(Draw) = ⅙
The probability of A winning 3 games and 1 game being a draw can be calculated as:
P(A wins 3 games and 1 game drawn) = C(4,3) * (⅓)^3 * (⅙) = 4 * (1/27) * (1/6) = 4/81
(iii) Probability that B wins exactly 3 games:
Similar to part (ii), we use the binomial coefficient to calculate the number of ways to choose 3 games out of 4 to be won by B. The binomial coefficient can be calculated as C(4,3) = 4.
The probability of B winning a game is 1 - P(A winning a game) - P(Draw) = 1 - (⅓) - (⅙) = 1 - 1/3 - 1/6 = 1/2. So,
P(B winning a game) = 1/2
The probability of B winning 3 games can be calculated as:
P(B wins exactly 3 games) = C(4,3) * (1/2)^3 * (1/2) = 4 * (1/8) * (1/2) = 4/16 = 1/4
Therefore, the probabilities are:
(i) Probability that none is drawn = 1/81
(ii) Probability that A wins 3 games with the other drawn = 4/81
(iii) Probability that B wins exactly 3 games = 1/4