Hager takes part in a chess competition - Junior Cycle Mathematics - Question 4 - 2018

The possible outcomes for Hager's first two games can be outlined as follows:

Game 1	Game 2
W	W
W	D
W	L
D	W
D	D
D	L
L	W
L	D
L	L

There are a total of 9 possible outcomes for two games. The specific outcome where Hager wins Game 1 and loses Game 2 is one of these outcomes. Therefore, we can find the probability as:

P(W, L) = rac{1}{9}

To find the probability that Hager wins at least one game, we can alternatively calculate the probability that she does not win any games and subtract from 1. The only outcome where she does not win is the case where she either draws or loses both games. This results in:

The outcomes where she wins at least one game are:

• WW
• WL
• WD
• DW
• DL
• LW So, we have 6 successful outcomes for winning at least one game out of 9 total outcomes, leading to:

P(at \, least \, one) = 1 - P(none) = 1 - rac{3}{9} = rac{6}{9} = rac{2}{3}

Each game has 3 possible outcomes: W, D, or L. Therefore, for 3 games, the total number of different possible outcomes can be calculated as:

$Total \, outcomes = 3 imes 3 imes 3 = 27$

The only outcomes where Hager does not win any of her games are when she either draws or loses all games. The outcomes are: DDD, DLL, DLD, LDD, LDL, LLD, LLL.

Thus, there are 8 favorable outcomes. The probability is given by:

P(doesn't \, win) = rac{8}{27}