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

To find this probability, first consider the outcomes that result in winning Game 1 and losing Game 2, which is WWL. Since each outcome has a probability of $\frac{1}{3}$, we find:

$P(Win\, Game\, 1\, and\, Lose\, Game\, 2) = P(W) \times P(L) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}.$

To find the probability that Hager wins at least one of her first two games, we can use the complementary approach. First, calculate the probability that she does not win any of the games:

The possible outcomes where she doesn’t win are: LL, LD, and DL. The total number of outcomes is 9. Thus, the probability she does not win is:

$P(No\, Win) = \frac{3}{9} = \frac{1}{3}.$

Therefore, the probability that she wins at least one game is:

$P(At\, least\, one\, win) = 1 - P(No\, Win) = 1 - \frac{1}{3} = \frac{2}{3}.$

For 3 games, each with 3 possible outcomes (W, D, or L), the total number of different possible outcomes is calculated by:

$Total\, outcomes = 3^3 = 27.$

The outcomes where she does not win are limited to 3 possible outcomes: D (Draw) and L (Lose). Therefore, for 3 games, the non-winning outcomes can be calculated:

Number of ways to not win: 2 outcomes (D or L) for each of the 3 games:

$P(Doesn’t\, Win) = \frac{2^3}{3^3} = \frac{8}{27}.$