A team plays two matches. The probabilities for a single match are P(W) = 0.6, P(L) = 0.3, and P(D) = 0.1. Draw a tree diagram showing all possible outcomes and calculate the probability of each outco

a) Let W denote a win, L denote a loss, and D denote a draw.
Given probabilities for a single match:
$P(W) = 0.6$
$P(L) = 0.3$
$P(D) = 0.1$

The team plays two matches. Here is the tree diagram:

Match 1

• Start with a single point.
• Branch 1: Win (W) with probability 0.6
• Branch 2: Lose (L) with probability 0.3
• Branch 3: Draw (D) with probability 0.1

Match 2

• From each outcome of Match 1, draw three new branches for Match 2:
  • If Match 1 was W:
    • Win (W) with probability 0.6 (Path: WW)
    • Lose (L) with probability 0.3 (Path: WL)
    • Draw (D) with probability 0.1 (Path: WD)
  • If Match 1 was L:
    • Win (W) with probability 0.6 (Path: LW)
    • Lose (L) with probability 0.3 (Path: LL)
    • Draw (D) with probability 0.1 (Path: LD)
  • If Match 1 was D:
    • Win (W) with probability 0.6 (Path: DW)
    • Lose (L) with probability 0.3 (Path: DL)
    • Draw (D) with probability 0.1 (Path: DD)

The possible outcomes and their probabilities are:

• $P(WW) = 0.6 \times 0.6 = 0.36$
• $P(WL) = 0.6 \times 0.3 = 0.18$
• $P(WD) = 0.6 \times 0.1 = 0.06$
• $P(LW) = 0.3 \times 0.6 = 0.18$
• $P(LL) = 0.3 \times 0.3 = 0.09$
• $P(LD) = 0.3 \times 0.1 = 0.03$
• $P(DW) = 0.1 \times 0.6 = 0.06$
• $P(DL) = 0.1 \times 0.3 = 0.03$
• $P(DD) = 0.1 \times 0.1 = 0.01$

b) What is the probability that the team will:

i) Win the two matches
This corresponds to the outcome WW.

$$P(\text{Win two matches}) = P(W) \times P(W) = 0.6 \times 0.6 = 0.36$$

The probability of winning two matches is $\boxed{0.36}$.

ii) Either wins all the matches or loses all the matches
This means the outcomes WW or LL. These are mutually exclusive events.

$$P(\text{Wins all or loses all}) = P(WW) + P(LL)$$ $$P(WW) = 0.6 \times 0.6 = 0.36$$ $$P(LL) = 0.3 \times 0.3 = 0.09$$ $$P(\text{Wins all or loses all}) = 0.36 + 0.09 = 0.45$$

The probability of either winning all matches or losing all matches is $\boxed{0.45}$.

iii) Wins one match and loses one
This means the outcomes WL or LW. These are mutually exclusive events.

$$P(\text{Wins one and loses one}) = P(WL) + P(LW)$$ $$P(WL) = P(W \text{ in match 1}) \times P(L \text{ in match 2}) = 0.6 \times 0.3 = 0.18$$ $$P(LW) = P(L \text{ in match 1}) \times P(W \text{ in match 2}) = 0.3 \times 0.6 = 0.18$$ $$P(\text{Wins one and loses one}) = 0.18 + 0.18 = 0.36$$

The probability of winning one match and losing one is $\boxed{0.36}$.

iv) Tie in one match
This means exactly one match is a draw. The other match must not be a draw.
Let $P(\text{not D}) = P(W) + P(L) = 0.6 + 0.3 = 0.9$.
The possible outcomes are (D and not D) or (not D and D).

$$P(\text{Tie in one match}) = P(D \text{ in match 1 and not D in match 2}) + P(\text{not D in match 1 and D in match 2})$$ $$P(\text{Tie in one match}) = (P(D) \times P(\text{not D})) + (P(\text{not D}) \times P(D))$$ $$P(\text{Tie in one match}) = (0.1 \times 0.9) + (0.9 \times 0.1)$$ $$P(\text{Tie in one match}) = 0.09 + 0.09 = 0.18$$

Alternatively, this includes outcomes WD, LD, DW, DL:

$$P(\text{Tie in one match}) = P(WD) + P(LD) + P(DW) + P(DL)$$ $$P(\text{Tie in one match}) = (0.6 \times 0.1) + (0.3 \times 0.1) + (0.1 \times 0.6) + (0.1 \times 0.3)$$ $$P(\text{Tie in one match}) = 0.06 + 0.03 + 0.06 + 0.03 = 0.18$$

The probability of a tie in one match is $\boxed{0.18}$.