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1. It has been established that the probabilities of Exater High School (EHS) and Kab Day School (KDS) soccer teams scoring 0, 1, 2, or 3 goals against each other are as shown below:

\begin{tabular}{|l|l|l|}
\hline
\begin{tabular}{l}
No. of \\
goals
\end{tabular} & \multicolumn{2}{|c|}{ Probabilities of scoring } \\
\hline
0 & [tex]$3.0 \times 10^{-1}$[/tex] & [tex]$2.0 \times 10^{-1}$[/tex] \\
1 & [tex]$3.0 \times 10^{-1}$[/tex] & [tex]$4.0 \times 10^{-1}$[/tex] \\
2 & [tex]$3.0 \times 10^{-1}$[/tex] & [tex]$3.0 \times 10^{-1}$[/tex] \\
3 & [tex]$1.0 \times 10^{-1}$[/tex] & [tex]$1.0 \times 10^{-1}$[/tex] \\
\hline
\end{tabular}

a) Using a tree diagram or otherwise, show all the possible outcomes.
(3 Marks)

b) Calculate the probability of:
(i) EHS winning
(3 Marks)
(ii) A draw
(2 Marks)
(iii) KDS winning
(2 Marks)



Answer :

GinnyAnswer
Absolutely, let's delve into the given problem step by step.

### Part a) Showing All Possible Outcomes

To show all possible outcomes of the number of goals scored by both teams, let’s enumerate them. Both teams can score 0, 1, 2, or 3 goals.

This gives us a total of 4 (EHS possible goals) × 4 (KDS possible goals) = 16 possible outcomes. They can be listed as follows:

```
(0, 0), (0, 1), (0, 2), (0, 3)
(1, 0), (1, 1), (1, 2), (1, 3)
(2, 0), (2, 1), (2, 2), (2, 3)
(3, 0), (3, 1), (3, 2), (3, 3)
```

### Part b) Calculating Probabilities

To calculate the probability of each outcome, multiply the probability of EHS scoring a certain number of goals by the probability of KDS scoring a certain number of goals for each pairing.

The probabilities of each pair of outcomes are:

| Outcome | Probability |
|---------|--------------------|
| (0, 0) | 0.06 |
| (0, 1) | 0.12 |
| (0, 2) | 0.09 |
| (0, 3) | 0.03 |
| (1, 0) | 0.06 |
| (1, 1) | 0.12 |
| (1, 2) | 0.09 |
| (1, 3) | 0.03 |
| (2, 0) | 0.06 |
| (2, 1) | 0.12 |
| (2, 2) | 0.09 |
| (2, 3) | 0.03 |
| (3, 0) | 0.02 |
| (3, 1) | 0.04 |
| (3, 2) | 0.03 |
| (3, 3) | 0.01 |

Now, let's use these values to calculate the probabilities for each scenario.

#### Part b) (i) Probability of EHS Winning

EHS wins in the following cases:
- EHS scores 1, KDS scores 0
- EHS scores 2, KDS scores 0 or 1
- EHS scores 3, KDS scores 0, 1, or 2

Adding up these probabilities:
[tex]\[ P(\text{EHS wins}) = 0.06 + 0.06 + 0.12 + 0.02 + 0.04 + 0.03 = 0.33 \][/tex]

So, the probability of EHS winning is:

[tex]\[ \boxed{0.33} \][/tex]

#### Part b) (ii) Probability of a Draw

A draw occurs in the cases:
- Both teams score 0
- Both teams score 1
- Both teams score 2
- Both teams score 3

Adding up these probabilities:
[tex]\[ P(\text{Draw}) = 0.06 + 0.12 + 0.09 + 0.01 = 0.28 \][/tex]

So, the probability of a draw is:

[tex]\[ \boxed{0.28} \][/tex]

#### Part b) (iii) Probability of KDS Winning

KDS wins in the following cases:
- KDS scores 1, EHS scores 0
- KDS scores 2, EHS scores 0 or 1
- KDS scores 3, EHS scores 0, 1, or 2

Adding up these probabilities:
[tex]\[ P(\text{KDS wins}) = 0.12 + 0.09 + 0.09 + 0.03 + 0.03 + 0.03 = 0.39 \][/tex]

So, the probability of KDS winning is:

[tex]\[ \boxed{0.39} \][/tex]

These results should thoroughly answer the question.

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