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Kiah plays a game.

The game involves throwing a coin onto a circular board. Points are scored based on where the coin lands on the board.

If the coin lands on part of a line or misses the board, 0 points are scored. The table shows the probabilities of Kiah scoring points on the board with one throw.

\begin{tabular}{|l|c|c|c|c|}
\hline Points scored & 20 & 10 & 5 & 0 \\
\hline Probability & [tex]$x$[/tex] & 0.2 & 0.3 & 0.45 \\
\hline
\end{tabular}

(a) Find the value of [tex]$x$[/tex] :
[tex]\[ x = \][/tex]

(b) Kiah throws a coin fifty times. Work out the expected number of times she scores 5 points.
[tex]\[ \text{Expected number of times} = \][/tex]

(c) Kiah throws a coin two times. Calculate the probability that:
(i) she scores either 5 or 0 with her first throw:
[tex]\[ P(\text{either 5 or 0 on first throw}) = \][/tex]

(ii) she scores 0 with her first throw and 5 with her second throw:
[tex]\[ P(\text{0 on first throw and 5 on second throw}) = \][/tex]



Answer :

GinnyAnswer
Certainly! Let's go through each part of the question step by step.

### (a) Finding the value of [tex]\( x \)[/tex]

Given the probabilities of scoring different points:
- Probability of scoring 10 points = 0.2
- Probability of scoring 5 points = 0.3
- Probability of scoring 0 points = 0.45

The sum of all probabilities must equal 1. Therefore, we have:
[tex]\[ x + 0.2 + 0.3 + 0.45 = 1 \][/tex]

We can solve for [tex]\( x \)[/tex]:
[tex]\[ x = 1 - (0.2 + 0.3 + 0.45) \][/tex]
[tex]\[ x = 1 - 0.95 \][/tex]
[tex]\[ x = 0.05 \][/tex]

Thus, the value of [tex]\( x \)[/tex] is [tex]\( 0.05 \)[/tex].

### (b) Expected number of times she scores 5 points in 50 throws

The probability of scoring 5 points is given as 0.3. If Kiah throws a coin 50 times, the expected number of times she scores 5 points is calculated by multiplying the number of throws by the probability of scoring 5 points:
[tex]\[ \text{Expected number of 5-point scores} = 50 \times 0.3 = 15 \][/tex]

Thus, Kiah is expected to score 5 points [tex]\( 15 \)[/tex] times in 50 throws.

### (c) Probability calculations for two throws

#### (i) Probability that she scores either 5 or 0 with her first throw

The probability of scoring 5 points is 0.3, and the probability of scoring 0 points is 0.45. To find the probability that she scores either 5 or 0 with her first throw, we add these probabilities together:
[tex]\[ \text{Probability of scoring 5 or 0} = 0.3 + 0.45 = 0.75 \][/tex]

Thus, the probability that she scores either 5 or 0 with her first throw is [tex]\( 0.75 \)[/tex].

#### (ii) Probability that she scores 0 with her first throw and 5 with her second throw

The probability of scoring 0 points with the first throw is 0.45, and the probability of scoring 5 points with the second throw is 0.3. Since these events are independent, the combined probability is the product of the two probabilities:
[tex]\[ \text{Probability of scoring 0 first and 5 second} = 0.45 \times 0.3 = 0.135 \][/tex]

Thus, the probability that she scores 0 with her first throw and 5 with her second throw is [tex]\( 0.135 \)[/tex].

In summary:

(a) [tex]\( x = 0.05 \)[/tex]

(b) The expected number of times Kiah scores 5 points in 50 throws is 15.

(c) (i) The probability that she scores either 5 or 0 with her first throw is [tex]\( 0.75 \)[/tex].

(c) (ii) The probability that she scores 0 with her first throw and 5 with her second throw is [tex]\( 0.135 \)[/tex].

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