**Question 8**:

Diagram 4 shows a dart’s target board at a dart game booth in a funfair.

**Diagram 4**

The booth offers 3 darts per game. The customers have to pay RM5 to play a game. A toy bear will be given to customers who are able to hit the bullseye for the three darts’ throws in a game. Bob is a dart player. By average, he hits the bullseye 7 times out of 10 darts thrown.

**(a)**Bob would play the game if he had at least 90% chance to win at least one toy bear by spending RM30.

By mathematical calculation, suggest to Bob whether he should play the game or otherwise. [4 marks]

**(b)**What is the minimum number of games that Bob needed so that he can get 4 toy bears? [4 marks]

**Solution**:**(a)**

$\begin{array}{l}X~B\left(3,\text{}0.7\right)\\ P\left(X=r\right)={}^{3}C{}_{r}{\left(0.7\right)}^{r}{\left(0.3\right)}^{3-r}\\ P\left(\text{wonatoybear}\right)=P\left(x=3\right)\\ \text{}={}^{3}C{}_{3}{\left(0.7\right)}^{3}{\left(0.3\right)}^{0}\\ \text{}=0.343\\ \\ Y=\text{NumberofgamesinwhichBobwinsatoybear}\text{.}\\ Y~B\left(n,\text{}0.343\right)\\ \text{NumberofgamesplayedbyBob}=\frac{\text{RM}30}{\text{RM}5}\\ \text{}=6\\ \end{array}$

$\begin{array}{l}P\left(Y\ge 1\right)\ge 0.90\\ 1-P\left(Y=0\right)\ge 0.90\\ P\left(Y=0\right)\le 0.10\\ {}^{n}C{}_{0}{\left(0.343\right)}^{0}{\left(0.657\right)}^{n}\le 0.10\\ {\left(0.657\right)}^{n}\le 0.10\\ n{\mathrm{log}}_{10}\left(0.657\right)\le {\mathrm{log}}_{10}0.10\\ \text{}n\ge 5.481\\ \text{}n=6\\ \end{array}$

By playing 6 games, Bob had at least 90% chance to win at least one toy bear. Hence, Bob should play the game.

**(b)**

*np*≥ 4

*n*(0.343) ≥ 4

*n*≥ 11.66

*n*= 12

Minimum number of games = 12