Sec 4 E Maths: Probability, practice questions & worked solutions

(a)

$$P(\text{walks to school})=1-\frac15$$

$$=\frac45$$

(b)

$$\text{No. of days}=\frac45(190)$$

$$=152$$

Answer: $\text{(a) }\dfrac45\quad\text{(b) }152\text{ days}$

(a)

$$\begin{aligned}&1-\dfrac59-\dfrac2{15}\\&=\dfrac{14}{45}\end{aligned}$$

(b)

$$\begin{aligned}&\textit{3 parts}\to10+8\ \textit{balls}\\&\textit{1 part}\to6\ \textit{balls}\\&\textit{2 parts}\to12\ \textit{balls}\\&n=12-7\\&n=5\ \textit{balls}\end{aligned}$$

OR

$$\begin{aligned}&\dfrac{7+n}{10+7+8+n}=\dfrac25\\&3n=15\\&n=5\end{aligned}$$

OR

$$\begin{aligned}&\text{P(pick red and blue balls)}=\dfrac35\\&\textit{3 parts}\to10+8\ \textit{balls}\\&\textit{1 part}\to6\ \textit{balls}\\&\textit{2 parts}\to12\ \textit{balls}\\&n=12-7\\&n=5\ \textit{balls}\end{aligned}$$

Answer: (a) $\dfrac{14}{45}$ (b) $n=5$

(a)

$$\text{Probability}=\frac59\times\frac48+\frac49\times\frac38=\frac49$$

(b)

$$\text{Probability}=\frac59\times\frac48+\frac49\times\frac58=\frac59$$

Answer: $\text{(a) }\dfrac49\quad\text{(b) }\dfrac59$

(a)

Probability of all good apple

$$\begin{aligned}&=\frac{17}{20}\times\frac{16}{19}\times\frac{15}{18}\\[9pt]&=0.59649\end{aligned}$$

Probability of at least 1 bad

$$\begin{aligned}&=1-0.59649\\[9pt]&=0.404\quad\text{or}\quad\frac{23}{57}\end{aligned}$$

(b)

No. $\dfrac{17}{20}\times\dfrac{16}{19}\times\dfrac3{18}$ refers to the probability of (Good, Good, Bad).

It can also be (Good, Bad, Good) or (Bad, Good, Good).

Probability should be

$$\begin{aligned}&\frac{17}{20}\times\frac{16}{19}\times\frac3{18}+\frac{17}{20}\times\frac3{19}\times\frac{16}{18}+\frac3{20}\times\frac{17}{19}\times\frac{16}{18}\\[9pt]&=3\left(\frac{17}{20}\times\frac{16}{19}\times\frac3{18}\right)\end{aligned}$$

Answer: (a) $\dfrac{23}{57}$ or $0.404$   (b) disagree: the rotten apple can be first, second or third, so the probability is $3\times\dfrac{17}{20}\times\dfrac{16}{19}\times\dfrac3{18}$

(a)

(b)

$P(\text{total is 6})=\dfrac2{16}=\dfrac18$

(c)

$=\dfrac6{64}=\dfrac3{32}$

Answer: (b) $\dfrac18$  (c) $\dfrac3{32}$

(i)

$$\begin{aligned}P(\text{Juline wins both days})&=\frac25\times\frac25\\&=\frac4{25}\end{aligned}$$

(ii)

$$\begin{aligned}P(\text{Juline wins one day only})&=P(J)\times P(\text{Not }J)+P(\text{Not }J)\times P(J)\\&=\left(\frac25\times\frac35\right)\times2\\&=\frac{12}{25}\end{aligned}$$

(iii)

$$\begin{aligned}P(\text{Draw})&=1-\frac25-\frac16\\&=\frac{13}{30}\end{aligned}$$

Answer: (i) $\dfrac4{25}$  (ii) $\dfrac{12}{25}$  (iii) $\dfrac{13}{30}$

(i)

(ii)

P($1 and 50 cents or 50 cents and $1)

$$\begin{aligned}&=\left(\frac6{14}\times\frac8{13}\right)+\left(\frac8{14}\times\frac6{13}\right)\\&=\frac{48}{91}\end{aligned}$$

(iii)

P($1 and 50 cents and 50 cents or 50 cents and $1 and 50 cents or 50 cents and 50 cents and $1)

$$\begin{aligned}&=\left(\frac6{14}\times\frac8{13}\times\frac7{12}\right)\times3\\&=\frac6{13}\end{aligned}$$

Answer: (ii) $\dfrac{48}{91}$ (iii) $\dfrac6{13}$

(a)

(b)(i)

$\left(\dfrac{12}{25}\right)\left(\dfrac{11}{24}\right)=\left(\dfrac{11}{50}\right)$

(b)(ii)

$\left(\dfrac{13}{25}\right)\left(\dfrac{12}{24}\right)\times2=\left(\dfrac{13}{25}\right)$

(c)

$$1-\left(\frac{12}{25}\right)\left(\frac{11}{24}\right)\left(\frac{10}{23}\right)-\left(\frac{13}{25}\right)\left(\frac{12}{24}\right)\left(\frac{11}{23}\right)$$

or
$\mathrm P(\mathrm{BBG})+\mathrm P(\mathrm{BGB})+\mathrm P(\mathrm{GBB})+\mathrm P(\mathrm{GGB})+\mathrm P(\mathrm{GBG})+\mathrm P(\mathrm{BGG})$

$$\begin{aligned}&=\left(\frac{12}{25}\right)\left(\frac{11}{24}\right)\left(\frac{13}{23}\right)\times3+\left(\frac{13}{25}\right)\left(\frac{12}{24}\right)\left(\frac{12}{23}\right)\times3\\&=0.78\text{ or }\left(\frac{39}{50}\right)\end{aligned}$$

Answer: (b)(i) $\dfrac{11}{50}$ (ii) $\dfrac{13}{25}$ (c) $\dfrac{39}{50}$

(a)

$$\begin{aligned}\left(\frac8{8+5+x}\right)\left(\frac7{7+5+x}\right)&=\frac4{33}\\\left(\frac8{13+x}\right)\left(\frac7{12+x}\right)&=\frac4{33}\\33(8)(7)&=4(13+x)(12+x)\\462&=x^2+25x+156\\x^2+25x-306&=0\\(x-9)(x+34)&=0\\x=9\ \ \text{or}\ \ x&=-34\ \text{(rej)}\end{aligned}$$

(b)

P(second marble is green when three marbles are drawn)

$$\begin{aligned}&=\left(\frac{17}{22}\right)\left(\frac5{21}\right)\left(\frac{16}{20}\right)\\&=\frac{34}{231}\end{aligned}$$

Answer: (a) $x=9$ (b) $\dfrac{34}{231}$

(a)

$$\begin{aligned}\text{probability}&=0.7\times0.7\times0.7\\&=0.343\end{aligned}$$

(b)

$$\begin{aligned}\text{probability}&=(0.7\times0.3)+(0.3\times0.8)\\&=0.45\end{aligned}$$

(c)(i)

$$\begin{aligned}\text{probability}&=(0.7\times0.3)+(0.3\times0.2)\\&=0.27\end{aligned}$$

(c)(ii)

$$\begin{aligned}\text{probability}&=1-(0.3\times0.8)\\&=0.76\end{aligned}$$

or

$$\begin{aligned}\text{probability}&=0.27+(0.7\times0.7)\\&=0.76\end{aligned}$$

Answer: (a) $0.343$ (b) $0.45$ (c)(i) $0.27$ (ii) $0.76$