**Question 1**:

In a school examination, 2 students out of 5 students failed Chemistry.

**(a)**

**If 6 students are chosen at random, find the probability that not more than 2 students failed Chemistry.**

**(b)**If there are 200 Form 4 students in that school, find the mean and standard deviation of the number of students who failed Chemistry.

*Solution:***(a)**

$\begin{array}{l}X-\text{NumberofstudentswhofailedChemistry}\text{.}\\ X~B\left(n,p\right)\\ X~B\left(6,\text{}\frac{2}{5}\right)\\ \\ P(X=r)={}^{n}c{}_{r}.{p}^{r}.{q}^{n-r}\\ P(X\le 2)\\ =P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)\\ ={}^{6}C{}_{0}{\left(\frac{2}{5}\right)}^{0}{\left(\frac{3}{5}\right)}^{6}+{}^{6}C{}_{1}{\left(\frac{2}{5}\right)}^{1}{\left(\frac{3}{5}\right)}^{5}+{}^{6}C{}_{2}{\left(\frac{2}{5}\right)}^{2}{\left(\frac{3}{5}\right)}^{4}\\ =0.0467+0.1866+0.3110\\ =0.5443\end{array}$

**(b)**

$\begin{array}{l}X~B\left(n,p\right)\\ X~B\left(200,\text{}\frac{2}{5}\right)\\ \text{Meanof}X\\ =np=200\times \frac{2}{5}=80\\ \\ \text{Standarddeviationof}X\\ =\sqrt{npq}\\ =\sqrt{200\times \frac{2}{5}\times \frac{3}{5}}\\ =\sqrt{48}\\ =6.93\end{array}$

**Question 2**:

5% of the supply of mangoes received by a supermarket are rotten.

**(a)**If a sample of 12 mangoes is chosen at random, find the probability that at least two mangoes are rotten.

**(b)**

**Find the minimum number of mangoes that have to be chosen so that the probability of obtaining at least one rotten mango is greater than 0.85.**

*Solution:***(a)**

*X*~

*B*(12, 0.05)

1 –

*P*(*X ≤*1)= 1 – [

*P*(*X =*0) +*P*(*X*=*1)]*= 1 – [
$C{}_{0}$
(0.05)

^{0}(0.95)^{12}+ $C{}_{1}$ (0.05)^{1}(0.95)^{11}]= 1 – 0.8816

**= 0.1184**

**(b)**

*P*(

*X ≥*1) > 0.85

1 –

*P*(*X =*0) > 0.85*P*(

*X =*0) < 0.15

$C{}_{0}$
(0.05)

^{0}(0.95)^{n}^{ }< 0.15*n lg*0.95 <

*lg*0.15

*n*> 36.98

**n =****37**

Therefore, the minimum number of mangoes that have to be chosen so that the probability of obtaining at least one rotten mango is greater than 0.85 is

**37**.

**Question 3**:

The result of a study shows that 20% of students failed the Form 5 examination in a school. If 8 students from the school are chosen at random, calculate the probability that

**(a)**exactly 2 of them who failed,

**(b)**less than 3 of them who failed.

*Solution:***(a)**

*p*= 20% = 0.2,

*q*= 1 – 0.2 = 0.8

*X*~

*B*(8, 0.2)

*P*(

*X =*2)

=
$C{}_{2}$
(0.2)

^{2}(0.8)^{6}=

**0.2936**

(b)

(b)

*P*(

*X <*3)

=

*P*(*X =*0) +*P*(*X*=*1) +**P*(*X*=*2)*=
$C{}_{0}$
(0.2)

^{0}(0.8)^{8 }+ $C{}_{1}$ (0.2)^{1}(0.8)^{7}+ $C{}_{2}$ (0.2)^{2}(0.8)^{6}= 0.16777 + 0.33554 + 0.29360

=

**0.79691**

**Question 4**:

In a survey carried out in a particular district, it is found that three out of five families own a LCD television.

If 10 families are chosen at random from the district, calculate the probability that at least 8 families own a LCD television.

*Solution:*$\begin{array}{l}\text{Let}X\text{betherandomvariablerepresentingthenumberoffamilies}\\ \text{whoownaLCDtelevision}\text{.}\\ X~B\left(n,p\right)\\ X~B\left(10,\text{}\frac{3}{5}\right)\\ p=\frac{3}{5}=0.6\\ q=1-0.6=0.4\\ \\ P(X=r)={}^{n}c{}_{r}.{p}^{r}.{q}^{n-r}\\ P\left(\text{atleast8familiesownaLCDtelevision}\right)\\ P(X\ge 8)\\ =P\left(X=8\right)+P\left(X=9\right)+P\left(X=10\right)\\ ={}^{10}C{}_{8}{\left(0.6\right)}^{8}{\left(0.4\right)}^{2}+{}^{10}C{}_{9}{\left(0.6\right)}^{9}{\left(0.4\right)}^{1}+{}^{10}C{}_{10}{\left(0.6\right)}^{10}{\left(0.4\right)}^{0}\\ =0.1209+0.0403+0.0060\\ =0.1672\end{array}$