Probability - Definitions

Random Experiment

A random experiment (for brevity, we will simply call them experiment instead) is an experiment or a process for which the exact result of this experiment cannot be predicted.

မည်သည့်ရလဒ် ရမည်ကို တိကျစွာ ကြိုတင်ခန့်မှန်းနိုင်ခြင်း မရှိသော စမ်းသပ်ချက်ကို ကျပမ်း စမ်းသပ်ချက် (random experiment) ဟုခေါ်သည်။

•The tossing of a coin - ပုံမှန်ဖြစ်သော (ဘက်မလိုက်သော) အကြွေစေ့ကို ပစ်မြှောက်ခြင်း

• rolling a die ပုံမှန်ဖြစ်သော (ဘက်မလိုက်သော) အံစာတုံးခေါက်ခြင်း

Outcome

A single specific result of an experiment is called an outcome.

စမ်းသပ်ချက်တစ်ခု၏ ရလဒ် တစ်ခုချင်းစီကို Outcom ဟုခေါ်သည်။

၆ မျက်နှာရှိသော အံစာတုံးတစ်ခုကို လှိမ့်လိုက်သည်ဆိုပါစို့။ 1 မှာ 6 အထိ မည်သည့်ဂဏန်းမဆို ကျနိုင်သည်။ အဆိုပါ 1 မှာ 6 အထိ ကျရောက်နိုင်သော ဂဏန်းတစ်ခုချင်းစီကို outcome ဟုခေါ်သည်

Sample Space

The set of all possible outcomes of an experiment is called the sample space.

စမ်းသပ်မှုတစ်ခု၏ ဖြစ်နိုင်သော ရလဒ်များအားလုံးပါဝင်သော အစုကို sample space ခေါ်သည်။ Sample Space ကို S ဖြင့် သတ်မှတ်သည်။

ဥပမာ အံစာတုံးခေါက်သော စမ်းသပ်ချက်၏ sample space မှာ $ \displaystyle S=\left\{ {1,2,3,4,5,6} \right\}$ ဖြစ်သည်။

Event

An event is a subset of a sample space. Sample Space ထဲရှိ လိုချင်သော outcomes များပါသော အစိပ်အပိုင်း (sample space ၏ အစုပိုင်း) ကို event ဟုခေါ်သည်။

ဥပမာ အံစာတုံးခေါက်ခြင်းတွင် sample space မှာ $ \displaystyle S=\left\{ {1,2,3,4,5,6} \right\}$ ဖြစ်သည်။ အံစာတုံးခေါက်သူက စုံကိန်းများသာ ကျလိုသည်ဆိုပါစို့။ ထိုအခါ event မှာ $ \displaystyle E=\left\{ {2,4,6} \right\}$ ဖြစ်မည်။

Probability

Let $S$ be a finite sample space for an experiment such that all outcomes are equally likely which means that they are random and have an equal likelihood of occurrence. Then the probability of an event $E$, denoted by $𝑃 (E)$, in the sample space $S$, is defined by
$ \begin{array}{|c|} \hline P(E)=\displaystyle \frac{{\text{number of outcomes in event }E}}{{\text{number of outcomes in sample space }S}}\\ \hline \end{array}$
In symbols,
$ \begin{array}{|c|} \hline P(E)=\displaystyle \frac{n(E)}{n(S)}\\ \hline \end{array}$
where $n(E)$ denotes the number of elements in the event $E$ and $n(S)$ denotes the number of outcomes in the sample space $S$.

လိုချင်သော ဖြစ်ရပ် တွင်ပါဝင်သော အစုဝင်အရေအတွက် (events ၏ အစုဝင်အရေအတွက်) နှင့် ဖြစ်နိုင်သော အစုဝင်အရေအတွက်စုစုပေါင်း (sample space ၏ အစုဝင်အရေအတွက်) တို့၏ အချိုးကို event တစ်ခု၏ probability (ဖြစ်တန်စွမ်း) ဟုခေါ်သည်။

Note

• For any event $E$, $P(E)$ is a real number such that $0 \le P(E) \le 1$.
• $P(\varnothing )=0$ and $P(S)=1$ (This means that the probability of an impossible event is 0 and that of a sure event is 1 .)
• For any event $E$, $P(E) + P(\text{not}\ E) = 1$.

$\begin{array}{l} S = \text{sample space}\\ E= \text{required event}\\ E'= \text{not}\ E\\ P(E) + P(\text{not}\ E) = 1\ (\text{or})\ P(E) + P(E') = P(S) \end{array}$

Independent Events

Two events are independent if the occurrence of any one of them does not affect the probability of the other.

ဖြစ်ရပ်တစ်ခု (event) ဖြစ်ခြင်း၊ မဖြစ်ခြင်းသည် အခြားဖြစ်ရပ်တစ်ခု ဖြစ်ခြင်း၊ မဖြစ်ခြင်းနှင့် ပတ်သက်မှု၊ သက်ဆိုင်မှုမရှိလျှင် ၎င်းဖြစ်ရပ်နှစ်ခု ကို independent events ဟုခေါ်သည်။ independent events များသည် တပြိုင်နက်ဖြစ်နိုင်သည်။

$E_1$ နှင့် $E_2$ သည် independent events များဖြစ်ကြလျှင် အဆိုပါ event နှစ်ခု တပြိုင်နက်ဖြစ်ရန် probability မှာ

$ \begin{array}{|c|} \hline P(E_1\ \text{and}\ E_2)= P(E_1)\times P(E_2) \\ \hline \end{array}$
ဖြစ်သည်။

ဥပမာ အံစာတုံးနှစ်တုံးကို တပြိုင်နက်ခေါက်သည် ဆိုပါစို့။ ပထမအံစာတုံး 6 ကျခြင်းမကျခြင်းသည် ဒုတိယအံစာတုံး 6 ကျခြင်းမကျခြင်းနှင့် သက်ဆိုင်မှုမရှိပေ။ ထိုကြောင့် အံစာတုံးနှစ်ခုလုံးတွင် 6 ကျ နိုင်သည်။ အံစာတုံးနှစ်ခုလုံးတွင် 6 ကျရန် probability ကို အောက်ပါအတိုင်း တွက်ယူနိုင်သည်။

$ \begin{array}{l} E_1= \text{6 appears on first die}\\ E_2= \text{6 appears on first die}\\ \therefore P(E_1)=P(E_2)=\displaystyle \frac{1}{6}\\ P(E_1\ \text{and}\ E_2)= P(E_1)\times P(E_2)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{6}\times \displaystyle \frac{1}{6}=\displaystyle \frac{1}{36} \\ \end{array}$

Dependent Events

Two events are dependent when the outcome of the first event influences the outcome of the second event.

ဖြစ်ရပ်တစ်ခု (event) ဖြစ်ခြင်း၊ မဖြစ်ခြင်းသည် အခြားဖြစ်ရပ်တစ်ခု ဖြစ်ခြင်း၊ မဖြစ်ခြင်းနှင့် ပတ်သက်မှု၊ သက်ဆိုင်မှုရှိပြီး ၎င်းဖြစ်ရပ်နှစ်ခု တပြိုင်နှက် (သို့မဟုတ်) ဆက်တိုက် ဖြစ်နိုင်လျှင် ၎င်းတို့ကို dependent events ဟုခေါ်သည်။ dependent events များသည်လည်း တပြိုင်နှက်ဖြစ်နိုင်သည်။

$E_1$ နှင့် $E_2$ သည် dependent events များဖြစ်ကြလျှင် အဆိုပါ event နှစ်ခု တပြိုင်နက်ဖြစ်ရန် probability မှာ

$ \begin{array}{|c|} \hline P(E_1\ \text{and}\ E_2)= P(E_1)\times P(E_2\ \text{after}\ E_1) \\ \hline \end{array}$
ဖြစ်သည်။

ဥပမာ ပုံးတစ်ပုံးထဲတွင် အနီရောင်ဂေါ်လီ 5 လုံးနှင့် အနက်ရောင်‌ဂေါ်လီ 5 လုံး ရှိသည်ဆိုပါစို့။ ထိုပုံးထဲမှ ဂေါ်လီနှစ်လုံးကို တပြိုင်နက် ယူလိုက်လျှင် ရရှိလာသော ဂေါ်လီနှစ်လုံး နှစ်ခုလုံး အနီရောင် ဖြစ်နိုင်သည်။ သို့ရာတွင် ဒုတိယဂေါ်လီလုံး အနီရောင်ဖြစ်ရန် probability သည် ပထမဂေါ်လီလုံး အနီရောင်ဖြစ်သော probability အပေါ်တွင် ပတ်သက်မှုရှိနေသည်။ ဂေါ်လီနှစ်ခုလုံး အနီရောင်ဖြစ်ရန် probability ကို အောက်ပါအတိုင်း တွက်ယူနိုင်သည်။

$ \begin{array}{l} \text{For choosing first marble},\\ S=\left\{\text{5 red marbles and 5 black marbles}\right\}\\ n(S)=10\\ E_1= \left\{\text{red marbles in a box}\right\}\\ n(E_1)=5\\ P(E_1)=\displaystyle \frac{5}{10}=\frac{1}{2}\\ \text{For choosing second marble},\\ S=\left\{\text{4 red marbles and 5 black marbles}\right\}\\ n(S)=9\\ E_2= \left\{\text{red marbles in a box after choosing first marble}\right\}\\ n(E_2)=4\\ P(E_2)=\displaystyle \frac{4}{9}\\ \therefore P(E_1\ \text{and}\ E_2)= P(E_1)\times P(E_2)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{2}\times \displaystyle \frac{4}{9}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{2}{9} \\ \end{array}$

မှတ်ချက်။ ။အကယ်၍ ပထမဂေါ်လီယူ အရောင်မှတ်သားပြီး ပြန်ထည့်၊ ထိုနောက်မှ ဒုတိယဂေါ်လီကို ယူလျှင် $E_1$ နှင့် $E_2$ မှာ independent events များ ဖြစ်သွားမည်။

Mutually Exclusive Events

Two events $E_1$ and $E_2$ are mutually exclusive if they cannot occur jointly, that is, they do not have common outcomes.

ဖြစ်ရပ်တစ်ခု ဖြစ်သည့်အခါ အခြားဖြစ်ရပ်တစ်ခု လုံး၀မဖြစ်နိုင်တော့လျှင် အဆိုပါဖြစ်ရပ်နှစ်ခုကို mutually exclusive events ဟုခေါ်သည်။ $E_1$ နှင့် $E_2$ သည် mutually exclusive events များဖြစ်‌ကြလျှင် $E_1$ သို့မဟုတ် $E_2$ တစ်ခုသာ ဖြစ်နိုင်သည်။

$E_1$ နှင့် $E_2$ သည် mutually exclusive events များဖြစ်ကြလျှင် အဆိုပါ event နှစ်ခု တစ်ခုမဟုတ် တစ်ခုဖြစ်ရန် probability မှာ

$ \begin{array}{|c|} \hline P(E_1\ \text{or}\ E_2)= P(E_1) + P(E_2) \\ \hline \end{array}$
ဖြစ်သည်။

ဥပမာ အံစာတုံးတစ်ခု ခေါက်သည် ဆိုပါစို့။ sample space မှာ $S=\left\{1,2,3,4,5,6\right\}$ ဖြစ်ပြီး outcomes တစ်ခုချင်းစီသည် mutually exclusive ဖြစ်ကြသည်။ အကယ်၍ အံစာခေါက်သူသည် 1 သို့မဟုတ် 6 ကျရန်အလိုရှိလျှင် အဆိုပါ ဖြစ်ရပ်အတွက် probability ကို အောက်ပါအတိုင်း တွက်ယူနိုင်သည်။

$ \begin{array}{l} S=\left\{1,2,3,4,5,6\right\}\\ E_1= \left\{1\right\}\\ E_2= \left\{6\right\}\\ \therefore P(E_1)=P(E_2)=\displaystyle \frac{1}{6}\\ P(E_1\ \text{or}\ E_2)= P(E_1) + P(E_2)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{6}+\displaystyle \frac{1}{6}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{2}{6}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{3} \\ \end{array}$

Mutually Inclusive Events

Two events $E_1$ and $E_2$ are mutually inclusive if they can happen at the same time.

$E_1$ နှင့် $E_2$ သည် mutually inclusive events များဖြစ်‌ကြလျှင် အဆိုပါဖြစ်ရပ်နှစ်ခုသည် တစ်ခုမဟုတ် တစ်ခုဖြစ်နိုင်သလို တပြိုင်နက်လည်း ဖြစ်နိုင်သည်။

$E_1$ နှင့် $E_2$ သည် mutually inclusive events များဖြစ်ကြလျှင် အဆိုပါ event နှစ်ခု တစ်ခုမဟုတ် တစ်ခုဖြစ်ရန် probability မှာ

$ \begin{array}{|c|} \hline P(E_1\ \text{or}\ E_2)= P(E_1) + P(E_2) - P(E_1\ \text{and}\ E_2)\\ \hline \end{array}$
ဖြစ်သည်။

ဥပမာ အံစာတုံးတစ်ခု ခေါက်သည် ဆိုပါစို့။ sample space မှာ $S=\left\{1,2,3,4,5,6\right\}$ ဖြစ်သည်။ အံစာခေါက်သူသည် စုံကိန်း သို့မဟုတ် 6 ၏ဆခွဲကိန်းများ ကျရန်အလိုရှိသည်ဟုယူဆမည်။ စုံကိန်းအတွက် event $(E_1)$မှာ $\left\{2,4,6\right\}$ ဖြစ်ပြီး 6 ၏ဆခွဲကိန်းအတွက် event $(E_2)$ မှာ $\left\{1,2,3,6\right\}$ ဖြစ်သည်။ ဖြစ်ရပ်နှစ်ခုတွင် ထပ်နေသော outcomes ပါသောကြောင့် $E_1$ သို့မဟုတ် $E_2$ ဖြစ်ရန် probability ကို အောက်ပါအတိုင်း တွက်ယူနိုင်သည်။

$ \begin{array}{l} S=\left\{1,2,3,4,5,6\right\}\\ E_1= \left\{\text{even numbers}\right\}=\left\{2,4,6\right\}\\ E_2= \left\{\text{factors of 6}\right\}=\left\{1,2,3,6\right\}\\ \therefore P(E_1)=\displaystyle \frac{3}{6}\\ \ \ \ \ P(E_2)=\displaystyle \frac{4}{6}\\ \ \ \ \ P(E_1\ \text{and}\ E_2)=P(E_1) \times P(E_2)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \displaystyle \frac{3}{6}\times \displaystyle\frac{4}{6}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \displaystyle\frac{1}{3} \\ \ \ \ \ P(E_1\ \text{or}\ E_2)= P(E_1) + P(E_2) - P(E_1\ \text{and}\ E_2)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{3}{6}+\displaystyle \frac{4}{6}-\displaystyle \frac{1}{3}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{5}{6} \end{array}$

Probability : Exercise (7.2) - Solutions

Independent Events

Two events are independent if the occurrence of any one of them does not affect the probability of the other.

ဖြစ်ရပ်တစ်ခု ဖြစ်ခြင်း၊ မဖြစ်ခြင်းသည် အခြားဖြစ်ရပ်တစ်ခု ဖြစ်ခြင်း၊ မဖြစ်ခြင်းနှင့် သက်ဆိုင်မှုမရှိပါက ၎င်းတို့ကို လွတ်လပ်သော အမှီခိုမဲ့ ဖြစ်ရပ်များ (independent events) ဟု ခေါ်သည်။

Note : Independent Events နှစ်ခုသည် တပြိုင်နက် ဖြစ်ပေါ်လာနိုင်သည်။

Multiplication Rule

$A$ and $B$ are independent events in the sample space if and only if $\mathrm{P}(A\ \text{and}\ B) = \mathrm{P}(A)\times\mathrm{P}(B)$.

Addition Rule 1

For any two events $A$ and $B$ in a sample space,

$\begin{array}{l}\text{P}(A\;\text{or}\;B)=\text{P}(A)+\text{P}(B)-\text{P}(A\;\text{and}B)\\\\\text{P}(A\cup B)=\text{P}(A)+\text{P}(B)-\text{P}(A\cap B)\end{array}$

တပြိုင်နက်ဖြစ်နိုင်သော ဖြစ်ရပ်နှစ်ခု တစ်ခုမဟုတ်တစ်ခုဖြစ်ရန် ဖြစ်တန်စွမ်းကို တွက်ယူလိုလျှင် ထိုဖြစ်ရပ်နှစ်ခု၏ တစ်ခုစီဖြစ်နိုင်သော ဖြစ်တန်စွမ်းများ ပေါင်းလဒ်မှ ဖြစ်ရပ်နှစ်ခုလုံး တပြိုင်နက်ဖြစ်နိုင်သော ဖြစ်တန်စွမ်းကို နုတ်ရသည်။

Mutually Exclusive Events

Two events $A$ and $B$ are mutually exclusive if they cannot occur jointly, that is, they do not have common outcomes. If $A$ and $B$ are mutually exclusive events, then P($A$ and $B$) = 0.

Addition Rule 2

If $A$ and $B$ are mutually exclusive events in a sample space, then

$\begin{array}{l} \text{P}(A\;\text{or}\;B)=\text{P}(A)+\text{P}(B)\\\\ \text{P}(A\cup B)=\text{P}(A)+\text{P}(B)\end{array}$

1.           At a conference, there are $7$ mathematics instructors, $5$ computer science instructors, $3$ statistics instructors and $4$ science instructors. If an instructor is selected, find the probability of getting a science instructor or a math instructor.

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$\begin{array}{|l|c|} \hline \text { mathematics instructor} & 7 \\ \hline \text { computer science instructor} & 5 \\ \hline \text { stastics instructor} & 3 \\ \hline \text { science instructor} & 4 \\ \hline \text { Total } & 19 \\ \hline \end{array}$

$\therefore\ P(\text { a science instructor }) = \displaystyle\frac{4}{19}\\ \ \ \ \ P(\text { aa mathematics instructor }) = \displaystyle\frac{7}{19}$
$\begin{aligned} & \ \ \ \ \ \ P(\text { a science instructor or a math instructor })\\\\ &=P(\text { a science instructor })+P(\text { a mathematics instructor })\\\\ &=\displaystyle \frac{4}{19}+\displaystyle \frac{7}{19}\\\\ &=\displaystyle \frac{11}{19} \end{aligned}$

2.           Two dices are rolled. Find the probability of getting

(a) a sum greater than $8$ or a sum less than $3$.

(b) a product greater than $9$ or a product less than $16$.

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$\begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\ \ \ \ \ \ \ \ \ \ \ {{\text{2}}^{{\text{nd}}}}\ \text{die}\\ \ {{\text{1}}^{{\text{st}}}}\ \text{die}\ \ \ \begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & {(1,1)} & {(1,2)} & {(1,3)} & {(1,4)} & {(1,5)} & {(1,6)} \\ \hline 2 & {(2,1)} & {(2,2)} & {(2,3)} & {(2,4)} & {(2,5)} & {(2,6)} \\ \hline 3 & {(3,1)} & {(3,2)} & {(3,3)} & {(3,4)} & {(3,5)} & {(3,6)} \\ \hline 4 & {(4,1)} & {(4,2)} & {(4,3)} & {(4,4)} & {(4,5)} & {(4,6)} \\ \hline 5 & {(5,1)} & {(5,2)} & {(5,3)} & {(5,4)} & {(5,5)} & {(5,6)} \\ \hline 6 & {(6,1)} & {(6,2)} & {(6,3)} & {(6,4)} & {(6,5)} & {(6,6)} \\ \hline\end{array}\end{array}$

The sample space consists of $36$ outcomes.

Let $E_1 = $ the sum greater than $8$.

$\therefore\ E_1=\{(3,6),(4,5),(4,6),(5,4),(5,5),\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5,6), (6,3), (6,4), (6,5), (6,6) \},\\\\ \therefore\ \mathrm{n}(E_1)=10\ \text{and}\ \mathrm{P}(E_1)=\displaystyle\frac{10}{36} $
Let $E_2 = $the sum less than $3$.

$\therefore\ E_2=\{ (1,1) \},\\\\ \therefore\ \mathrm{n}(E_2)=1\ \text{and}\ \mathrm{P}(E_2)=\displaystyle\frac{1}{36} $
Let $E_3=$ the product greater than $9$.

$\therefore\ E_3=\{(2,5), (2,6), (3,4), (3,5), (3,6), (4,3), \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4,4), (4,5), (4,6), (5,2), (5,3), (5,4), (5,5), \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5,6), (6,2), (6,3), (6,4), (6,5), (6,6) \},\\\\ \therefore\ \mathrm{n}(E_3)=19\ \text{and}\ \mathrm{P}(E_3)=\displaystyle\frac{19}{36} $

Let $E_4=$ the product less than $16$ .

$\therefore\ E_4=\{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2,1), (2,2), (2,3), (2,4), (2,5), (2,6),\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3,1), (3,2), (3,3), (3,4), (3,5),(4,1), \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4,2), (4,3), (5,1), (5,2), (5,3), (6,1), (6,2)\},\\\\ \therefore\ \mathrm{n}(E_4)=25\ \text{and}\ \mathrm{P}(E_4)=\displaystyle\frac{25}{36} $

Let $E_5= 9 < \text{the product} < 16$ .

$\therefore\ E_5=\{(2,5), (2,6),(3,4), (3,5),(4,3), (5,2), (5,3), (6,2)\},\\\\ \therefore\ \mathrm{n}(E_5)=8\ \text{and}\ \mathrm{P}(E_5)=\displaystyle\frac{8}{36} $

(a)    P (a sum greater than $8$ or a sum less than $3$)

$\ \ \ \ \ \ \ =\mathrm{P}(E_1) +\mathrm{P} (E_2 ) =\displaystyle\frac{10}{36}+\displaystyle\frac{1}{36}= \displaystyle\frac{11}{36}$

(b)    P (a product greater than $9$ or a product less than $16$)

$\ \ \ \ \ \ \ =\mathrm{P}(E_3) +\mathrm{P} (E_4 )- \mathrm{P} (E_5 )=\displaystyle\frac{19}{36}+\displaystyle\frac{25}{36}-\displaystyle\frac{8}{36}= \displaystyle\frac{36}{36}=1$

3.           In a survey about a change in public policy, 100 people were asked if they favor the change, oppose the change, or have no opinion about the change. The responses are indicated as below:

$$\begin{array}{|l|c|c|c|} \hline & \text{The} & \text{Senior} & \text{Total}\\ & \text{Youth} & \text{Citizens} & \\ \hline \text{Favor} & 18 & 9 & 27 \\ \hline \text{Oppose } & 12 & 25 & 37 \\ \hline \text{No opinion} & 20 & 16 & 36 \\ \hline \text{Total} & 50 & 50 & 100 \\ \hline \end{array}$$

Find the probability that a randomly selected respondent to this survey oppose or has no opinion about the change policy.

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\begin{aligned} & \ \ \ \ \ P(\text { oppose or no opinion })\\\\ &= P(\text { oppose })+P(\text { no opinion }) \\\\ &=\frac{37}{100}+\frac{36}{100} \\\\ &=\frac{73}{100} \end{aligned}

4.           A bag contains 15 discs of which 3 are white, 5 are red and 7 are blue. Two discs are to be drawn at random, in succession, each being replaced after its colour has been noted. Calculate the probability that the two discs will be of the same colour.

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$\begin{array}{l}\text{The sample space consists of 15 outcomes}\text{.}\\\\\text{Let}\ W\ \text{denote white discs}\text{.}\\\\\therefore n(W)=3\ \text{and}\ P(W)=\displaystyle \frac{3}{{15}}\ \ \ \ \\\\\text{Let}\ R\ \text{denote red discs}\text{.}\\\\\therefore n(R)=5\ \text{and}\ P(R)=\displaystyle \frac{5}{{15}}\ \\\\\text{Let}\ B\ \text{denote blue discs}\text{.}\\\\\therefore n(B)=7\ \text{and}\ P(B)=\displaystyle \frac{7}{{15}}\ \\\\\ \ \ \ P(\text{two discs}\ \text{of the same color})\\\\=P\left[ {(W,W)\ \text{or}\ (R,R)\ \text{or}\ (B,B)\ } \right]\\\\=P(W,W)+P(R,R)\ +(B,B)\\\\=\left( {\displaystyle \frac{3}{{15}}\ \times \displaystyle \frac{3}{{15}}} \right)\ +\left( {\displaystyle \frac{5}{{15}}\times \displaystyle \frac{5}{{15}}} \right)+\left( {\displaystyle \frac{7}{{15}}\times \displaystyle \frac{7}{{15}}} \right)\\\\=\displaystyle \frac{9}{{225}}+\displaystyle \frac{{25}}{{225}}+\displaystyle \frac{{49}}{{225}}\\\\=\displaystyle \frac{{83}}{{225}}\end{array}$

5.           The probabilities that the student $A$ and $B$ pass an examination are $\displaystyle\frac{2}{3}$ and $\displaystyle\frac{3}{4}$ respectively. Find the probabilities that:

(a) both $A$ and $B$ pass the examination.

(b) exactly one of $A$ and $B$ passes the examination.

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$\begin{array}{l}\ \ \ \ \ P(A\ \text{will pass the exam})=\displaystyle \frac{2}{3}\\\\\therefore \ P(A\ \text{will fail the exam})=1-\displaystyle \frac{2}{3}=\displaystyle \frac{1}{3}\\\\\ \ \ \ \ P(B\ \text{will pass the exam})=\displaystyle \frac{3}{4}\\\\\therefore \ P(A\ \text{will fail the exam})=1-\displaystyle \frac{3}{4}=\displaystyle \frac{1}{4}\\\\(\text{a)}\ \ \ \ P(\text{both}\ A\ \text{and }B\ \text{pass the exam})\\\\\ \ \ =P(A\ \text{will pass the exam})\times P(B\ \text{will pass the exam})\\\\\ \ \ =\displaystyle \frac{2}{3}\times \displaystyle \frac{3}{4}\\\\\ \ \ =\displaystyle \frac{1}{2}\\\\(\text{a)}\ \ \ \ P(\text{exactly one of }A\ \text{and }B\ \text{passes the examination})\\\\\ \ \ =P(A\ \text{will pass and }B\text{ will fail})+P(B\ \text{will pass and }A\text{ will fail})\\\\\ \ \ =\left( {\displaystyle \frac{2}{3}\times \displaystyle \frac{1}{4}} \right)+\left( {\displaystyle \frac{3}{4}\times \displaystyle \frac{1}{3}} \right)\\\\\ \ \ =\displaystyle \frac{2}{{12}}+\displaystyle \frac{3}{{12}}\\\\\ \ \ =\displaystyle \frac{5}{{12}}\end{array}$

6.           Three groups of children consist of 3 boys and 1 girl, 2 boys and 2 girls, and 1 boy and 3 girls respectively. If a child is chosen from each group, find the probability that 1 boy and 2 girls are chosen.

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$$\begin{array}{|c|c|c|c|} \hline \text { Group } & \text { Boy } & \text { Girl } & \text { Total } \\ \hline \text { I } & 3 & 1 & 4 \\ \hline \text { II } & 2 & 2 & 4 \\ \hline \text { III } & 1 & 3 & 4 \\ \hline \end{array}$$

$\begin{array}{l}\text{Let the boy from group I, group II}\\\text{and group III be }{{B}_{1}}\text{,}\ {{B}_{2}}\text{ and }{{B}_{3}}\ \text{respectively}\text{.}\\\\\text{Let the girl from group I, group II}\\\text{and group III be }{{G}_{1}}\text{,}\ {{G}_{2}}\text{ and }{{G}_{3}}\ \text{respectively}\text{.}\\\\\therefore \ \ \ \ \ P(1\ \text{boy and 2 girls})\\\\\ \ \ =P({{B}_{1}}{{G}_{2}}{{G}_{3}}\ \text{or}\ {{B}_{2}}{{G}_{1}}{{G}_{3}}\ \text{or}\ {{B}_{3}}{{G}_{1}}{{G}_{2}})\\\\\ \ \ =P({{B}_{1}}{{G}_{2}}{{G}_{3}})+P({{B}_{2}}{{G}_{1}}{{G}_{3}})+P({{B}_{3}}{{G}_{1}}{{G}_{2}})\\\\\ \ \ =\left( {\displaystyle \frac{3}{4}\times \displaystyle \frac{2}{4}\times \displaystyle \frac{3}{4}} \right)+\left( {\displaystyle \frac{2}{4}\times \displaystyle \frac{1}{4}\times \displaystyle \frac{3}{4}} \right)+\left( {\displaystyle \frac{1}{4}\times \displaystyle \frac{1}{4}\times \displaystyle \frac{2}{4}} \right)\\\\\ \ \ =\displaystyle \frac{{18}}{{64}}+\displaystyle \frac{6}{{64}}+\displaystyle \frac{2}{{64}}\\\\\ \ \ =\displaystyle \frac{{26}}{{64}}\\\\\ \ \ =\displaystyle \frac{{13}}{{32}}\end{array}$

Probability : Problems using Tree Diagrams and Tables

1.     A family has $ \displaystyle 4$ children. Draw a tree diagram to list all possible outcomes. If each outcomes is equally likely to occur, find the probability that the last two children are girls. Find also the probability that exactly two children are boys.

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           Let Boy be denoted by $ \displaystyle B$ and Girl be by $ \displaystyle G.$

 

$ \displaystyle \therefore\ \ $Number of possible outcomes $ \displaystyle =16$

      Set of favourable outcomes for the last two children are girls

$ \displaystyle \ \ \ \ \ \ \ \ = \{(B,B,G,G),(B,G,G,G),(G,B,G,G),(G,G,G,G)\} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle =4$

$ \displaystyle \therefore\ \ P$( the last two children are girls) $ \displaystyle = \frac{4}{16}= \frac{1}{4}$


      Set of favourable outcomes for exactly two children are boys

$ \displaystyle \ \ \ \ \ \ \ \ = \{(B,B,G,G),(B,G,B,G),(B,G,G,B),(G,B,B,G),(G,B,G,B),(G,G,B,B)\} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle =6$

$ \displaystyle \therefore\ \ P$( the last two children are girls) $ \displaystyle = \frac{6}{16}= \frac{3}{8}$

2.      How many $ \displaystyle 2$-digit numerals can you form from $ \displaystyle 2,3,5,6$ without repeating any digit? Find the probability of a numeral which is divisible by $ \displaystyle 2$. If one of these numerals is chosen at random, find the probability that it is divisible by $ \displaystyle 13.$

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$ \displaystyle \therefore\ \ $Number of possible outcomes $ \displaystyle =12$

      Set of favourable outcomes for a numeral which is divisible by $ \displaystyle 2$

$ \displaystyle \ \ \ \ \ \ \ \ = \{26,32,36,52,56,62\} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle =6$

$ \displaystyle \therefore\ \ P$( a numeral which is divisible by $ \displaystyle 2$) $ \displaystyle = \frac{6}{12}= \frac{1}{2}$


      Set of favourable outcomes for a numeral which is divisible by $ \displaystyle 13$

$ \displaystyle \ \ \ \ \ \ \ \ = \{26,52,65\} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle =3$

$ \displaystyle \therefore\ \ P$( a numeral which is divisible by $ \displaystyle 13$) $ \displaystyle = \frac{3}{12}= \frac{1}{4}$

3.      How many $ \displaystyle 3$-digit numerals can you form from $ \displaystyle 1, 0 , 5$ and $ \displaystyle 6$ without repeating any digit? Find the probability of numeral which is divisible by $ \displaystyle 5.$

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$ \displaystyle \therefore\ \ $Number of possible outcomes $ \displaystyle =18$

      Set of favourable outcomes for a numeral which is divisible by $ \displaystyle 5$

$ \displaystyle \ \ \ \ \ \ \ \ = \{105,150,160,165,510,560,605,610,615,650\} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle =10$

$ \displaystyle \therefore\ \ P$( a numeral which is divisible by $ \displaystyle 5$) $ \displaystyle = \frac{10}{18}= \frac{5}{9}$

4.      A fair coin is tossed three times. Draw a tree diagram to determine the set of all possible outcomes. Hence find the probability of

(a) getting three heads.

(b) getting two heads and one tail in any order.

(c) getting at least one head.

(d) getting no head.

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$ \displaystyle \therefore\ \ $ Set of possible outcomes $ \displaystyle = \{HHH,HHT,HTH, HTT,THH,THT,TTH,TTT\} $

$ \displaystyle \therefore\ \ $Number of possible outcomes $ \displaystyle = 8$

(a)       Set of favourable outcomes for getting three heads $ \displaystyle = \{HHH \}$

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle = 1$

$ \displaystyle \therefore\ \ P$(getting three heads) $ \displaystyle = \frac{1}{8}$

(b)       Set of favourable outcomes for getting two heads and one tail in any order

$ \displaystyle \ \ \ \ \ \ = \{HHT,HTH,THH \}$

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle = 3$

$ \displaystyle \therefore\ \ P$(getting two heads and one tail in any order) $ \displaystyle = \frac{3}{8}$

(c)       Set of favourable outcomes for getting at least one head

$ \displaystyle \ \ \ \ \ \ = \{HHH,HHT,HTH, HTT,THH,THT,TTH \} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle = 7$

$ \displaystyle \therefore\ \ P$(getting at least one head) $ \displaystyle = \frac{7}{8}$


(d)    Set of favourable outcomes for getting no head
$ \displaystyle \ \ \ \ \ \ = \{TTT \}$

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle = 1$

$ \displaystyle \therefore\ \ P$(getting no head) $ \displaystyle = \frac{1}{8}$

5.      Draw a table of all posssible outcomes for throwing two fair dice. Hence, calculate the probability that

(a) the product of the scores on the two dice is divisible by $ \displaystyle 6$ or $ \displaystyle 9.$

(b) both dice showing the same number.

(c) the sum of the score is a multiple of $ \displaystyle 3.$

(d) the total score on the two dice is prime.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\ \ \ \ \ \ \ \ \ \ \ {{\text{2}}^{{\text{nd}}}}\ \text{die}\\ \ {{\text{1}}^{{\text{st}}}}\ \text{die}\ \ \ \begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & {(1,1)} & {(1,2)} & {(1,3)} & {(1,4)} & {(1,5)} & {(1,6)} \\ \hline 2 & {(2,1)} & {(2,2)} & {(2,3)} & {(2,4)} & {(2,5)} & {(2,6)} \\ \hline 3 & {(3,1)} & {(3,2)} & {(3,3)} & {(3,4)} & {(3,5)} & {(3,6)} \\ \hline 4 & {(4,1)} & {(4,2)} & {(4,3)} & {(4,4)} & {(4,5)} & {(4,6)} \\ \hline 5 & {(5,1)} & {(5,2)} & {(5,3)} & {(5,4)} & {(5,5)} & {(5,6)} \\ \hline 6 & {(6,1)} & {(6,2)} & {(6,3)} & {(6,4)} & {(6,5)} & {(6,6)} \\ \hline\end{array}\end{array}$

$ \displaystyle \therefore\ \ $Number of possible outcomes $ \displaystyle =36$

(a)     Set of favourable outcomes for the product of the scores on the two dice is divisible by $ \displaystyle 6$ or $ \displaystyle 9$

$ \displaystyle \ \ \ \ \ \ \ \ = \{(1,6),(2,3), (2,6), (3,2), (3,3), (3, 4), (3, 6),(4,3), \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4, 6), (5,6), (6,1), (6,2), (6, 3), (6, 4), (6, 5), (6, 6)\} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle =16$

$ \displaystyle \therefore\ \ P$(the product of the scores on the two dice is divisible by $ \displaystyle 6$ or $ \displaystyle 9$) $ \displaystyle = \frac{16}{36}= \frac{4}{9}$

(b)     Set of favourable outcomes for both dice showing the same number

$ \displaystyle \ \ \ \ \ \ \ \ = \{(1,1),(2,2), (3,3), (4,4), (5,5), (6,6) \} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle = 6$

$ \displaystyle \therefore\ \ P$(both dice showing the same number) $ \displaystyle = \frac{6}{36}= \frac{1}{6}$

(c)     Set of favourable outcomes for the sum of the score is a multiple of $ \displaystyle 3$

$ \displaystyle \ \ \ \ \ \ \ \ = \{(1,2),(1,5), (2,1), (2,4), (3,3),(3,6),\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4,2), (4,5), (5,1), (5,4), (6,3), (6,6) \} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle = 12$

$ \displaystyle \therefore\ \ P$(both dice showing the same number) $ \displaystyle = \frac{12}{36}= \frac{1}{3}$


(d)     Set of favourable outcomes for the total score on the two dice is prime

$ \displaystyle \ \ \ \ \ \ \ \ = \{(1,1),(1,2), (1,4),(1,6),(2,1), (2,3), (2,5), \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3,2), (3,4), (4,1), (4,3), (5,2), (5,6), (6,1), (6,5) \} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle = 15$

$ \displaystyle \therefore\ \ P$(both dice showing the same number) $ \displaystyle = \frac{15}{36}= \frac{5}{12}$