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.
Show/Hide Solution
$\begin{array}{|l|c|} \hline \text { mathematics instructor} & 7 \\ \hline \text { computer science instructor} & 5 \\ \hline \text { stastics instructor} & 3 \\ \hline \text { science instructor} & 4 \\ \hline \text { Total } & 19 \\ \hline \end{array}$
$\therefore\ P(\text { a science instructor }) = \displaystyle\frac{4}{19}\\ \ \ \ \ P(\text { aa mathematics instructor }) = \displaystyle\frac{7}{19}$
$\begin{aligned} & \ \ \ \ \ \ P(\text { a science instructor or a math instructor })\\\\ &=P(\text { a science instructor })+P(\text { a mathematics instructor })\\\\ &=\displaystyle \frac{4}{19}+\displaystyle \frac{7}{19}\\\\ &=\displaystyle \frac{11}{19} \end{aligned}$
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 $16$.
Show/Hide Solution
$\begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\ \ \ \ \ \ \ \ \ \ \ {{\text{2}}^{{\text{nd}}}}\ \text{die}\\ \ {{\text{1}}^{{\text{st}}}}\ \text{die}\ \ \ \begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & {(1,1)} & {(1,2)} & {(1,3)} & {(1,4)} & {(1,5)} & {(1,6)} \\ \hline 2 & {(2,1)} & {(2,2)} & {(2,3)} & {(2,4)} & {(2,5)} & {(2,6)} \\ \hline 3 & {(3,1)} & {(3,2)} & {(3,3)} & {(3,4)} & {(3,5)} & {(3,6)} \\ \hline 4 & {(4,1)} & {(4,2)} & {(4,3)} & {(4,4)} & {(4,5)} & {(4,6)} \\ \hline 5 & {(5,1)} & {(5,2)} & {(5,3)} & {(5,4)} & {(5,5)} & {(5,6)} \\ \hline 6 & {(6,1)} & {(6,2)} & {(6,3)} & {(6,4)} & {(6,5)} & {(6,6)} \\ \hline\end{array}\end{array}$
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.
Show/Hide Solution
\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.
Show/Hide Solution
$\begin{array}{l}\text{The sample space consists of 15 outcomes}\text{.}\\\\\text{Let}\ W\ \text{denote white discs}\text{.}\\\\\therefore n(W)=3\ \text{and}\ P(W)=\displaystyle \frac{3}{{15}}\ \ \ \ \\\\\text{Let}\ R\ \text{denote red discs}\text{.}\\\\\therefore n(R)=5\ \text{and}\ P(R)=\displaystyle \frac{5}{{15}}\ \\\\\text{Let}\ B\ \text{denote blue discs}\text{.}\\\\\therefore n(B)=7\ \text{and}\ P(B)=\displaystyle \frac{7}{{15}}\ \\\\\ \ \ \ P(\text{two discs}\ \text{of the same color})\\\\=P\left[ {(W,W)\ \text{or}\ (R,R)\ \text{or}\ (B,B)\ } \right]\\\\=P(W,W)+P(R,R)\ +(B,B)\\\\=\left( {\displaystyle \frac{3}{{15}}\ \times \displaystyle \frac{3}{{15}}} \right)\ +\left( {\displaystyle \frac{5}{{15}}\times \displaystyle \frac{5}{{15}}} \right)+\left( {\displaystyle \frac{7}{{15}}\times \displaystyle \frac{7}{{15}}} \right)\\\\=\displaystyle \frac{9}{{225}}+\displaystyle \frac{{25}}{{225}}+\displaystyle \frac{{49}}{{225}}\\\\=\displaystyle \frac{{83}}{{225}}\end{array}$
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.
Show/Hide Solution
$\begin{array}{l}\ \ \ \ \ P(A\ \text{will pass the exam})=\displaystyle \frac{2}{3}\\\\\therefore \ P(A\ \text{will fail the exam})=1-\displaystyle \frac{2}{3}=\displaystyle \frac{1}{3}\\\\\ \ \ \ \ P(B\ \text{will pass the exam})=\displaystyle \frac{3}{4}\\\\\therefore \ P(A\ \text{will fail the exam})=1-\displaystyle \frac{3}{4}=\displaystyle \frac{1}{4}\\\\(\text{a)}\ \ \ \ P(\text{both}\ A\ \text{and }B\ \text{pass the exam})\\\\\ \ \ =P(A\ \text{will pass the exam})\times P(B\ \text{will pass the exam})\\\\\ \ \ =\displaystyle \frac{2}{3}\times \displaystyle \frac{3}{4}\\\\\ \ \ =\displaystyle \frac{1}{2}\\\\(\text{a)}\ \ \ \ P(\text{exactly one of }A\ \text{and }B\ \text{passes the examination})\\\\\ \ \ =P(A\ \text{will pass and }B\text{ will fail})+P(B\ \text{will pass and }A\text{ will fail})\\\\\ \ \ =\left( {\displaystyle \frac{2}{3}\times \displaystyle \frac{1}{4}} \right)+\left( {\displaystyle \frac{3}{4}\times \displaystyle \frac{1}{3}} \right)\\\\\ \ \ =\displaystyle \frac{2}{{12}}+\displaystyle \frac{3}{{12}}\\\\\ \ \ =\displaystyle \frac{5}{{12}}\end{array}$
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.
Show/Hide Solution
$$\begin{array}{|c|c|c|c|} \hline \text { Group } & \text { Boy } & \text { Girl } & \text { Total } \\ \hline \text { I } & 3 & 1 & 4 \\ \hline \text { II } & 2 & 2 & 4 \\ \hline \text { III } & 1 & 3 & 4 \\ \hline \end{array}$$
$\begin{array}{l}\text{Let the boy from group I, group II}\\\text{and group III be }{{B}_{1}}\text{,}\ {{B}_{2}}\text{ and }{{B}_{3}}\ \text{respectively}\text{.}\\\\\text{Let the girl from group I, group II}\\\text{and group III be }{{G}_{1}}\text{,}\ {{G}_{2}}\text{ and }{{G}_{3}}\ \text{respectively}\text{.}\\\\\therefore \ \ \ \ \ P(1\ \text{boy and 2 girls})\\\\\ \ \ =P({{B}_{1}}{{G}_{2}}{{G}_{3}}\ \text{or}\ {{B}_{2}}{{G}_{1}}{{G}_{3}}\ \text{or}\ {{B}_{3}}{{G}_{1}}{{G}_{2}})\\\\\ \ \ =P({{B}_{1}}{{G}_{2}}{{G}_{3}})+P({{B}_{2}}{{G}_{1}}{{G}_{3}})+P({{B}_{3}}{{G}_{1}}{{G}_{2}})\\\\\ \ \ =\left( {\displaystyle \frac{3}{4}\times \displaystyle \frac{2}{4}\times \displaystyle \frac{3}{4}} \right)+\left( {\displaystyle \frac{2}{4}\times \displaystyle \frac{1}{4}\times \displaystyle \frac{3}{4}} \right)+\left( {\displaystyle \frac{1}{4}\times \displaystyle \frac{1}{4}\times \displaystyle \frac{2}{4}} \right)\\\\\ \ \ =\displaystyle \frac{{18}}{{64}}+\displaystyle \frac{6}{{64}}+\displaystyle \frac{2}{{64}}\\\\\ \ \ =\displaystyle \frac{{26}}{{64}}\\\\\ \ \ =\displaystyle \frac{{13}}{{32}}\end{array}$
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