Probability

Chapter 7

Probability

Probability as a general concept can be defined as the chance of an event occuring. Many people are famaliar with Probability from observation or playing games of chance, such as card games, slot machines of lotteries. In addition to being used in games of chance, Probability theory is used in the fields of insurance, investments, weather forecasting and in various other areas.Blaise Pascal (1623-1662) and Pierre de Fermat (1601-1665) laid the foundation of Probability theory.

Probability၏ယေဘုယျအယူအဆမှာဖြစ်ရပ်တစ်ခုထွက်ပေါ်ခြင်းအခွင့်အလမ်းဟုသတ်မှတ်နိုင်သည်။လူအများစုသည်အခွင့်အလမ်းရှိသည့်ကစားနည်းများ(ဖဲရိုက်ကစားခြင်း၊အကြွေစေ့ထည့်စက်များ၊ထီများ)အစရှိသည့်ကစားနည်းများမှအတွေ့အကြုံများအရ'ဖြစ်တန်စွမ်း'နှင့်ရင်းနှီးကျွမ်းဝင်ပြီးဖြစ်သည်။ဤအခွင့်အလမ်းရှိသည့်ကစားနည်းများအပြင်အာမခံလုပ်ငန်း၊ရင်းနှီးမြုပ်နှံမှုလုပ်ငန်း၊ရာသီဉတုခန့်မှန်းခြင်းနှင့်များစွာသောနယ်ပယ်တို့တွင်'ဖြစ်တန်စွမ်းသီအိုရီ'ကိုအသုံးပြုလာကြသည်။Blaise Pascal(1623-1662)နှင့်Pierre de Fermat(1601-1665)တို့က'ဖြစ်တန်စွမ်းသီအိုရီ'၏အခြေခံကိုချမှတ်ခဲ့ကြသည်။

$7.1$ Calculating Probalility

In this Chapter, we are concerned with experiments such as tossing a coin, rolling a die, drawing a ball from a bag and so on. These experiments are said to be random experiments or chance experiments (for brevity, we will simply call them experiments instead), since exact results of these experiments cannot be predicted. A single specific result of an experiment is called an outcome. The set of all possible outcomes of an experiment is called the sample space. An event is a subset of a sample space. ဤသင်ခန်းစာတွင်အကြွေစေ့မြှောက်ခြင်း၊အန်စာတုံးလှိမ့်ခြင်း၊အိတ်ထဲမှဘောလုံးတစ်လုံးနှိုက်ခြင်းစသဖြင့်လက်တွေ့စမ်းသပ်ချက်များနှင့်ရလဒ်များကိုလေ့လာပါမည်။ထိုသို့စမ်းသပ်မှုပြုလုပ်ရာတွင်တိကျသောရလဒ်ကိုမခန့်မှန်းနိုင်သောကြောင့်ဤစမ်းသပ်ချက်များကိုကျပန်းစမ်းသပ်ချက်များဟုခေါ်သည်။ စမ်းသပ်မှုတစ်ခုအတွက်သီးခြားရလဒ်တစ်ခုကို an outcome ဟုခေါ်သည်။ စမ်းသပ်ချက်တစ်ခု၏ဖြစ်နိုင်သမျှသောရလဒ်များ (outcomes) အားလုံး၏အစုကို sample space ဟုခေါ်သည်။ An event (the set of favourable outcomes) ဖြစ်ရပ်တစ်ခုသည် sample space ၏ subset တစ်ခုဖြစ်သည်။ 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 likehood of occurrence. Then the Probability of an event $A$, denoted by $P(A)$, in the sample space $S$, is defined by sample space ($S$) သည်ကန့်သတ်ရှိအစုဖြစ်ပါစေ။စမ်းသပ်မှုတစ်ခုအတွက်ရလဒ်များအားလုံးသည်အညီအမျှဖြစ်နိုင်ချေရှိသည်။ဆိုလိုသည်မှာ၎င်းတို့သည်ကျပန်းဖြစ်ပြီးရလဒ်တစ်ခုအတွက်ဖြစ်နိုင်ချေအညီအမျှရှိသည်ဟုဆိုလိုသည်။ The sample space $S$ ထဲရှိဖြစ်ရပ် $A$ ၏ဖြစ်တန်စွမ်းကို $P(A)$ ဟုဖော်ပြပြီးအောက်ပါအတိုင်းအဓိပ္ပါယ်သတ်မှတ်နိုင်သည်။ In symbols, where n$(A)$ denotes the number of elements in the event $A$ and n$(S)$ denotes the number of outcomes in the sample space $S$. Notice that : For any event $A$, $P(A)$ is a real number such that $0 \leq P(A) \leq 1$. $P(\phi) = 0$(This means that the probability of an impossible event is $0$.) $P(S) = 1$. (This means that the probability of a sure event is $1$.) For any event $A$, $P(A) + P(\text{ not } A) = 1$. (Rule for complementary events)

For the experiment "tossing a coin", there are two possible outcomes: head and tail (which will be denoted by H snd T respectively). The sample space is $\{H,T\}$, and the events are $\phi$, $\{H\}$, $\{T\}$ and $\{H,T\}$. But for the experiment "tossing two coins", the sample space is $\{(H,H), (H,T), (T,H), (T,T)\}$, which can also be expressed as $\{HH,HT, TH, TT\}$. The event $\{(H,T)\}$ means that "the first toss shows head and the second toss shows tail."

အကြွေစေ့တစ်စေ့မြှောက်သည့် စမ်းသပ်ချက်အတွက် ဖြစ်နိုင်သမျှသောရလဒ်နှစ်ခု Head(H) နှင့် Tail(T) ရှိသည်။
ထို့ကြောင့် the sample space $=\{H,T\}$
events များမှာ $\phi$, $\{H\}$, $\{T\}$ နှင့် $\{H,T\}$

အကြွေစေ့နှစ်စေ့မြှောက်သည့် စမ်းသပ်ချက်အတွက်
the set of sample space ={(H,H), (H,T), (T,H), (T,T)}
{HH,HT, TH, TT} ဟုလည်းဖော်ပြနိုင်သည်။
The event $\{(H,T)\}$ ၏အဓိပ္ပါယ်မှာ ပထမအကြွေစေ့မှ head ကျပြီး၊ဒုတိယအကြွေစေ့မှ tail ကျသည့်ဖြစ်ရပ်ဟုဆိုလိုသည်။
အကြွေစေ့နှစ်စေ့ မြှောက်သည့်စမ်းသပ်ချက်တွက် event အရေအတွက် $16$ ရှိသည်။

For the experiment "rolling a die", there are six possible outcomes: $1$, $2$, $3$, $4$, $5$, $6$. The sample space is $\{1, 2, 3, 4, 5, 6\}$. Some of the events are $\{3\}$, $\{4, 5, 6\}$ and $\{2, 3, 5\}$, which can respectively be described as "the result is $\{3\}$", "the result is at least $\{4\}$" and "the result is a prime number".

အန်စာတုံးတစ်တုံးလှိမ့်သည့်စမ်းသပ်ချက်အတွက် ဖြစ်နိုင်သမျှသောရလဒ်ခြောက်ခု ($1$, $2$, $3$, $4$, $5$, $6$) ရှိသည်။
The sample space $=\{1, 2, 3, 4, 5, 6\}$
events အချို့မှာ $\{3\}$, $\{4, 5, 6\}$ and $\{2, 3, 5\}$ စသဖြင့်။
$\{3\}$  'the result is $3$' (ရလဒ် $3$ ဖြစ်သည်။)
$\{4, 5, 6\}$  'the result is at least $4$'  (ရလဒ်အနည်းဆုံး $4$ ဖြစ်သည်။)
$\{2, 3, 5\}$  'the result is a prime' (ရလဒ်သည် သုဒ္ဓကိန်းတစ်ခုဖြစ်သည်။)

The following table represents the sample space for rolling two dice. The first part of an ordered pair in the table represents the number appears on the first die and the second part represents the corresponding number on the second die. Sample spaces of similar experiments can also be obtained by constructing such tables.

အောက်ပါဇယားသည် အန်စာတုံးနှစ်တုံးလှိမ့်သည့်စမ်းသပ်ချက်အတွက် sample space များကိုဖော်ပြသည်။ ordered pair တစ်ခုစီ၏ပထမကိန်းသည် ပထမအန်စာတုံးပေါ်မှကိန်းများကိုကိုယ်စားပြုပြီး၊ဒုတိယကိန်းသည်ဒုတိယအန်စာတုံးပေါ်မှသက်ဆိုင်ရာကိန်းများကိုကိုယ်စားပြုသည်။အလားတူစမ်းသပ်ချက်များအတွက်ဤကဲ့သို့သော table များပြု လုပ်၍ sample space များကိုရရှိနိုင်သည်။



Sometimes, it is convenient to use a tree diagram to list all possible outcomes in a sample space. The following tree diagram displays the sample space for tossing a coin three times.

တစ်ခါတရံ tree diagram ကိုအသုံးပြု၍ sample space တစ်ခုအတွင်းရှိ ဖြစ်နိုင်သမျှသောရရဒ်များအားလုံးကိုစီစဉ်နိုင်သည်။အောက်ပါ tree diagram သည်အကြွေစေ့တစ်စေ့ကိုသုံးကြိမ်မြှောက်ခြင်းစမ်းသပ်ချက်မှ sample space ကိုပြသနေသည်။

Example $1$

Find the probability of randomly selecting a red pen from a box that contains $2$ red pens, $4$ blue pens and $3$ yellow pens.

$\text{ Let } A \text{ be the event of selecting a red pen.}$ $P(A)=\dfrac{\text{ number of outcomes in } A}{\text{ number of outcomes in the sample space } S}=\dfrac{2}{9}$

Example $2$

If a whole number from $1$ to $20$ both inclusive is randomly selected, and if each number has an equal chance of being selected, what is the probability that the number will be
(a)  even?  (b)  greater than $1$    (c) prime?

$\text{ The sample space is }S=\{1, 2, 3, ...., 20\}$   $\text{ n }(S)=20$ $\text{ (a) The event is } E_1=\{2, 4, 6, ..., 20\}$   $\text{ n }(E_1)=10$   $\text{ P }(E_1)=\dfrac{\text{ n }(E_1)}{\text{ n }(S)}=\dfrac{10}{20}=\dfrac{1}{2}$ $\text{ (b) The event is }E_2=\{2, 3, 4, ..., 20\}$   $\text{ n }(E_2)=19$   $\text{ P }(E_2)=\dfrac{\text{ n }(E_2)}{\text{ n }(S)}=\dfrac{19}{20}$ $\text{ (c) The event is } E_3=\{2, 3, 5, 7, 11, 13, 17, 19 \}$   $\text{ n }(E_3)=8$   $\text{ P }(E_3)=\dfrac{\text{ n }(E_3)}{\text{ n }(S)}=\dfrac{8}{20}=\dfrac{2}{5}$

Example $3$

In a sample of $50$ people, $21$ had type O blood, $22$ had type A blood, $5$ had type B blood and, $2$ had type AB blood. Find the following probabilities.
(a)  A person has type O blood.
(b)  A person has type A or type B blood.
(c)  A person has neither type A nor type O blood.
(d)  A person does not have type AB blood.

$\text{ (a) P(O) }=\dfrac{21}{50}$ $\text{ (b) P(A or B) }=\dfrac{27}{50}$ $\text{ (c) P(neither A nor O) }=\dfrac{7}{20}$ $\text{ (d) P(not AB) }=\dfrac{48}{50}=\dfrac{24}{25}$

Example $4$

A family has three children. Find the probabilities of
(a)  all boys,      (b) exactly two boys,
(c)  at most two boys,  (d) at least one girl,
(e)  at least one boy and at least one girl.

$\text{ The sample space consists of } 8 \text{ outcomes. }$ $\text{ (a) The event is }\{BBB\}$   $\text{ P(all boys) }=\dfrac{1}{8}$ $\text{ (b) The event is }\{BBG, BGB, GBB\}$   $\text{ P(exactly two boys) }=\dfrac{3}{8}$ $\text{ (c) The event is }\{BBG, BGB, BGG, GBB, GBG, GGB, GGG\}$   $\text{ P(at most two boys) }=\dfrac{7}{8}$ $\text{ (d) The event is }\{BBG, BGB, BGG, GBB, GBG, GGB, GGG\}$   $\text{ P(at least one girl) }=\dfrac{7}{8}$  (or)   $\text{ P(at least one girl) } =1- \text{ P(all boys) }=1-\dfrac{1}{8}=\dfrac{7}{8}$  $\text{ (by using rule for complementary events) }$ $\text{ (e) The outcomes is }\{BBG, BGB, BGG, GBB, GBG, GGB \}$   $\text{ P(at least one boy and at least one girl) }=\dfrac{6}{8}=\dfrac{3}{4}$

Example $5$

Draw a tree diagram to list all possible two-digit numerals which can be formed by using the digits $2$, $3$, $5$ and $6$ without repeating any digit. If one of these numerals is chosen at random, find the probability that it is divisible by $13$. Find also the probability that it is either a prime or a perfect square. Find the probability that none of its digits is $6$.

$\text{ The sample space consists of }12 \text{ outcomes. }$ $\text{ The event containing numerals divisible by }13 \text{ is }\{26, 52, 65\}$ $\text{ P(a numeral which is divisible by }13 \text{ ) }=\dfrac{3}{12}=\dfrac{1}{4}$ $\text{ The event containing numerals which are primes or perfect square is }\{23, 25, 36, 53\}$ $\text{ P(a numeral which is prime or a perfect square) }=\dfrac{4}{12}=\dfrac{1}{3}$ $\text{ The event containing numerals which do not have digit }6 \text{ is }\{23, 25, 32, 35, 52, 53\}$ $\text{ P(a numeral none of its digits is $6$) }=\dfrac{6}{12}=\dfrac{1}{2}$

Example $6$

Two fair dice are thrown and the numbers appeared on top faces are recorded.
Find the probability of each event:
(a)  The first die shows $5$.
(b)  The sum of the numbers on the dice is $7$.
(c)  The product of the numbers on the two dice is greater than $24$.

$\text{ The sample space consists of } 36 \text{ outcomes. }$ $\text{ (a) The event is }\{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$   $\text{ P( } 5 \text{ on the first die) }=\dfrac{6}{36}=\dfrac{1}{6}$ $\text{ (b) The event is }\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$   $\text{ P(the sum is } 7 \text{ ) }=\dfrac{6}{36}=\dfrac{1}{6}$ $\text{ (c) The event is }\{(5,5), (5,6), (6,5), (6,6)\}$   $\text{ P(product is greater than } 24 \text{ ) }=\dfrac{4}{36}=\dfrac{1}{9}$

$7.2$ Probabilities of Combined Events

Two events are independent if the occurrence of any one of them does not affect the probability of the other. $A$ နှင့် $B$ ဖြစ်ရပ်နှစ်ခု independent ဖြစ်သည်ဆိုသည်မှာ ဖြစ်ရပ် $A$ ဖြစ်ခြင်း၊မဖြစ်ခြင်းသည် ကျန်ဖြစ်ရပ် $B$ ဖြစ်ခြင်း၊မဖြစ်ခြင်းအပေါ် အကျိုးသက်ရောက်မှုမရှိခြင်း(တစ်နည်းအားဖြင့် မသက်ဆိုင်ခြင်း) ကိုဆိုလိုသည်။ For example, when two fair coins are tossed, the event of getting head on the first coin and the event of getting tail on the second coin are independent. ဉပမာ။ အကြွေစေ့နှစ်စေ့ကိုမြှောက်လိုက်သည်ဆိုပါစို့။ ရလဒ်ကိုလေ့လာကြမည်။ ပထမအကြွေစေ့၌ခေါင်းရရှိခြင်းဖြစ်ရပ်နှင့် ဒုတိယအကြွေစေ့၌ပန်းရရှိခြင်းဖြစ်ရပ်တို့သည် မသက်ဆိုင်သောဖြစ်ရပ်များဖြစ်သည်။ Suppose an experiment consists of choosing a marble from a jar containing $3$ red, $5$ green, $2$ blue and $6$ yellow marbles, and rolling a die. Let us consider the events: getting a blue marble from the jar getting a prime number on the die getting both a blue marble from the jar and a prime number on the die The probability of the first event is $\dfrac{1}{8}$. The probability of the second event is $\dfrac{1}{2}$. The probability of the third event is $\dfrac{1}{16}$. ဖန်ဘူးတစ်လုံးတွင် အနီရောင်ဂေါ်လီလုံး $3$ လုံး၊အစိမ်းရောင်ဂေါ်လီလုံး $5$ လုံး၊အပြာရောင်ဂေါ်လီလုံး $2$ လုံး၊အဝါရောင်ဂေါ်လီလုံး $6$ လုံးရှိပြီး၊အန်စာတုံးတစ်တုံးကိုလှိမ့်မည်ဆိုပါစို့။ ဤတွင် ဖန်ဘူးထဲမှဂေါ်လီလုံးမည်မျှနှိုက်သည်ဖြစ်စေ၊ မည်သည့်အရောင်ရွေးချယ်သည်ဖြစ်စေ၊ ထိုဖြစ်ရပ်များသည် အန်စာတုံးလှိမ့်သည့်အခါ ထွက်ပေါ်လာမည့်ဂဏန်းများနှင့် သက်ဆိုင်မှုမရှိကြောင်းသိထားရမည်ဖြစ်သည်။ သို့သော် ဖန်ဘူးထဲမှဂေါ်လီလုံးတစ်လုံးနှိုက်ခြင်းနှင့် အန်စာတုံးလှိမ့်သည့်အခါထွက်ပေါ်လာသော ဂဏန်းတစ်ခုရွေးချယ်ခြင်းတို့သည်တစ်ပြိုင်နက်တည်းဖြစ်ပေါ်နိုင်ပေသည်။ ဤသို့ ဖြစ်ရပ် $A$ နှင့်ဖြစ်ရပ် $B$ တို့တစ်ပြိုင်နက်ဖြစ်နိုင်မည့်ဖြစ်တန်စွမ်းကို $P(A)\times P(B)$ ဖြင့်တွက်ချက်နိုင်သည်။ The probability of getting a blue marble from the jar $=\dfrac{2}{16}=\dfrac{1}{8}$  (ဖန်ဘူးထဲမှအပြာရောင်ဂေါ်လီလုံးတစ်လုံးရရှိမည့်ဖြစ်တန်စွမ်း) The probability of getting a prime number on the die $=\dfrac{3}{6}=\dfrac{1}{2}$  (အန်စာတုံးပေါ်မှသုဒ္ဓကိန်းတစ်ခုရရှိမည့်ဖြစ်တန်စွမ်း) The probability of getting both a blue marble from the bag and a prime number on the die $=\dfrac{6}{96} =\dfrac{1}{16}$   (ဖန်ဘူးထဲမှအပြာရောင်ဂေါ်လီလုံးတစ်လုံးနှင့်အန်စာတုံးပေါ်မှသုဒ္ဓကိန်းတစ်ခုရရှိမည့်ဖြစ်တန်စွမ်း)  (by using definition of probability) The probability of getting both a blue marble from the bag and a prime number on the die $=\dfrac{1}{8} \times \dfrac{1}{2} = \dfrac{1}{16}$ (combined events by using Multiplication Rule ) The first two events are independent. In fact, the product of the probabilities of the first two events is the same as the probability of the third (combined) event. The general rule for independent events is as follow: ပထမဖြစ်ရပ်နှစ်ခုသည် မသက်ဆိုင်သောဖြစ်ရပ်များဖြစ်ကြသည်။ ပထမဖြစ်ရပ်နှစ်ခုတို့၏ဖြစ်တန်စွမ်းများမြှောက်လဒ်သည် တတိယဖြစ်ရပ်(ပေါင်းစပ် ဖြစ်ရပ်)၏ဖြစ်တန်စွမ်းနှင့်တူညီနေကြောင်းတွေ့ရသည်။ထို့ကြောင့် independent events များအတွက် general rule မှာအောက်ပါအတိုင်းဖြစ်သည်။ Multiplication Rule: A and B are independent events in a sample space if and only if P(A and B) = P(A) x P(B)

Example $7$

A bag contains $3$ red balls, $2$ blue balls and $5$ white balls. A ball is selected at random and its colour noted. Then it is replaced. A second ball is selected and its colour noted. Find the probabilities of :
(a)  selecting $2$ blue balls,
(b)  selecting $1$ blue ball and then $1$ white ball,
(c)  selecting $1$ red ball and then and then $1$ blue ball.

$\begin{aligned} \text{ (a) }\quad\text{ P }(2 \text{ blue balls })&=\text{ P }(1\text{st} \text{ ball is blue and } 2\text{nd ball is blue})\\ &=\text{ P }(1\text{st} \text{ ball is blue })\times \text{ P }(2\text{nd} \text{ ball is blue })\\ &=\dfrac{2}{10}\times \dfrac{2}{10}\\ &=\dfrac{1}{25} \end{aligned}$ $\begin{aligned} \text{ (b) }\quad&\text{ P }(1 \text{ blue ball and then } 1 \text{ white ball })\\ &=\text{ P }(1\text{st} \text{ ball is blue })\times \text{ P }(2\text{nd} \text{ ball is white })\\ &=\dfrac{2}{10}\times \dfrac{5}{10}\\ &=\dfrac{1}{10} \end{aligned}$ $\begin{aligned} \text{ (c) }\quad&\text{ P }(1 \text{ red ball then } 1 \text{ blue ball })\\ &=\text{ P }(1\text{st} \text{ ball is red and } 2\text{nd} \text{ ball is blue })\\ &=\text{ P }(1\text{st} \text{ ball is red })\times \text{ P }(2\text{nd} \text{ ball is blue })\\ &=\dfrac{3}{10}\times \dfrac{2}{10}\\ &=\dfrac{3}{50} \end{aligned}$

Example $8$

A bag contains $9$ red marbles and $3$ green marbles. For each case belo, find the probability of randomly selecting a red marble on the first draw and a green marble on the second draw.
(a)  The first marble is replaced.
(b)  The first marble is not replaced.

$\text{ There are } 12 \text{ marbles (} 9 \text{ red and } 3 \text{ green). }$ $\begin{aligned} \text{ (a) }\quad &\text{ P }(1\text{st} \text{ marble is red and } 2\text{nd marble is green })\\ &=\text{ P }(1\text{st} \text{ marble is red })\times \text{ P }(2\text{nd} \text{ marble is green })\\ &=\dfrac{9}{12}\times \dfrac{3}{12}\\ &=\dfrac{3}{16} \end{aligned}$ $\begin{aligned} \text{ (b) }\quad &\text{ P }(1\text{st} \text{ marble is red and } 2\text{nd marble is green })\\ &=\text{ P }(1\text{st} \text{ marble is red })\times \text{ P }(2\text{nd} \text{ marble is green })\\ &=\dfrac{9}{12}\times \dfrac{3}{11}\\ &=\dfrac{9}{44} \end{aligned}$

Example $9$

At a teachers' conference, there are $4$ English teachers, $3$ Mathematics teachers, and $5$ Science teachers.If $4$ teachers are selected for a committee, find the probability that at least one is a science teacher.

$\text{ are } 12 \text{ teachers (} 4 \text{ English, } 3 \text{ Mathematics and } 5 \text{ Science).}$ $\begin{aligned} \text{ P( at least } & \text{ one is a Science teacher ) }\\ = 1-\text{ P(} & \text{none of the } 4 \text{ teacher) }\\ =1-\text{ P(} & 1\text{st teacher is not a science teacher },\\ & 2\text{nd teacher is not a science teacher },\\ & 3\text{rd teacher is not a science teacher and }\\ & 4\text{th teacher is not a science teacher )}\\ =1-\dfrac{7}{12}&\times \dfrac{6}{11}\times\dfrac{5}{10}\times\dfrac{4}{9}\\ =1-\dfrac{7}{99}&=\dfrac{92}{99} \end{aligned}$

Addition Rule $1$:

တစ်ပြိုင်နက်ဖြစ်နိုင်သောဖြစ်ရပ်နှစ်ခု တစ်ခုမဟုတ်တစ်ခုဖြစ်ရန် ဖြစ်တန်စွမ်းကိုတွက်ယူလိုလျှင် ထိုဖြစ်ရပ်နှစ်ခု၏ တစ်ခုစီဖြစ်နိုင်သောဖြစ်တန်စွမ်းများ ပေါင်းလဒ်မှဖြစ်ရပ်နှစ်ခုလုံးတစ်ပြိုင်နက်ဖြစ်နိုင်သော ဖြစ်တန်စွမ်းကိုနုတ်ရသည်။ set theory သဘောတရားအရ, $P(A$ ⋃ $B) = P(A) + P(B) - P(A$ ⋂ $B)$ Addition Rule 1: For any two events A and B in a sample space, P(A or B) = P(A) + P(B) - P(A and B).

Example $10$

In a hospital unit, there are $8$ nurses and $5$ doctors. Among them, there are $7$ nurses and $3$ doctors are females. If a staff person is selected, find the probability that the staff is a nurse or a male.

$\begin{aligned} \text{ P(nurse or male) }&= \text{ P(nurse)} + \text{ P(male)} - \text{ P(male nurse)}\\ &=\dfrac{8}{13} + \dfrac{3}{13} - \dfrac{1}{13}\\ &=\dfrac{10}{13} \end{aligned}$

Addition Rule $2$:

Two events $A$ and $B$ are mutually exclusive if they cannot occur jointly, that is, they do not have common outcomes. For example, the occurrence of head and the occurrence of tail, in a single flip of a coin, are mutually exclusive. ဖြစ်ရပ် $A$ နှင့်ဖြစ်ရပ် $B$၊ ဖြစ်ရပ်နှစ်ခု mutually exclusive ဖြစ်သည်ဆိုသည်မှာ ဖြစ်ရပ်နှစ်ခုတို့သည်တစ်ခုနှင့်တစ်ခုမှီခိုဆက်စပ်နေသည်ဟုဆိုလိုသည်။ဖြစ်ရပ် $A$ ဖြစ်လျှင် ဖြစ်ရပ် $B$ မဖြစ်နိုင်။ ဉပမာ။ အကြွေပြားတစ်စေ့ကိုမြှောက်လိုက်သောအခါ ခေါင်းနှင့်ပန်းတို့သည် တစ်ပြိုင်နက်တည်းမကျနိုင်ချေ။အလားတူပင် ကျောင်းသားတစ်ယောက်စာမေးပွဲအောင်ခြင်းနှင့် ရှုံးခြင်းသည်တစ်ပြိုင်နက်တည်းမဖြစ်နိုင်ချေ။ တစ်နည်းအားဖြင့် ဖြစ်ရပ် $A$ နှင့် ဖြစ်ရပ် $B$ တို့တွင် ဘုံထွက်ပေါ်ချက်မရှိဟုဆိုလိုသည်။ set theory သဘောတရားအရ $A$ ⋂ $B=\phi$ ဖြစ်သည်။ ထိုသို့ ဖြစ်ရပ် $A$ နှင့်ဖြစ်ရပ် $B$ တစ်ခုမဟုတ်တစ်ခုဖြစ်နိုင်သောဖြစ်တန်စွမ်းကိုတွက်ချက်ရန် $P(A)+P(B)$ ကိုသုံးသည်။ If $A$ and $B$ are mutually exclusive events, then $P(A \text{ and } B)=0$. Addition Rule 2: If A and B are mutually exclusive events in a sample space, then P(A or B) = P(A) + P(B).

Example $11$

A box contains $3$ strawberry doughnuts, $4$ jelly doughnuts, and $5$ chocolate doughnuts. If a doughnut is selected at random, find the probability that it is either a strawberry doughnut or a chocolate doughnut.

$\text{ The box contains (} 12 \text{ doughnuts, } 3 \text{ strawberry, } 4 \text{ jelly and } 5 \text{ chocolate). }$ $\begin{aligned} &\text{ P(strawberry doughnut or chocolate doughnut) }\\ &= \text{ P(strawberry doughnut) } + \text{ P(chocolate doughnut) }\\ &=\dfrac{3}{12} + \dfrac{5}{12}\\ &=\dfrac{8}{12}\\ &=\dfrac{2}{3} \end{aligned}$

$7.3$ Calculation of Expected Frequency

The expected frequency of an event is the number of times that we predict the event will occur, in a given number of trials, based on a calculation using probabilities. It can be calculated by the formula: ဖြစ်ရပ်တစ်ခု၏မျှော်မှန်းထပ်ကြိမ်ဆိုသည်မှာ ပေးထားသောစမ်းသပ်ကြိမ်များအတွက် ဖြစ်တန်စွမ်းများကိုအသုံးပြုတွက်ချက်ထားသော၊ ဖြစ်ရပ်တစ်ခုဖြစ်ပေါ်ရန်ကျွန်ုပ်တို့မျှော်မှန်းထားသောအကြိမ်အရေအတွက်ဖြစ်သည်။ ၎င်းကိုတွက်ချက်ရန်ပုံသေနည်းမှာ Consider tossing a coin $1000$ times. Since each toss is a trial, the number of trials is $1000$. Since the probability of getting head in each trial is $0.5$, the expected frequency of getting head is $500$ $(=0.5\times 1000)$.This means that we can expect $500$ heads occurring in $1000$ trials. The expected frequency is based on probability. On the other hand, you can actually flip a coin $1000$ times and observe the frequency of head. Do you think that these two frequencies are the same? အကြွေစေ့တစ်စေ့ကို အကြိမ် $1000$ မြှောက်မည်ဆိုပါစို့။တစ်ကြိမ်မြှောက်ခြင်းသည် စမ်းသပ်ကြိမ် $1$ ဖြစ်သောကြောင့် number of trials $=1000$ ဖြစ်သည်။တစ်ကြိမ်မြှောက်တိုင်း head ကျရန်ဖြစ်တန်စွမ်းမှာ $P(\text{ head })=0.5$ ဖြစ်သောကြောင့် head ကျရန်မျှော်မှန်းထပ်ကြိမ်မှာ $500$ ဖြစ်သည်။ဆိုလိုသည်မှာ စမ်းသပ်ကြိမ်ပေါင်း $1000$ အတွက်ခေါင်းကျရန်အကြိမ် $500$ မျှော်မှန်းနိုင်သည်ဟုဆိုလိုသည်။ The expected frequency သည် probability ပေါ်တွင်အခြေခံသည်။အခြားတစ်ဖက်တွင်အကြွေစေ့ကိုအကြိမ် $1000$ မြှောက်၍ head ကျမည့်အကြိမ်အရေအတွက်ကိုရေတွက်ကြည့်ပါ။ဤနည်းလမ်းနှစ်ခုအတွက် မျှော်မှန်းထပ်ကြိမ်တို့သည်တူညီသည်ဟုသင်ထင်ပါသလား?

Example $12$

Traffic analysis found that the probability that a motorist will turn right at the intersection is $\dfrac{1}{3}$. Out of $300$ motorists, how many would you expect to turn right at that intersection?

$\text{ P(turning right) }= \dfrac{1}{3}$ $\text{ Number of trials or motorists }= 300$ $\begin{aligned} \text{ Expected number of motorists turning right at the intersection }& =\dfrac{1}{3}\times 300\\ &=100 \end{aligned}$

Example $13$

A spinner is equally to point to any one of the numbers $1$, $2$, $3$, $4$, $5$, $6$, $7$. What is the probability of scoring a number divisible by $3$? If the arrow is spun $700$ times, how many would you expect a number not divisible by $3$?

$\text{ Among the given numbers, the numbers }3 \text{ and }6 \text{ are divisible by }3$. $\text{ P(a number divisible by } 3 \text{ ) }=\dfrac{2}{7}$ $\text{ P(a number not divisible by }3 \text{ ) }= 1-\dfrac{2}{7} = \dfrac{5}{7}$ $\text{ The number of trials }= 700$ $\begin{aligned} \text{ Expected frequency of a number which is not divisible by }3 &= \dfrac{5}{7}\times 700\\ &=500 \end{aligned}$

Problems

Exercise $7.1$

1. A letter is chosen at random from the letters of the word ORANGE.
   What is the probability that it is a vowel?





$\text{ The sample space } S = \{O, R, A, N, G, E\}$ $\text{ n(} S \text{ ) }= 6$ $\text{ The event of the letter is a vowel is } E = \{O, A, E\}$ $\text{ n( }E \text{ ) }= 3$ $\text{ P(it is a vowel) } = \dfrac{\text{n}(E)}{\text{n}(S)} = \dfrac{3}{6} = \dfrac{1}{2}$



2. A bag contains $10$ red balls and $30$ black balls.
(a)  If a ball is drawn at random, what is the probability of getting a red ball?
(b)  Suppose the first ball drawn at random is red and is not replaced. If another ball is drawn at random, what is the probability that it will again be red?





$\text{ Let red ball }= R, \text{ black ball }= B$ $\text{ the sample space } S \text{ is }\{R_1, R_2,....., R_{10},B_1, B_2,....., B_{30}\}$ $\text{ n( }S \text{ ) } = 40, \text{ n( }E_1 \text{ ) } = 10$ $\text{ (a) P(getting a red ball) }= \dfrac{\text{ n } (E_1)}{\text{ n }(S)} = \dfrac{10}{40} = \dfrac{1}{4}$ $\text{ (b) (when the first red ball is not replaced, }$   $\text{ n( } S \text{ ) }=39$   $\text{ n( }E_2 \text{ ) }=9$   $\text{ P(the next red ball) }=\dfrac{\text{n}(E_2)}{\text{n}(S)} = \dfrac{9}{39} = \dfrac{3}{13}$



3. How many three-digit numerals can be formed from $1$, $5$ and $7$, without repeating any digit?
   Find the probability of a numeral which begins with $1$?





$\text{ The sample space } S \text{ contains } 6 \text{ outcomes. }$ $\text{ n( } S\text{ ) }=6$ $\text{ The event } E \text{ containing a numeral which begins with } 1 \text{ is }\{157, 175\}$. $\text{ n( } E \text{ ) }=2$ $\text{ P(a numeral which begins with } 1)=\dfrac{2}{6}=\dfrac{1}{3}$



4. A box contains five cards numbered as $2$, $3$, $4$, $5$ amd $9$. A card is chosen, the number is recorded, and the card is not replaced. Then another card is chosen and the number is recorded. Draw a tree diagram to get the possible outcomes. Find the probabilities of
(a)  getting two prime numbers,
(b) getting two odd numbers and
(c) getting a pair of numbers whose sum is a prime number.





$\text{ The samples space contains } 20 \text{ outcomes. }$   $\therefore \text{ n( } S \text{ ) }=20$ $\text{ (a)The event } E_1 \text{ is }\{(2,3), (2,5), (3,2), (3,5), (5,2), (5,3)\}$   $\therefore \text{ n( }E_1 \text{ ) }=6$   $\text{ P(getting two prime numbers) }=\dfrac{\text{n}(E_1)}{\text{n}(S)} = \dfrac{6}{20}=\dfrac{3}{10}$ $\text{ (b)The event $E_2$ is }\{(3,5), (3,9), (5,3), (5,9), (9,3), (9,5)\}$   $\therefore \text{ n( }E_2 \text{ ) }=6$   $\text{ P(getting two odd numbers) }=\dfrac{ \text{ n }(E_2) }{\text{ n }(S) } = \dfrac{6}{20}=\dfrac{3}{10}$ $\text{ (c) The event } E_3 \text{ is } \{(2,3), (2,5), (2,9), (3,2), (3,4), (4,3), (5,2), (9,2), (9,4), (4,9)\}$   $\therefore \text{ n }(E_3)=10$   $\text{ P(getting a pair of numbers whose sum is a prime number) }=\dfrac{\text{n}(E_3)}{\text{n}(S)} = \dfrac{10}{20}=\dfrac{1}{2}$



5. A box contains four marbles of two blue, one red and one yellow. A marble is chosen, the colour is recorded, and the marble is not replaced. Then another marble is chosen and the colour is recorded. Draw a tree diagram to determine possible outcomes. Hence, find the probabilities of
   
(a)  getting two blue marbles and
(b)  getting two different colours.





$\text{ The samples space contains } 12 \text{ outcomes. }$ $\therefore \text{ n( } S \text{ ) }=12$ $\text{ (a) The event } E_1 \text{ is }\{(B_1,B_2), (B_2,B_1)\}$   $\therefore \text{ n( } E_1\text{ ) }=2$   $\text{ P(getting two blue marbles) }=\dfrac{\text{ n }(E_1)}{\text{ n }(S)} = \dfrac{2}{12}=\dfrac{1}{6}$ $\text{ (b) The event } E_2 \text{ is } \{(B_1,R), (B_1,Y), (B_2,R), (B_2,Y), (R,B_1), (R,B_2), (R,Y), (Y,B_1), (Y,B_2), (Y,R)\}$   $\therefore \text{ n( } E_2 \text{ ) }=10$   $\text{ P(getting two different marbles) }=\dfrac{\text{ n }(E_2)}{\text{ n }(S)} = \dfrac{10}{12}=\dfrac{5}{6}$



6. A spinner is equally likely to point to any one of the numbers $2$, $3$, $4$ and $5$. Make a table of ordered pairs (first spin, second spin).Find the probability of
   
(a)  two odd numbers,
(b)  an even number followed by an odd number.







$\text{ The samples space contains } 16 \text{ outcomes. }$   $\therefore \text{ n( } S \text{ ) }=16$ $\text{ (a) For two odd numbers, }$   $\text{ the event } E_1 \text{ is } \{(3,3), (3,5), (5,3), (5,5)\}$   $\therefore \text{ n }(E_1)=4$   $\text{ P(two odd numbers) }=\dfrac{\text{n}(E_1)}{\text{n}(S)} = \dfrac{4}{16}=\dfrac{1}{4}$ $\text{ (b) For (even, odd) outcomes, }$   $\text{ the event } E_2 \text{ is } \{(2,3), (2,5), (4,3), (4,5)\}$   $\therefore \text{ n }(E_2)=4$   $\text{ P(an even number followed by an odd number) }=\dfrac{\text{n}(E_2)}{\text{n}(S)} = \dfrac{4}{16}=\dfrac{1}{4}$



7. A coin is tossed and then a die is thrown. Head or tail and the number turns up are recorded each time. Draw a tree diagram and list the possible outcomes. Hence, find the probability that head and $6$ turns up.





$\text{ The samples space contains } 12 \text{ outcomes. }$ $\therefore \text{ n( } S\text{ ) }=12$ $\text{ the event } E \text{ is }\{(H,6)\}$ $\therefore \text{ n( } E \text{ ) }=1$ $\text{ P(head and } 6 \text{ turns up })=\dfrac{\text{n}(E)}{\text{n}(S)} = \dfrac{1}{12}$

Exercise $7.2$

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





$\text{ There are } 19 \text{ instructors, } 7 \text{ Mathematics, } 5 \text{ computer science, } 3 \text{ stastics and } 4 \text{ science }.$ $\begin{aligned} \text{ P(getting a science instructor or a math instructor) }&=\text{ P(getting a science instructor) }+\text{ P(getting a math instructor) }\\ &=\dfrac{4}{19}+\dfrac{7}{19}\\ &=\dfrac{11}{19} \end{aligned}$



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 $16$.





$\text{ The sample space consists of } 36 \text{ outcomes. }$ $\text{ (a)The event containing a sum greater than } 8 \text{ is }$   $\{(3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6)\}$.   $\text{ The event containing a sum less than } 3 \text{ is }\{(1,1)\}$.   $\text{ P(a sum greater than } 8 \text{ or a sum less than } 3 \text{ ) }$   $=\text{ P(a sum greater than } 8\text{ ) }+ \text{ P(a sum less than } 3\text{ ) }$   $=\dfrac{10}{36}+\dfrac{1}{36}=\dfrac{11}{36}$ $\text{ (b) Let } A \text{ be the event containing a product greater than } 9 \text{ and }$   $B \text{ be the event containing a product less than } 16$. $\begin{aligned} \text{ event } A =&\{(2,5), (2,6), (3,4), (3,5), (3,6), (4,3), (4,4), (4,5), (4,6), (5,2), (5,3), (5,4),\\ &(5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6)\}\\ \text{ event } B =&\{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6),\\ &(3,1),(3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (6,1), (6,2)\} \end{aligned}$ $\text{ event } A \text{ and } B =\{(2,5), (2,6), (3,4), (3,5), (4,3), (5,2), (5,3), (6,2)\}$ $\begin{aligned} \text{ P(A or B) }&=\text{ P(A) }+ \text{ P(B) }- \text{ P(A and B) }\\ &=\dfrac{19}{36}+\dfrac{25}{36}-\dfrac{8}{36}\\ &=\dfrac{36}{36}\\ &=1 \end{aligned}$



3. 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.





$\text{ A bag contains } 15 \text{ discs (} 3 \text{ white,} 5 \text{ red, } 7 \text{ blue). }$ $\text{ Let } W = \text{ white, } R = \text{ red, } B = \text{ blue }$ $\text{ P(the two discs will be of the same colour) }$ $=\text{ P }( W, W) \text{ or }( R, R) \text{ or }( B, B)$ $=\text{ P }( W, W) + \text{ P }(R, R) + \text{ P }(B, B)$ $=\dfrac{3}{15}\times \dfrac{3}{15}+ \dfrac{5}{15}\times \dfrac{5}{15}+\dfrac{7}{15}\times \dfrac{7}{15}$ $=\dfrac{9+25+49}{225}$ $=\dfrac{83}{225}$



4. In a survey about a change in public policy, $100$ people were asked if they favour the change, oppose the change, or have no opinion about the change. The responses are indicated as below.
   
   
   
   Find the probability that a randomly selected respondent to this survey oppose or has no oponion about the change policy.





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



5. The probabilities that the student $A$ and $B$ pass an examination are $\dfrac{2}{3}$ and $\dfrac{3}{4}$ respectively. Find the probabilities that:
   (a)  both $A$ and $B$ pass the examination.
   (b)  exactly one of $A$ and $B$ passes the examination.





$\text{ P( } A \text{ will pass the exam) }=\dfrac{2}{3}$, $\text{ P( } A \text{ will fail the exam) }=1-\dfrac{2}{3}=\dfrac{1}{3}$ $\text{ P( } B \text{ will pass the exam) }=\dfrac{3}{4}$, $\text{ P( } B \text{ will fail the exam) }=1-\dfrac{3}{4}=\dfrac{1}{4}$ $\begin{aligned} \text{ (a) P(both A and B pass the exam) }&= \text{ P(A will pass the exam) }\times \text{ P(B will pass the exam) }\\ &=\dfrac{2}{3}\times \dfrac{3}{4}\\ &=\dfrac{1}{2} \end{aligned}$ $\begin{aligned} \text{ (b) P(both A and B pass the exam) }&= \text{ P(A will pass and B will fail) or P(A will fail and B will pass) }\\ &= \text{ P(A will pass) }\times \text{ P(B will fail) }+ \text{ P(A will fail) }\times \text{ P(B will pass) }\\ &=\dfrac{2}{3}\times \dfrac{1}{4}+\dfrac{1}{3}\times \dfrac{3}{4}\\ &=\dfrac{2+3}{12}\\ &=\dfrac{5}{12} \end{aligned}$



6. Three groups of children consists 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.





$\text{ Let } B = \text{ boy, } R = \text{ red, } G = \text{ girl }$ $\begin{aligned} \text{ P( } 1 \text{ boy and } 2 \text{ girls) }&= \text{ P(B, G, G) or ( G, B, G ) or ( G, G, B) }\\ &= \text{ P( B, G, G) + ( G, B, G ) + ( G, G, B) }\\ &=\dfrac{3}{4}\times \dfrac{2}{4}\times \dfrac{3}{4}+\dfrac{1}{4}\times\dfrac{2}{4}\times \dfrac{3}{4}+\dfrac{1}{4}\times\dfrac{2}{4}\times \dfrac{1}{4}\\ &=\dfrac{18+6+2}{64}\\ &=\dfrac{26}{64}\\ &=\dfrac{13}{32} \end{aligned}$

Exercise $7.3$

1. After a large number of tossing a pin, the probability of 'pin up' was estimated to be $0.3$. In $400$ more trials, how many times would 'pin up' be expected





$\text{ P(pin up) }=0.3$ $\text{ The number of trials }=400$ $\begin{aligned} \text{ The expected frequency of times would 'pin up' }&= \text{ P(pin up) }\times \text{ number of trials}\\ &= 0.3\times 400\\ &=120 \end{aligned}$



2. If a die is rolled $60$ times, what is the expected frequency of
   (a)  $1 \text{ turns up? }$
   (b)  $\text{ a number divisible by } 3 \text{ turns up? }$
   (c)  $\text{ a factor of } 6 \text{ turns up? }$





$\text{ The sample spaces contains } 6 \text{ outcomes }=\{1, 2, 3, 4, 5, 6\}$ $\text{ The number of trials }=60$ $\text{ (a) P( }1 \text{ turnsup) }=\dfrac{1}{6}$   $\text{ The expected frequency of } 1 \text{ turns up }=\dfrac{1}{6}\times 60=20$ $\text{ (b) The numbers } 3 \text{ and } 6 \text{ are divisible by } 3$.   $\text{ P(a nunmber divisibleby } 3 \text{ turns up) }=\dfrac{2}{6}=\dfrac{1}{3}$   $\text{ The expected frequency of a number divisible by } 3 \text{ turns up }=\dfrac{1}{3}\times 60=20$ $\text{ (c) The numbers } 1, 2, 3 \text{ and } 6 \text{ are factors of } 6$.   $\text{ P(a factor of } 6 \text{ turns up) }=\dfrac{4}{6}=\dfrac{2}{3}$   $\text{ The expected frequency of a factor of } 6 \text{ turns up }=\dfrac{2}{3}\times 60=40$



3. Two honest coins are tossed. How many times would you expect to obtain two heads in $200$ trials?





$\text{ Let } H = \text{ head }, T = \text{ tail }$ $\text{ The samples spaces is } \{HH, HT, TH, TT\}$. $\text{ P(two heads) }=\dfrac{1}{4}$ $\text{ The number of trials }=200$ $\text{ The expected frequency of times to obtain two head }=\dfrac{1}{4}\times 200=50$



4. The probability of scoring $12$ when throwing two dice at once is $\dfrac{1}{36}$. If such an experiment is repeated $720$ times, what is the expected frequency of the score not being $12$?





$\text{ P(scoring } 12 \text{ ) } =\dfrac{1}{36}$ $\text{ The number of trials }=720$ $\begin{aligned} \text{ P(the score not being }12) &=1-\text{ P(scoring }12)\\ &=1-\dfrac{1}{36}\\ &=\dfrac{35}{36} \end{aligned}$ $\text{ Expected frequency of the score not being } 12=\dfrac{35}{36}\times 720=700$



5. A spinner is equally lilely to point to any one of the numbers: $1$, $2$, $3$, ..., $10$.
   (a)  What is the probability of an odd number?
   (b)  What is the probability of an even number?
   (c)  If the arrow is spun $1000$ times, what final score would you expect if all the individual scores are added together?





$1$, $2$, $3$,...,$10$.   $\text{ There are } 10 \text{ numbers. }$ $\text{ (a) The number } 1, 3, 5, 7, 9 \text{ are odd numbers. }$   $\text{ P(an odd number) }=\dfrac{5}{10}=\dfrac{1}{2}$ $\text{ (b) The numbers } 2, 4, 6, 8, 10 \text{ are even numbers. }$   $\text{ P(an even nunmber) }=\dfrac{5}{10}=\dfrac{1}{2}$ $\text{ (c) The number of trials }=1000$.   $\text{ P(each number) }=\dfrac{1}{10}$   $\text{ The expected frequency of each number }=\dfrac{1}{10}\times 1000=100$ $\begin{aligned} \text{ If all the individual scores are added together },& \\ \text{ The expected final score }=&(1\times 100)+(2\times 100)+(3\times 100)+(4\times 100)+(5\times 100)\\ &+(6\times 100)+(7\times 100)+(8\times 100)+(9\times 100)+(10\times 100)\\ =&(1+2+3+4+5+6+7+8+9+10)\times 100\\ =&(55)\times 100\\ =& 5500 \end{aligned}$