12th Class Mathematics Probability Notes - Mathematics Olympiads - Probability

Notes - Mathematics Olympiads - Probability

Category : 12th Class

Probability

• Conditional Probability: The probability of event A is called the conditional probability of A given that event B has already occurred. It is written as $P(A|B)$or $P\left( \frac{A}{B} \right)$.Mathematically it is given by the formula $P(A|B)=\frac{P(A\bigcap B)}{P(B)}$

Event A is independent of B if the conditional probability of A given B is the same as the unconditional probability as A,i.e, they are independent if $P(A|B)=P(A)$

'Gender gap' in politics is a well-known example of conditional probability and independence in the real life.

Suppose candidate A receive 55% of the entire vote and the only 45% of the female vote. Let $P(R)=$ Probability that a random person voted for A But $P\left( \frac{W}{R} \right)=$ Probability of a random women voted for A

$\therefore P(R)=0.55$and $P\left( \frac{W}{R} \right)=0.45$

Then $P\left( \frac{W}{R} \right)$ is said to be the conditional probability of W given R.

• If $P(R)\ne P\left( \frac{W}{R} \right)$ then there exists a gender gap in politics. On the other hand. If $P(A)=P\left( \frac{A}{W} \right),$ then there is no gender gap. i.e. the probability that a person voted for A is independent of the gender gap.

• Independent Events: Let A and B be the two events. If$P(A\bigcap B)=P(A).\,P(B)$, then A and B are said to be the independent events.

• Mutually Exclusive Events: The events are said to be mutually exclusive, if the sets are disjoints i.e. $P(A\bigcap B)=0$i.e. $A\bigcap B=\phi$

In such cases, $P(A\bigcup B)=P(A)+P(B)$

• Disjoint events/sets: Two sets or events are said to be disjoint if they have no element in common.

• Properties of Conditional probability
1. If A and B be the events of a sample space Sand F is an event of S, then

$P(S|F)=P(F|F)=1$

1. If A and B are any two events of sample space S and F is an event of S such that $P(F)\ne 0,$ then

$P[(A\bigcap B)|F]=P(A|F)+P(B|F)-P[A\bigcap B)|F]$ If A and B are disjoint events, then

$P[(A\bigcap B)|F]=P(A|F)+P(B|F)$

1. $P(A'|B)=1-P(A|B)$

• Law of Total Probability: Suppose S is a sample space and the subsets be ${{A}_{1}},{{A}_{2}},{{A}_{3}},...{{A}_{k}},$ then any other event E is union of all the subsets. When $E\cap {{A}_{k}}$ are disjoint, then

$P(E)=P(E\cap {{A}_{1}})+P(E\cap {{A}_{2}})+P(E\cap {{A}_{k}})$

Using the conditional theorem of probability, we have

$P(E\cap {{A}_{k}})=P({{A}_{k}}\cap E)=P({{A}_{k}}).P\left( \frac{E}{{{A}_{k}}} \right)$

• Theorem of Total Probability: Let $\{{{A}_{1}},{{A}_{2}},...{{A}_{n}}\}$ be a partition of sample space S and each of the event has non-zero probability. Let E be any event associated with S, then

$P(E)=P({{A}_{1}}).P\left( \frac{E}{{{A}_{1}}} \right)+P({{A}_{2}}).P\left( \frac{E}{{{A}_{2}}} \right)+...+P({{A}_{n}}).P\left( \frac{E}{{{A}_{n}}} \right)$

$=\sum\limits_{i=1}^{n}{n}({{A}_{i}})P\left( \frac{E}{{{A}_{i}}} \right)$

This is known as theorem of total probability.

• Bayes' Theorem: Suppose the events ${{A}_{1}},{{A}_{2}},...{{A}_{n}}$ form the partitions of the sample space S, i.e., ${{A}_{1}},{{A}_{2}},...{{A}_{n}}$ are pairwise disjoint and ${{A}_{1}}\cup {{A}_{2}}\cup ...\cup {{A}_{n}}=S$ and E is any event. Then

$P\left( \frac{{{A}_{i}}}{E} \right)=\frac{P({{A}_{i}}).P\left( \frac{E}{{{A}_{i}}} \right)}{P({{A}_{1}}).P\left( \frac{E}{{{A}_{1}}} \right)+P({{A}_{2}}).P\left( \frac{E}{{{A}_{2}}} \right)+...P({{A}_{n}}).P\left( \frac{E}{{{A}_{n}}} \right)}$

• Random Variable: In probability, random variables play an important role. A random variable is a special kind of function. It is the real-valued function: S$\to$R, whose domain is the sample space of the random experiment.

It is always denoted by capital letters X, Y, Z... etc.

• Discrete Random Variable: A random variable which can take any only finite or countable infinite numbers of values is said to be the discrete random variable.

• Continuous Random Variable: A random variable which can take any value between given limits is called continuous random variable.

• Sum and Product of Random Variable: Let X and Y be the random variables on the some sample space S. Then X + Y, X + K, KX and XY where K is any real number, are the functions of S defined as

$(X+Y)(S)=X(S)+Y(S)$.

$(K\,X)(S)=K\,X(S).$

$(X+K)S=X(S)+K.$

$(X\,Y)(S)=X(S).Y(S).$

• Probability Distribution of Random Variable: Let X be a finite random variable on a sample space S. i.e. X is assigned or correspondent only to a finite number of values of S.

${{R}_{x}}=\{{{x}_{1}},{{x}_{2}},{{x}_{3}},...{{x}_{n}},\},$provided$({{x}_{1}}<{{x}_{2}}<{{x}_{3}}<{{x}_{4}}..{{x}_{n}}\}.$Then X induces a function f which assign the probability to the points in${{R}_{x}}.$

$\therefore$      $f({{x}_{k}})=P(X={{x}_{k}})=P\{\}s\in S:X(S)={{x}_{k}}\}$

$\therefore$      In the set of ordered pair. It can be written as $\{{{x}_{k}},f({{x}_{k}})\}.$

$\therefore$

 $x$ ${{x}_{1}}$ ${{x}_{2}}$ ${{x}_{3}}$ ${{x}_{4}}$ $f(x)$ $f({{x}_{1}})$ $f({{x}_{2}})$ $f({{x}_{3}}).....$ $f({{x}_{4}})$

Here, this function f is said to be the probability distribution of the random variable X.

$\therefore$      $f({{x}_{i}})\ge 0$and$\sum f({{x}_{k}})=1$

Solved Example

1. If two coins are tossed, then find the probability of occurring 0, 1 and 2 heads.

Sol. Let X be the event of occurring head.

Now,

$P(X=0)=P(T,T)=\frac{1}{2}.\frac{1}{2}=\frac{1}{4}$

$P(X=1)=P(H,T)+P(T\,H)=\left[ \frac{1}{2}\times \frac{1}{2} \right]+\left[ \frac{1}{2}\times \frac{1}{2} \right]=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

$P(X=2)=P(H\,H)=\frac{1}{2}.\frac{1}{2}=\frac{1}{4}$

In the table from it can be shown as follows:

 $x:$ 0 1 2 $p(x):$ $\frac{1}{4}$ $\frac{1}{2}$ $\frac{1}{4}$.

• Expectation of a finite random Variable: Let x be a finite random variable and suppose its distributions are:

 $x$ ${{x}_{1}}$ ${{x}_{2}}$ ${{x}_{3}}$ . …. … ${{x}_{n}}$ $f(x)$ $f({{x}_{1}})$ $f({{x}_{2}})$ $f({{x}_{3}})$ … $f({{x}_{n}})$

The mean expected value (expectation) of x is denoted as $E(x)$ and is defined by $E=E(x)={{x}_{1}}.f({{x}_{1}})+{{x}_{2}}$

$f({{x}_{2}})+...{{x}_{n}}+f({{x}_{n}}).=\sum {{x}_{i}}f({{x}_{i}})$

$E(x)=\sum {{x}_{i}}{{P}_{i}}={{x}_{1}}{{P}_{1}}+{{x}_{2}}{{P}_{2}}+...{{x}_{n}}.{{P}_{n}}+[\because f({{x}_{i}})={{P}_{i}}]$

Solved Example

1. If X and Y be the random variables with the distributions

 ${{X}_{i}}$ 2 3 6 10 ${{P}_{i}}$ 0.2 0.2 0.5 0.1

and

 ${{Y}_{i}}$ 0.8 0.2 0 3 7 ${{P}_{i}}$ 0.2 0.3 0.1 0.3 0.1

Then find the expected value of X and Y.

Sol.      $E(x)=\sum {{X}_{i}}{{P}_{i}}$

$=2\times (0.2)+3\times 0.2+6\times 0.5+10\times 0.1$

$=0.4+0.6+3.0+1.0=5$

$E(Y)=\sum {{Y}_{i}}{{P}_{i}}$

$=(0.8)\times (0.2)+(0.2)(0.3)+0\times 0.1+3\times 0.3+7\times 0.1$

$=0.16+0.06+0+0.9+0.7=0.38$

• Variance: For a random variable A whose possible values ${{a}_{1}},{{a}_{2}}$……...${{a}_{n}}$ occur with the probabilities respectively

$p({{a}_{1}}),$$p({{a}_{2}}),$-------$p({{a}_{n}})$respectively

Let $\mu =E(A)$ be the mean of A. Then the variance of A is given by

Var(A) or $\sigma _{a}^{2}=E{{(A-\mu )}^{2}}$

• Relation between Variance and Expectation

Var$(A)=E({{A}^{2}})-{{[E(A)]}^{2}}$,

where$E({{A}^{2}})=\underset{i=1}{\overset{n}{\mathop{\sum }}}\,a_{i}^{2}p({{a}_{i}})$

• Binomial Distribution: Consider a random experiment and an event E associated with it. Let p = probability of occurrence of event E in the one trial and $q=1-p=$ Probability of non-occurrence of event E in one trial.

If X denotes the number of success in n trials of the random experiment. Then $P(X=r)=$ Probability of r successes ${{=}^{n}}{{C}_{r}}{{p}^{r}}.{{q}^{n-r}}$

Note:

1. Probability of at most r successes inn trials $=\sum\limits_{n=0}^{r}{^{n}{{C}_{r}}.{{p}^{r}}.{{q}^{n-r}}}$
2. Probability of at least r successes in n trials $=\sum\limits_{n=r}^{r}{^{n}{{C}_{r}}.{{p}^{r}}.{{q}^{n-r}}}$
3. Probability of having first success at the rth trial$=p.{{q}^{r-1}}.$

• Important results for binomial distribution

For a binomial distribution B(n,p)

(i) Mea nor expected no. of successes, $\mu =np$

(ii) Variance, $\sigma =npq$

(iii) Standard deviation, $s=\sqrt{npq}$

Solved Example

1. A die is thrown 6 times. If getting an odd number is a success, then what is the probability of (i) 5 successes (ii) at least 5 successes.

Sol. In throwing a dice, odd numbers will be 1, 3 and 5.

$\therefore$ No. of favourable cases = 3

No. of exhaustive cases = 6

P = probability of getting an odd number $=\frac{3}{6}=\frac{1}{2}$

$\Rightarrow$ q = probability of not getting an odd number $=1-\frac{1}{2}=\frac{1}{2}$

$P(X=r)={}^{n}{{C}_{r}}.{{p}^{r}}.{{q}^{n-r}}={}^{6}{{C}_{r}}.{{p}^{r}}.{{q}^{6-r}}$

$={}^{6}{{C}_{r}}.{{\left( \frac{1}{2} \right)}^{r}}.{{\left( \frac{1}{2} \right)}^{6-r}}={}^{6}{{C}_{r}}.{{\left( \frac{1}{2} \right)}^{6}}=\frac{1}{64}.{}^{6}{{C}_{r}}$

(i)  Probability of 5 successes$=P(x=5)=\frac{1}{64}{}^{6}{{C}_{5}}=\frac{1}{64}\times 6=\frac{3}{32}$

(ii) Probability of at least 5 successes

$=P(X=5)+P(X=6)$

$=\frac{1}{64}{}^{6}{{C}_{5}}+\frac{1}{64}\times {}^{6}{{C}_{6}}=\frac{3}{32}+\frac{1}{64}\times 1\frac{6+1}{64}=\frac{7}{64}$

(iii) Probability at most 5 successes

$=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)$

$=1-P(X=6)=1-\frac{1}{64}=\frac{63}{64}$

1. A dice is rolled 20 times. If getting a number greater then 4 is considered as a success, then find the mean, variance and standard deviation of the success.

Sol. In rolling a dice, the numbers greater than 4 are 5 or 6.

So, probability of success, $p=\frac{2}{6}=\frac{1}{3}$

and probability of failure, $q=1-\frac{1}{3}=\frac{2}{3}$

$\therefore$ Mean$=np=20\times \frac{1}{3}=\frac{20}{3}$

Variance, ${{\sigma }^{2}}=npq=20\times \frac{1}{3}\times \frac{2}{3}=\frac{40}{9}$

Standard deviation $=\sqrt{npq}=\sqrt{20\times \frac{1}{3}\times \frac{2}{3}}=\sqrt{\frac{40}{9}}=\frac{2}{3}\sqrt{10}$

1. A fair coin is tossed 100 times. Find the probability p that head occurs less than 45 times.

Sol. This is the bionomial experiment $B(n,p)=$with $n=100$

$\therefore p=0.5$and $1-0.5=0.5$

Mean or expected no. of success $=np=100\times 0.5=50$

Variance, ${{\sigma }^{2}}=npq=100\times \frac{1}{2}\times \frac{1}{2}=25$

$\therefore \sigma =\sqrt{25}=5$

$\therefore$ We seek ${{B}_{p}}(k<45)={{B}_{p}}(k\le 44).$ or, Approximately, normal probability ${{N}_{p}}(x\le 44.5)$Transforming,$a=44.5$into standard unit

$\therefore {{z}_{1}}\frac{a-\mu }{6}=\frac{44.5-50.0}{5}=\frac{-5.5}{5}=-1.1$

$\therefore$ Here ${{z}_{1}}<0$

$\therefore p={{B}_{p}}(k\le 44)\approx {{N}_{p}}(x\le 44.5)={{N}_{p}}(x=1.1)=0.5-f(1.1)$

$=0.5-0.3643=6.1357$

1. A fair coin is tossed 6 times. If calls a head is considered as a success, then find the probability that

(b) at least 4 heads occurs.

(c) at least one head occur.

Sol. Here no of trials, $n=6$

Probability of success, $p=\frac{1}{2}$

$\therefore$ probability of failure, $q=1-\frac{1}{2}=\frac{1}{2}$

(a) Probability of getting exactly 2 heads

$=P(2)={}^{6}{{C}_{2}}{{\left( \frac{1}{2} \right)}^{2}}.{{\left( \frac{1}{2} \right)}^{6-2}}={}^{6}{{C}_{2}}{{\left( \frac{1}{2} \right)}^{6}}$

$=15\times \frac{1}{64}=\frac{15}{64}$

(b) Probability of getting at least 4 heads $=P(4)+P(5)+P(6)$

$={}^{6}{{C}_{4}}{{\left( \frac{1}{2} \right)}^{2}}.{{\left( \frac{1}{2} \right)}^{4}}+{}^{6}{{C}_{5}}{{\left( \frac{1}{2} \right)}^{5}}.\left( \frac{1}{2} \right)=15.{{\left( \frac{1}{2} \right)}^{6}}+6.{{\left( \frac{1}{2} \right)}^{6}}+1{{\left( \frac{1}{2} \right)}^{6}}+{}^{6}{{C}_{6}}{{\left( \frac{1}{2} \right)}^{0}}.{{\left( \frac{1}{2} \right)}^{6}}$

$={{\left( \frac{1}{2} \right)}^{6}}(15+6+1)=\frac{22}{64}=\frac{11}{32}$

(c) The probability of getting no head (i.e. all failures) $={{q}^{6}}={{\left( \frac{1}{2} \right)}^{6}}=\frac{1}{64}$

$\therefore$ Probability of 1 or more head

$=1-{{q}^{4}}=1-\frac{1}{64}=\frac{63}{64}$

1. Two man A and B fire at a target. Suppose $P(A)=\frac{1}{3},$$P(B)=\frac{1}{5},$ denote their proabilities of hitting the target. (Consider A and B be independent events). Find the probability that

(a) A does not hit the target                      (b) Both hit the target

(c) One of them hit the target                    (d) Neither hits the target

Sol. (a) P (A does not hit the target) $=P(\bar{A})=P({{A}^{c}})=1-P(A)=1-\frac{1}{3}=\frac{2}{3}$

(b) Since A and Bare independent events

$\therefore$ P(A and B)$=P(A\cap B)=P(A).P(B)=\frac{1}{3}.\frac{1}{5}=\frac{1}{15}$

(c) Probability of one of them hit the target

= P(A or B) $=P(A\cup B)$

$=P(A)+P(B)-P(A\cap B)$(By addition rule)

$=\frac{1}{3}+\frac{1}{5}-\frac{1}{15}=\frac{8-1}{15}=\frac{7}{15}$

(d) Probability of neither hits the target.

$\therefore$ (Neither A nor B) $=P\{{{(A\cup B)}^{c}}\}=1-P(A\cup B)=1-\frac{7}{15}=\frac{8}{15}$

1. Three fair coins – a penny, a nickel and a dime are tossed. Find the Probability p that they are all heads if (a) the penny is head (b) at least one of the coin is head (c) the dice is tails.

Sol. Sample Space, S = {HHH, HHT, HTT, HTH, THT, TTH, THH, TTT}

(a) E = penny is head = {HHH, HHT, HTH, HTT}

$P(E)=\frac{1}{4}$

(b) P(at least one of the coin is head)

$=\frac{1}{7}$

(c)

$P=0$

1. A fair coin is tossed untill a head or five tails occurs. Find the expected number E of tosses of the coin.

Sol. The sample space S consists of six points.

H, TH, TTH, TTTH, TTTTH, TTTTT

Their respective probabilities (independent trails) are

$\frac{1}{2},{{\left( \frac{1}{2} \right)}^{2}},{{\left( \frac{1}{2} \right)}^{3}},{{\left( \frac{1}{2} \right)}^{4}},{{\left( \frac{1}{2} \right)}^{5}}$

and

${{\left( \frac{1}{2} \right)}^{5}}$

$\therefore X(H)=1[X(S)\to$

1. of tosses in each outcome]

$X(TH)=2$

$X(TTH)=3$

$X(TTTH)=4$

$X(TTTTH)=5$

$X(TTTTT)=5$

$\therefore P(1)=P(H)=\frac{1}{2}$                    $P(2)=P(\text{TH})={{\left( \frac{1}{2} \right)}^{2}}$

$P(3)=P(\text{TTH})={{\left( \frac{1}{2} \right)}^{3}}$                        $P(4)=P(\text{TTTH})={{\left( \frac{1}{2} \right)}^{4}}$

$P(5)=P(\text{TTTTH})={{\left( \frac{1}{2} \right)}^{5}}$        $P(5)=P(\text{TTTTT})={{\left( \frac{1}{2} \right)}^{5}}$

$E(X)=1.\left( \frac{1}{2} \right)+2.{{\left( \frac{1}{2} \right)}^{2}}+3{{\left( \frac{1}{2} \right)}^{3}}+4{{\left( \frac{1}{2} \right)}^{4}}+5{{\left( \frac{1}{2} \right)}^{5}}+5{{\left( \frac{1}{2} \right)}^{5}}$

$=\left( \frac{1}{2} \right)+2\left( \frac{1}{4} \right)+3\left( \frac{1}{8} \right)+4\left( \frac{1}{16} \right)+5\left( \frac{1}{32} \right)+5\left( \frac{1}{32} \right)$

$=\frac{31}{16}\approx 1.9$

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Notes - Mathematics Olympiads - Probability

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