Notes  Mathematics Olympiads  Probability
Category : 11th Class
Probability
i.e. \[0\le P(E)\le 1\]
Where P (E) = Probability of occurrence of the event E.
Random experiment of Trial: An experiment of event or trial of event does not follow any rule of system is said to be random experiments, e.g. throwing a dice in which one of {1, 2, 3, 4, 5, 6} will be occurred. We cannot predict that in it throwing if integer 4 is occurred then the next throwing dice. 3 or 4 or any certain number will be occurred. It cannot be predicted. It is the random experiment and not to consider, throwing dice follows any rule/system.
Generally, it is denoted by S. e.g. when a coin is tossed then, whatever, H (Head) or T (Tale) is occurred, i.e. S = {H,T}.
To throw a dice then Sample space
S = {1, 2, 3, 4, 5, 6} and an event E = {6}.
Note: Here, we will study the probability which is based on set theory.
\[=\frac{Number\,\,of\,\,cases\,\,favourable\,\,to\,\,A}{Number\,\,of\,possible\,\,outcome}\]
e.g. To throw a dice, what is the probability of occurrence of even numbers. Usually, someone can ask this type of question. Then Sample Space, S = {1, 2, 3, 4, 5, 6} E = event of occurrence of even numbers = {2, 4, 6} n(S) = total o. of element/member of sample space = 6
n (E) = no. of element of event = 3
So, P (E) = Probability of occurrence of even no.
\[=\frac{n(E)}{n(S)}=\frac{3}{6}=\frac{1}{2}\]
Sol. When, a coin is tossed, as usually, head (H) or Tail (T) can be appeared, i.e. Net consider sample other things that coin will be standard strictly. Only we have to think fruitfully possible outcome.
Here, S= {H, T} and n (S) = 2
Probability of occurrence of tail: P (T) = \[\frac{1}{2}\]
Sol. 1st of all we have to know how many kings are there in the wellshuffled deck
There are four kings in wellshuffled deck.
Let E = event of occurrence of a king
\[\therefore \,\,\,(E)=4=\] Total no. of kings in deek of 52 cards
\[\therefore \,\,\,\,P(E)=\frac{n(E)}{n(S)}\] (By mathematical deinitin of prob.)
\[=\frac{4}{52}=\frac{1}{13}\]
(a) Probability is always greater and equal to 0
\[\therefore \,\,\,\,\,\,P(E)\ge 0\]
(b) Probability of any event lies between 0 and 1
\[\therefore \,\,\,\,\,\,0\le P(E)\le 1\]
(c) Probability of sample space/certainity event is P(s) =1
(d) Probability of null event/empty set/event is zero.
\[\therefore \,\,\,\,\,P(\phi )=0\]
Mutually exclusive events: Let two events A and B is said to mutually exclusive events. If two events A and B are disjoint, i.e. there are no common elements between A and B.
i.e. \[A\bigcap B=\phi .\] Probability of occurrence of A and B is written as
\[P(\cup B)=P(A)+P(B)\]
\[\therefore \] As we mention easily that probability theory is based an set theory. Maximum results of set theory are used in probability.
Probability of occurrence of two event A and B is written
\[P(A\cup B)=P(A)+P(B)P(A\cap B)\]
If A and B be disjoint then
\[P(A\cap B)=\phi \]
i.e. \[P(A\cup B)=P(A)+P(B)\]
\[=\frac{Number\,\,o\text{f}\,\,\text{f}avourable\,\,outcome}{Number\,\,o\text{f}\,\,un\text{f}avourable\,\,outcome}=\frac{a}{b}.\] Similarly odd against of an event
\[=\frac{Number\,\,o\text{f}\,\,un\text{f}avourable\,\,outcome}{Number\,\,o\text{f}\,\,\text{f}avourable\,\,outcome}=\frac{b}{a}.\] Through, the odd in favour of an event and odd against of an event. We can obtain the probability of an event P (E).
Probability of an event, \[P(E)=\frac{a}{a+b}\]
Probability of Nonoccurrence of \[E=\frac{b}{a+b}\]
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 count is head)\[=\frac{1}{7}\]
(c) \[P=0\]
(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) \[p(not\,A)=P({{A}^{c}})=1P(A)=1\frac{1}{3}=\frac{2}{3}\]
(b) Since A and Bare independent
\[\therefore \,\,\,P(A\,and\,B)=P(A\cap B)=P(A).P(B)\]
\[=\frac{1}{3}.\frac{1}{5}=\frac{1}{15}\]
(c) One of the hit the target
\[=P(A\,or\,B)=P(A\cup B)\]
\[=P(A)\,+P(B)P(A\cap B)\] (By addition rule)
\[=\frac{1}{3}+\frac{1}{5}+\frac{1}{15}=\frac{81}{15}=\frac{7}{15}\]
(d) Neither hits the target.
\[\therefore \,\,\,\,P(Niether\,\,A\,\,nor\,\,B)=P\{{{(A\cup B)}^{c}}\}\]
\[=1P(A\cup B)\]
\[=1\frac{7}{15}=\frac{8}{15}\]
The wellknown example of conditional probability and independence in the real life according is
Suppose candidate A receive 55% of the entire vote and the only 45% of the female vote.
Let P (A) = Probability that a random person voted for A
But \[P\left( \frac{A}{W} \right)=\] Probability of a random women voted for A
\[\therefore \,\,\,P(A)=0.55\,\,P\left( \frac{A}{W} \right)=0.45\]
Then \[\,P\left( \frac{A}{W} \right)\] is said to be the conditional probability of A given W.
If \[P(A)\ne P\left( \frac{A}{W} \right)\] is said to be gender gap in politics. On the other hand \[P(A)=P\left( \frac{A}{W} \right)\] then, there is no gender gap. i.e. the probability that a person voted for A is independent of the gender gap.
\[\Rightarrow P(A\cap B)=P(A).P\left( \frac{B}{A} \right)\]
If \[P(A\bigcap B)=P(A).P(B).\] Then A and B be said to be independent events.
We have to be no confussion about the mutually exclussive and independent events. Then the events/set is said to be mutually exclussive. If the sets are disjoints
i.e. \[P(A\bigcap B)=0\] i.e. \[A\bigcap B=\phi \]
\[\therefore \,\,\,P(A\bigcup B)=P(A)+P(B)\]
When it is said that the events A and B be independent the \[\,P(A\bigcap B)=P(A).P(B)\] Key point to remember:
\[\Rightarrow \,\,P(A\cap B)=P(A).P\left( \frac{B}{A} \right)=P(B).P\left( \frac{A}{B} \right)\]
Similarly, \[B\ne 0\] Then \[P\left( \frac{A}{B} \right)+P\left( \frac{\overline{A}}{B} \right)=1\]
\[P(E)=P(E\cap {{A}_{1}})+P(E\cap {{P}_{2}})+....P(E\cap {{A}_{K}})\]
Using the conditional theorem of probability, we have
\[P(E\cap {{A}_{K}})=P({{A}_{K}}\cap E)=P({{A}_{K}}).P\left( \frac{E}{{{A}_{K}}} \right)\]
\[P(E)=P({{A}_{1}}).P\left( \frac{E}{{{A}_{1}}} \right)+P({{A}_{2}}).P\left( \frac{E}{{{A}_{2}}} \right)\]
\[+....P({{A}_{K}}).P\left( \frac{E}{{{A}_{K}}} \right)\]
This theorem is said to be the law of total probability.
\[P({{A}_{K}}\cup E)=P({{A}_{K}}).P\left( \frac{E}{{{A}_{K}}} \right)\]
\[\therefore \,\,\,P\left( \frac{{{A}_{K}}}{E} \right)=\frac{P({{A}_{K}}\cap E)}{P(E)}=\frac{P({{A}_{K}}).P\left( \frac{E}{{{A}_{K}}} \right)}{P(E)}\]
\[\frac{P({{A}_{K}}).P\left( \frac{E}{{{A}_{K}}} \right)}{P({{A}_{1}}).P\left( \frac{E}{{{A}_{1}}} \right)+P({{A}_{2}}).P\left( \frac{E}{{{A}_{2}}} \right)+....P({{A}_{n}}).P\left( \frac{E}{{{A}_{n}}} \right)}\]
\[(X+Y)(S)=X(S)+Y(S).\]
\[(KX)(S)=KX(S).\]
\[(X+K)S=X(S)+K.\]
\[(XY)(S)=X(S).Y(S).\]
\[\therefore \,\,\,\,\,\text{f}({{x}_{k}})=P(X={{x}_{k}})=P\{\}s\in S:X(S)={{x}_{k}}\}\]
\[\therefore \] In the set of ordered pair. It can be written as \[\{{{x}_{k}},\text{f}({{x}_{k}})\}.\]
Here, this function f is said to be the probability distribution or simple distribution of the random variable X.
\[\therefore \,\,\,\text{f}({{x}_{i}})\ge 0\] and \[\,\sum{\text{f(}{{\text{x}}_{\text{k}}}\text{)=1}}\]
e.g. \[\therefore \] If two coin are tossed then the probability of 0, 1 and 2 heads occurred is given by
\[P(X)=0=P(T,T)=\frac{1}{2}.\frac{1}{2}=\frac{1}{4}\]
\[P(X=1)=P(H\,T)+P(T\,H)=\frac{1}{2}.\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\]
\[P(x=2)=P(H\,H)=\frac{1}{2}.\frac{1}{2}=\frac{1}{4}\]
In the table form it can be shown.
X: 
0 
1 
2 
p(x): 
\[\frac{1}{4}\] 
\[\frac{1}{2}\] 
\[\frac{1}{4}\] 
\[x\] 
\[{{x}_{1}}\] 
\[{{x}_{2}}\] 
\[{{x}_{3}}\] 
. 
…. …. 
\[{{x}_{4}}\] 
\[\text{f(}x)\] 
\[\text{f(}x{{)}_{1}}\] 
\[\text{f(}x{{)}_{2}}\] 
\[\text{f(}x{{)}_{3}}\] 

… 
\[\text{f(}x{{)}_{n}}\]. 
The mean, expected value (expectation) of x is denoted as E(X). and is defined by \[E=E(X)={{x}_{1}}.\text{f}({{x}_{1}})+{{x}_{2}}\text{f}({{x}_{2}})+...{{x}_{n}}+\text{f}({{x}_{n}}).=S{{x}_{i}}f({{x}_{i}})\]
\[E(X)=S{{x}_{i}}{{P}_{i}}={{x}_{1}}{{P}_{1}}+{{x}_{2}}{{P}_{2}}+....{{x}_{n}}.{{P}_{n}}\,\,[\because f({{x}_{1}})=Pi]\]
e.g.: A pair of fair dice are tossed. Let X and \[Y\,\,{{X}^{be}}\] the random variable and X denotes the maximum of numbers \[X({{a}_{1}}b)\] and Y denotes the find the expectation of X and Y.
\[{{X}_{i}}\] 
2 
3 
6 
10 
\[{{P}_{i}}\] 
0.2 
0.2 
0.5 
0.1 
and
\[{{Y}_{i}}\] 
0.8 
2 
0 
3 
7 
\[{{P}_{i}}\] 
0.2 
0.2 
0.1 
0.3 
0.7 
Then
Expected value of X
\[=E(X)=S{{x}_{i}}{{P}_{i}}\]
\[=2\times (0.2)+3\times 0.2+6\times 0.5+10\times 0.1\]
\[=0.4+0.6+3.0+1.0=5\]
\[EY=S{{y}_{i}}{{p}_{i}}=(8)\times (0.2)+(2).0.3+0\times 0.1+3\times 0.03+7\times 0.1=0.16+(0.6)+0+0.9+0.7=1.61.60.6=0.6\]
Consider a random experiment ad an event E associated with it. Let p = probability of occurrence of event E in the one trial and q = 1  p= Probability of nonoccurrence 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 successeds \[{{=}^{n}}{{C}_{r}}{{\operatorname{P}}^{r}}.{{q}^{nr}}\]
Note:
\[=\sum\limits_{n=0}^{r}{^{n}Cr.{{p}^{r}}.{{q}^{nr}}}\]
\[=\sum\limits_{n=r}^{r}{^{n}Cr.{{p}^{r}}.{{q}^{nr}}}\]
\[=p.{{q}^{r1}}.\]
e.g. A die is thrown 6 times getting an odd number is a success. What is the probability of (i) 5 successes (ii) at least 5 successes (iii) at most 5 successes.
Sol. In throwing a dice, odd number be \[{}^{3}/{}_{4}\] 1, 3 and 5
\[\therefore \] No. of favourable case = 3
No. of exhaustive case =6
P= probability of getting an odd no. \[=\frac{3}{6}=\frac{1}{2}\]
\[\Rightarrow q=\] probability of not getting odd no. \[=1\frac{1}{2}=\frac{1}{2}\]
Now \[P{{(x=r)}^{n}}{{C}_{r}}.{{p}^{r}}.{{q}^{nr}}{{=}^{6}}{{C}_{r}}.{{p}^{r}}.{{q}^{6r}}\]
\[{{=}^{6}}{{C}_{r}}.{{\left( \frac{1}{2} \right)}^{r}}.{{\left( \frac{1}{2} \right)}^{6r}}{{=}^{6}}{{C}_{r}}.{{\left( \frac{1}{2} \right)}^{6}}=\frac{1}{64}{{.}^{6}}{{C}_{r}}\]
(i) Probability that there are 5 successes
\[P(x=5)={{\frac{1}{64}}^{6}}{{C}_{5}}=\frac{1}{64}\times 6=\frac{3}{32}\]
(ii) Probability that there are 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 that there are almost 5 successes
\[=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\]
\[=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)P(X=6)\]
\[=1P(X=6)=1\frac{1}{64}=\frac{63}{64}\]
Binomial Distribution B (n, P)
Mean or expected no. of successes u = np
Variance, \[{{s}^{2}}=npq\]
Standard deviation \[s=\sqrt{npq}\]
e.g. In rolling a row.
The numbers greater then 4 be 5 or 6.
Given,
\[\therefore \,\,\,\,n=20\]
\[p=\frac{2}{6}=\frac{1}{3}\]
\[q=1\frac{1}{3}=\frac{2}{3}\]
\[\therefore Mean=np=20\times \frac{1}{3}=\frac{20}{3}\]
Variance, \[{{s}^{2}}=npq=20\times \frac{1}{3}\times \frac{2}{9}=\frac{40}{9}\]
Standard deviation \[=\sqrt{npq}=\sqrt{20\times \frac{1}{3}\times \frac{2}{3}}\]
\[=\sqrt{\frac{40}{9}}=\frac{2}{3}\sqrt{10}\]
Sol. This is the binomial experiment B (n, p) = with n = 100
\[\therefore \] p = 0.5 and q =1 0.5 = 0.5
Mean or expected no. of success \[=np=100\times 0.5=50,\]Varience \[{{s}^{2}}=npq=100\times \frac{1}{2}\times \frac{1}{2}=25\]
\[\therefore \,\,\,\sigma =\sqrt{25}=5\]
\[\therefore \] We seek \[Bp(k<45)=Bp(k\le 44).\]
Transforming, a = 44.5 into standard unit
\[\therefore \,\,\,\,{{z}_{1}}=\frac{a\mu }{6}=\frac{44.55.0}{5}=\frac{5.5}{5}=1.1\]
\[\therefore \] Here \[\therefore \,\,\,\,{{z}_{1}}<0\]
\[\therefore \,\,\,p=BP(k\le 44)\approx NP(x\le 44.5)\]
\[NP(x=1.1)=0.5\text{f}(1.1)=0.50.3643=0.1357\]
[f (1.1=0.3643) By normal distribution Table]
(a) Exactly 2 head occur
(b) At least 4 head occurs.
(c) At least one head occur.
Sol. n=6
probability of success, \[p=\frac{1}{2}\]
\[q=1\frac{1}{2}=\frac{1}{2}\]
(a) Here k = exactly 2 head
\[P(2){{=}^{6}}{{C}_{2}}{{\left( \frac{1}{2} \right)}^{2}}.{{\left( \frac{1}{2} \right)}^{62}}{{=}^{6}}{{C}_{2}}{{\left( \frac{1}{2} \right)}^{6}}=15\times \frac{1}{64}=\frac{15}{64}\]
(b) Probability of getting at least 4 head and
i.e. k= 4, 5, 6
\[P(4)+P(5)+P(6){{=}^{6}}{{C}_{4}}{{\left( \frac{1}{1} \right)}^{2}}.{{\left( \frac{1}{2} \right)}^{4}}{{+}^{6}}{{C}_{5}}{{\left( \frac{1}{2} \right)}^{5}}.\left( \frac{1}{2} \right){{+}^{6}}{{C}_{6}}{{\left( \frac{1}{2} \right)}^{0}}.{{\left( \frac{1}{2} \right)}^{6}}\]
\[=15.{{\left( \frac{1}{2} \right)}^{6}}+6.{{\left( \frac{1}{2} \right)}^{6}}+1{{\left( \frac{1}{2} \right)}^{6}}={{\left( \frac{1}{2} \right)}^{6}}(15+6+1)=\frac{32}{64}=\frac{11}{32}\]
(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 \[=1{{q}^{4}}=1\frac{1}{64}=\frac{63}{64}\]
(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) \[p(not\,A)=P({{A}^{c}})=1\frac{1}{3}=\frac{2}{3}\]
(b) Since A and Bare independent
\[\therefore \,\,\,\,\,\,P(A\,and\,B)=P(A\,\,B)=P(A).P(B)=\frac{1}{3}.\frac{1}{5}=\frac{1}{15}\]
(c) One of the hit the target
\[\therefore \,\,\,\,\,\,=P(A\,or\,B)=P(A\,\grave{E}\,B)=P(A)+P(B)P(A\,\,\,B)\](By addition rule)
\[=\frac{1}{3}+\frac{1}{5}\frac{1}{15}=\frac{81}{15}=\frac{7}{15}\]
(d) Neither hits the target.
\[\therefore \,\,\,\,\,P(Neither\,\,A\,\,nor\,\,B)=P\{{{(A\,\grave{E}\,B)}^{c}}\}\]
\[=1P(A\,\grave{E}\,B)=1\frac{7}{15}=\frac{8}{15}\]
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