Notes - Mathematics Olympiads - Probability
Category : 12th Class
Probability
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.
In such cases, \[P(A\bigcup B)=P(A)+P(B)\]
\[P(S|F)=P(F|F)=1\]
\[P[(A\bigcap B)|F]=P(A|F)+P(B|F)-P[A\bigcap B)|F]\] If A and B are disjoint events, then
\[P[(A\bigcap B)|F]=P(A|F)+P(B|F)\]
\[P(E)=P(E\cap {{A}_{1}})+P(E\cap {{A}_{2}})+P(E\cap {{A}_{k}})\]
Using the conditional theorem of probability, we have
\[P(E\cap {{A}_{k}})=P({{A}_{k}}\cap E)=P({{A}_{k}}).P\left( \frac{E}{{{A}_{k}}} \right)\]
\[P(E)=P({{A}_{1}}).P\left( \frac{E}{{{A}_{1}}} \right)+P({{A}_{2}}).P\left( \frac{E}{{{A}_{2}}} \right)+...+P({{A}_{n}}).P\left( \frac{E}{{{A}_{n}}} \right)\]
\[=\sum\limits_{i=1}^{n}{n}({{A}_{i}})P\left( \frac{E}{{{A}_{i}}} \right)\]
This is known as theorem of total probability.
\[P\left( \frac{{{A}_{i}}}{E} \right)=\frac{P({{A}_{i}}).P\left( \frac{E}{{{A}_{i}}} \right)}{P({{A}_{1}}).P\left( \frac{E}{{{A}_{1}}} \right)+P({{A}_{2}}).P\left( \frac{E}{{{A}_{2}}} \right)+...P({{A}_{n}}).P\left( \frac{E}{{{A}_{n}}} \right)}\]
It is always denoted by capital letters X, Y, Z... etc.
\[(X+Y)(S)=X(S)+Y(S)\].
\[(K\,X)(S)=K\,X(S).\]
\[(X+K)S=X(S)+K.\]
\[(X\,Y)(S)=X(S).Y(S).\]
\[{{R}_{x}}=\{{{x}_{1}},{{x}_{2}},{{x}_{3}},...{{x}_{n}},\},\]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
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 |
\[p(x):\] |
\[\frac{1}{4}\] |
\[\frac{1}{2}\] |
\[\frac{1}{4}\]. |
\[x\] |
\[{{x}_{1}}\] |
\[{{x}_{2}}\] |
\[{{x}_{3}}\] |
. |
…. … |
\[{{x}_{n}}\] |
\[f(x)\] |
\[f({{x}_{1}})\] |
\[f({{x}_{2}})\] |
\[f({{x}_{3}})\] |
|
… |
\[f({{x}_{n}})\] |
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
\[{{X}_{i}}\] |
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. \[E(x)=\sum {{X}_{i}}{{P}_{i}}\]
\[=2\times (0.2)+3\times 0.2+6\times 0.5+10\times 0.1\]
\[=0.4+0.6+3.0+1.0=5\]
\[E(Y)=\sum {{Y}_{i}}{{P}_{i}}\]
\[=(0.8)\times (0.2)+(0.2)(0.3)+0\times 0.1+3\times 0.3+7\times 0.1\]
\[=0.16+0.06+0+0.9+0.7=0.38\]
\[p({{a}_{1}}),\]\[p({{a}_{2}}),\]-------\[p({{a}_{n}})\]respectively
Let \[\mu =E(A)\] be the mean of A. Then the variance of A is given by
Var(A) or \[\sigma _{a}^{2}=E{{(A-\mu )}^{2}}\]
Var\[(A)=E({{A}^{2}})-{{[E(A)]}^{2}}\],
where\[E({{A}^{2}})=\underset{i=1}{\overset{n}{\mathop{\sum }}}\,a_{i}^{2}p({{a}_{i}})\]
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:
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
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
\[=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\]
\[=1-P(X=6)=1-\frac{1}{64}=\frac{63}{64}\]
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 \[=\sqrt{npq}=\sqrt{20\times \frac{1}{3}\times \frac{2}{3}}=\sqrt{\frac{40}{9}}=\frac{2}{3}\sqrt{10}\]
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 \[{{z}_{1}}<0\]
\[\therefore p={{B}_{p}}(k\le 44)\approx {{N}_{p}}(x\le 44.5)={{N}_{p}}(x=1.1)=0.5-f(1.1)\]
\[=0.5-0.3643=6.1357\]
(a) exactly 2 heads occur
(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
\[=1-{{q}^{4}}=1-\frac{1}{64}=\frac{63}{64}\]
(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) \[=P\{{{(A\cup B)}^{c}}\}=1-P(A\cup B)=1-\frac{7}{15}=\frac{8}{15}\]
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)
\[P=0\]
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
\[{{\left( \frac{1}{2} \right)}^{5}}\]
\[\therefore X(H)=1[X(S)\to\]
\[X(TH)=2\]
\[X(TTH)=3\]
\[X(TTTH)=4\]
\[X(TTTTH)=5\]
\[X(TTTTT)=5\]
\[\therefore P(1)=P(H)=\frac{1}{2}\] \[P(2)=P(\text{TH})={{\left( \frac{1}{2} \right)}^{2}}\]
\[P(3)=P(\text{TTH})={{\left( \frac{1}{2} \right)}^{3}}\] \[P(4)=P(\text{TTTH})={{\left( \frac{1}{2} \right)}^{4}}\]
\[P(5)=P(\text{TTTTH})={{\left( \frac{1}{2} \right)}^{5}}\] \[P(5)=P(\text{TTTTT})={{\left( \frac{1}{2} \right)}^{5}}\]
\[E(X)=1.\left( \frac{1}{2} \right)+2.{{\left( \frac{1}{2} \right)}^{2}}+3{{\left( \frac{1}{2} \right)}^{3}}+4{{\left( \frac{1}{2} \right)}^{4}}+5{{\left( \frac{1}{2} \right)}^{5}}+5{{\left( \frac{1}{2} \right)}^{5}}\]
\[=\left( \frac{1}{2} \right)+2\left( \frac{1}{4} \right)+3\left( \frac{1}{8} \right)+4\left( \frac{1}{16} \right)+5\left( \frac{1}{32} \right)+5\left( \frac{1}{32} \right)\]
\[=\frac{31}{16}\approx 1.9\]
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