Notes - Mathematics Olympiads - Probability

Category : 11th Class

 

                                                                                             Probability

 

  • Probability: Actually, Probability is the mathematical modelling of chances or outcome of the events. In other, it is the branch of mathematics in which we study the occurrence of any element in the numerically form. It always lies between 0 & 1.

i.e.        \[0\le P(E)\le 1\]

Where P (E) = Probability of occurrence of the event E.

 

  • Some basic terms and its concepts

 

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.

 

  • Outcome and Sample Space: A possible result of a random experiment is said to be its outcome/results.

 

  • Sample space: The set of all possible outcome of an experiment is called a sample space.

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}.

 

  • Event: An event is the subset of the sample space.
  • g. when a dice is thrown the 6 is appeared then i.e. the occurrence of 6 is an event. In other word

 

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.

 

  • Probability of an event: Here, we will define the probability into two ways:
  1. Mathematical (or a priori) definition
  2. Statistical (or empirical) definition.

 

  • Mathematical definition of Probability: Probability of an event A, denoted as P(A), is defined as P(A)

 

\[=\frac{Number\,\,of\,\,cases\,\,favourable\,\,to\,\,A}{Number\,\,of\,possible\,\,outcome}\]

 

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.

           

\[=\frac{n(E)}{n(S)}=\frac{3}{6}=\frac{1}{2}\]

 

  1. A coin is tossed once, what are the all possible outcome? What is the probability of the coin coming of tails?

 

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) = \[\frac{1}{2}\]

 

  1. What is the probability of drawing a king from a well-shuffled deck of 52 cards.

 

Sol.      1st of all we have to know how many kings are there in the well-shuffled deck

There are four kings in well-shuffled 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.)

                       

\[=\frac{4}{52}=\frac{1}{13}\]

 

  • Axiom of Probability of the Event E

 

(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.        \[P(A\cup B)=P(A)+P(B)\]

 

  • Odd in favour and odd against of an event: Let us consider, there are a outcome favourable to a certain event & b outcome unfavourable to the event in a sample space. Then Odd in favour of an event

 

\[=\frac{Number\,\,o\text{f}\,\,\text{f}avourable\,\,outcome}{Number\,\,o\text{f}\,\,un\text{f}avourable\,\,outcome}=\frac{a}{b}.\] 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 Non-occurrence of \[E=\frac{b}{a+b}\]

 

  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 count is head)\[=\frac{1}{7}\]

(c)        \[P=0\]

 

  1. Two man A and B fire at target. Suppose \[\mathbf{P(A)=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{,}\] \[\mathbf{P(B)=}\frac{\mathbf{1}}{\mathbf{5}}\] denote their probabilities 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)        \[p(not\,A)=P({{A}^{c}})=1-P(A)=1-\frac{1}{3}=\frac{2}{3}\]

 

(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)

           

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

 

(d) Neither hits the target.

                        \[\therefore \,\,\,\,P(Niether\,\,A\,\,nor\,\,B)=P\{{{(A\cup B)}^{c}}\}\]

                        \[=1-P(A\cup B)\]

 

                        \[=1-\frac{7}{15}=\frac{8}{15}\]

 

  • Conditional Probability

 

The well-known example of conditional probability and independence in the real life according is

  1. Gender gap
  2. Insurance rates

 

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.

 

  • Mathematically Conditional probability is defined as: Suppose A is any event B is another event which is occurred after the occurrence of any event A. Then probability of occurrence of B after the occurrence of A is denoted by \[P\left( \frac{B}{A} \right).\] Mathematically, \[\frac{P\left( \frac{B}{A} \right)}{1}=\frac{P(A\cap B)}{P(A)}\]

 

\[\Rightarrow P(A\cap B)=P(A).P\left( \frac{B}{A} \right)\]

 

  • Independent Event: Let A and B two events

          

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:

           

  1. \[P\left( \frac{B}{A} \right)=\frac{P(A\cap B)}{P(A)}\]

           

\[\Rightarrow \,\,P(A\cap B)=P(A).P\left( \frac{B}{A} \right)=P(B).P\left( \frac{A}{B} \right)\]           

  1. If A and B be independent events. Then \[P\left( \frac{A}{B} \right)=P(A)\]
  2. If A and B are two events such that \[A\ne 0\] then \[P\left( \frac{B}{A} \right)+P\left( \frac{\overline{B}}{A} \right)=1\]

           

Similarly,         \[B\ne 0\] Then \[P\left( \frac{A}{B} \right)+P\left( \frac{\overline{A}}{B} \right)=1\]

 

  • Law of Total Probability: Suppose S is a sample space and the subsets be \[{{A}_{1}},\,{{A}_{2}},\,{{A}_{3}}....{{A}_{n}}\] An of E is an event. When E Ç \[{{A}_{K}}\]are disjoint, then

 

\[P(E)=P(E\cap {{A}_{1}})+P(E\cap {{P}_{2}})+....P(E\cap {{A}_{K}})\]

 

Using the conditional theorem of probability, we have

 

\[P(E\cap {{A}_{K}})=P({{A}_{K}}\cap E)=P({{A}_{K}}).P\left( \frac{E}{{{A}_{K}}} \right)\]

 

  • Theorem of Total Probability: Let E be an event in the sample space S and let be \[{{A}_{1}},\,{{A}_{2}},\,{{A}_{3}}....{{A}_{n}}\] mutually disjoint events whose union is S. Then

 

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

 

\[+....P({{A}_{K}}).P\left( \frac{E}{{{A}_{K}}} \right)\]

This theorem is said to be the law of total probability.

 

  • Bayes' Theorem: Suppose the events \[{{A}_{1}},\,{{A}_{2}},\,{{A}_{n}}\] do form the partitions of the sample space S, and E is any event. Then for K = 1, 2,.... n. the multiplication for conditional probability be written as

 

\[P({{A}_{K}}\cup E)=P({{A}_{K}}).P\left( \frac{E}{{{A}_{K}}} \right)\]

 

\[\therefore \,\,\,P\left( \frac{{{A}_{K}}}{E} \right)=\frac{P({{A}_{K}}\cap E)}{P(E)}=\frac{P({{A}_{K}}).P\left( \frac{E}{{{A}_{K}}} \right)}{P(E)}\]

 

\[\frac{P({{A}_{K}}).P\left( \frac{E}{{{A}_{K}}} \right)}{P({{A}_{1}}).P\left( \frac{E}{{{A}_{1}}} \right)+P({{A}_{2}}).P\left( \frac{E}{{{A}_{2}}} \right)+....P({{A}_{n}}).P\left( \frac{E}{{{A}_{n}}} \right)}\]

 

  • Random Variable: In Probability, random variables play in important role. A random variable is a special kind of function. It is the real-value 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 descrete 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 i random variable on the some sample space S. Then X + Y, X + K, K.X and Xy where K is any real number, are the function an S defined as:

 

\[(X+Y)(S)=X(S)+Y(S).\]

\[(KX)(S)=KX(S).\]

\[(X+K)S=X(S)+K.\]

\[(XY)(S)=X(S).Y(S).\]

 

  • Probability Distribution of Random Variable: Let X be a finite random variable an a sample spaces, i.e. X assigned or correspondent only a finite number of value to 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 \[{{R}_{x}}.\]

\[\therefore \,\,\,\,\,\text{f}({{x}_{k}})=P(X={{x}_{k}})=P\{\}s\in S:X(S)={{x}_{k}}\}\]

\[\therefore \]      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):

\[\frac{1}{4}\]

\[\frac{1}{2}\]

\[\frac{1}{4}\]

 

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

 

\[x\]

\[{{x}_{1}}\]

\[{{x}_{2}}\]

\[{{x}_{3}}\]

.

….    ….

\[{{x}_{4}}\]

\[\text{f(}x)\]

\[\text{f(}x{{)}_{1}}\]

\[\text{f(}x{{)}_{2}}\]

\[\text{f(}x{{)}_{3}}\]

 

\[\text{f(}x{{)}_{n}}\].

 

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.

 

  1. Let X and Y be the random variable with the following respective distribution.

           

\[{{X}_{i}}\]

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

            \[=E(X)=S{{x}_{i}}{{P}_{i}}\]

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

            \[=0.4+0.6+3.0+1.0=5\]

            \[EY=S{{y}_{i}}{{p}_{i}}=(-8)\times (0.2)+(-2).0.3+0\times 0.1+3\times 0.03+7\times 0.1=-0.16+(-0.6)+0+0.9+0.7=1.6-1.6-0.6=-0.6\]

 

  • Binomial Distribution

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 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 successeds \[{{=}^{n}}{{C}_{r}}{{\operatorname{P}}^{r}}.{{q}^{n-r}}\]

 

Note:

  1. Probability of at most r successes in n trials.

           

            \[=\sum\limits_{n=0}^{r}{^{n}Cr.{{p}^{r}}.{{q}^{n-r}}}\]

 

  1. Probability of at least r successes in n trials.

           

            \[=\sum\limits_{n=r}^{r}{^{n}Cr.{{p}^{r}}.{{q}^{n-r}}}\]

 

  1. Probability of having first success at the \[{{r}^{th}}\] trial

            \[=p.{{q}^{r-1}}.\]

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 \[P{{(x=r)}^{n}}{{C}_{r}}.{{p}^{r}}.{{q}^{n-r}}{{=}^{6}}{{C}_{r}}.{{p}^{r}}.{{q}^{6-r}}\]

 

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

 

(i)         Probability 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

           

\[=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\]

 

\[=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)-P(X=6)\]

 

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

 

  • Important Results

Binomial Distribution B (n, P)

Mean or expected no. of successes u = np

Variance, \[{{s}^{2}}=npq\]

Standard deviation \[s=\sqrt{npq}\]

 

  1. A die is rolled 20 times. Getting a number greater then 4 is a success. Find the mean, variance and standard deviation of the success.

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 \[=\sqrt{npq}=\sqrt{20\times \frac{1}{3}\times \frac{2}{3}}\]

 

\[=\sqrt{\frac{40}{9}}=\frac{2}{3}\sqrt{10}\]

 

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

 

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 \[Bp(k<45)=Bp(k\le 44).\]

  1. Approximately, normal probability \[Np(x\le 44.5)\]

Transforming, a = 44.5 into standard unit

            \[\therefore \,\,\,\,{{z}_{1}}=\frac{a-\mu }{6}=\frac{44.5-5.0}{5}=\frac{-5.5}{5}=-1.1\]

 

\[\therefore \]      Here \[\therefore \,\,\,\,{{z}_{1}}<0\]

 

\[\therefore \,\,\,p=BP(k\le 44)\approx NP(x\le 44.5)\]

                        \[NP(x=1.1)=0.5-\text{f}(1.1)=0.5-0.3643=0.1357\]

[f (1.1=0.3643) By normal distribution Table]

 

  1. A fair coin is tossed 6 times, call a heads a success. Find the probability that

(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

                       

\[P(2){{=}^{6}}{{C}_{2}}{{\left( \frac{1}{2} \right)}^{2}}.{{\left( \frac{1}{2} \right)}^{6-2}}{{=}^{6}}{{C}_{2}}{{\left( \frac{1}{2} \right)}^{6}}=15\times \frac{1}{64}=\frac{15}{64}\]

 

(b)        Probability of getting at least 4 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 \[=1-{{q}^{4}}=1-\frac{1}{64}=\frac{63}{64}\]

 

 

  1. Two man A and B fire at target. Suppose \[\mathbf{P(A)=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{,}\] \[\mathbf{P(B)=}\frac{\mathbf{1}}{\mathbf{5}}\] denote their probabilities 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)        \[p(not\,A)=P({{A}^{c}})=1-\frac{1}{3}=\frac{2}{3}\]

 

(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)

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

 

(d)        Neither hits the target.

            \[\therefore \,\,\,\,\,P(Neither\,\,A\,\,nor\,\,B)=P\{{{(A\,\grave{E}\,B)}^{c}}\}\]

                \[=1-P(A\,\grave{E}\,B)=1-\frac{7}{15}=\frac{8}{15}\]

Notes - Mathematics Olympiads - Probability
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