one

Note : Probability of impossible event will be zero.

Let A denotes the event that the batsman did not hit a boundary. As per question total number of trials are 36.                                               Trials in which the event A happened = 36 - 9 =  27.     

                      P(A) = $\frac{27}{36}$ =  0.75

Note that the year 1990 is a non-leap year. Out of two friends, say, Ankita's birthday can be any day of the 365 days in a (non - leap) year. Also Nagma's birthday can be any day of the 365 days of the same year. So, the total number of outcomes = 365 . 

 assume that all these 365 outcomes are equally likely . 

(i) Let E be the event 'Ankita and Nagma have same birthday', then the number of favourable outcomes to the event E = 1

 $\therefore$ P(E) = $\frac{1}{365}$ ,

(ii)  P(Ankita and Nagma  have different birthdays) = P(not E) , 

= P $\overline{\left(E\right)}$  = 1 - P(E)   (concept of complementary)

= 1 -  $\frac{1}{365}$ = $\frac{364}{365}$  

The cards are mixed thoroughly and a card is drawn at random from the box means that all the outcomes are equally likely . 

Sample space = {1, 2, 3,......., 17, 18}, which has 18 equally likely outcomes . 

(i) Let E be the event'the number of the card drawn is divisible by 3 and 2 both', then E = {6, 12, 18}. 

$\therefore$ The number of favourable outcomes to the event E = 3.

$\therefore$ P(divisible by 3 and 2 both) = $\frac{1}{6}$

(ii) Let F be the event'the number of the card drawn is divisible by 3 or 2 both', then F = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18}.

$\therefore$ The number of favourable outcomes to the event F = 12.

$\therefore$ P(divisible by 3 or 2) = $\frac{2}{3}$

The three digit numbers made from the digits 4, 5, 9 (without repetition) are 459, 495, 549, 594, 945, 954 . (using permutations: 3! = 3 x 2 x 1 = 6)

$\therefore$ The sample space S = {459, 495, 549, 594, 945, 954}. 

The sample space has 6 equally likely outcomes . 

(i) Let E be the event 'a multiple of 5' then E = {495, 945} and O(E) = 2 . 

$\therefore$ P(E) = P(a multiple of 5) = $\frac{2}{3} = \frac{1}{3}$ . 

(ii) Note that all the six numbers are divisible by 9.

$\therefore$ P(a multiple of 9) = $\frac{6}{6}$ = 1 , 

(iii) The numbers that end with 9 are 459, 549.

$\therefore$ P(end with 9) = $\frac{2}{6} = \frac{1}{3}$

(i) In  non - leap year , there are 365 days and 364 days make 52 complete weeks. Therefore, we have to find the probability of having a Sunday out of the remaining 1 day. Hence, probability (having 53 Sundays) = 

$\frac{1}{7}$. 

(ii)  In a leap year, there are 366 days and 364 days  make 52 weeks and in each week there is one Sunday. Therefore, we have to find the probability of having a Sunday out of the remaining 2 days . 

Now the 2 days can be (Sunday, Monday) or (Monday, Tuesday) or (Tuesday, Wednesday) or (Wednesday, Thursday) or (Thursday, Friday) or (Friday, Saturday) or (Saturday, Sunday). Note that Sunday occurs 2 times out of in these 7 pairs. 

Let the event be 'having a Sunday' then the number of favourable outcomes to the event = 2 .

$\therefore$  Probability (having 53 Sundays) = $\frac{2}{7}$

All outcomes are equally likely . 

(i) Total number of outcomes = 12. 

Let the event be 'drawing a black ball' , then the number of favourable outcome to the event = x . 

$\therefore$ P(drawing  a black ball) =  $\frac{x}{12}$ , 

(ii) Now, 6 more black balls are put in the bag . 

$\therefore$ Total number of balls in the bag = 12 + 6 = 18 , 

Number of black balls in the bag become x + 6 . 

Let the event be 'drawing a black ball', then the number of favourable outcomes to the event = x + 6 . 

$\therefore$ P(drawing a black ball) = $\frac{x + 6}{18}$ . 

According to the question, $\frac{x + 6}{18}$ = (2) x  $\frac{x}{12}$

 $\Rightarrow$ x + 6 = 3x $\Rightarrow$ 2x = 6 $\Rightarrow$ x = 3 ,

Hence, the value of x is 3 

A box contains 180 apples and one apple is taken out at random, so the sample space has 180 equally likely outcomes . 

Let the box contains x rotten apples, then 

P(rotten apple) =  0.05 (given), 

 $\Rightarrow$ x = 180 x 0.05 = 9. 

$\therefore$ The number of good apples in the box = 180 - 9 = 171 . 

Box contains 15 balls of which x are blue , so the number of red balls in the box = 15 - x , 

$\therefore$ P(red ball) = $\frac{15 - x}{15}$  . 

When 5 red balls are increased , then number of red balls in the box = (15 - x) + 5  = 20 - x . 

Total number of balls in the box = 15 + 5 = 20 . 

$\therefore$ Probability of drawing a red ball in this case = $\frac{20 - x}{20}$ . 

According to given , $\frac{20 - x}{20}$ = (2) x  $\frac{15 - x}{15}$    , 

 $\Rightarrow$  $\frac{20 - x }{4} = \frac{2}{3}$ (15 - x)   $\Rightarrow$  60 - 3x =  120 - 8x , 

$\Rightarrow$  5x =  60  $\Rightarrow$ x = 12 . 

Thus , originally the box contains 12 blue balls and 3 red balls 

We know that the sum of probabilities of all elementary events = 1 , 

$\Rightarrow$ P (red ball) + P(blue ball) + P(orange ball) = 1 , 

$\Rightarrow$ $\frac{1}{4}$ +  $\frac{1}{3}$ + P(orange ball) =  1 , 

$\Rightarrow$  P(orange ball) = 1 -  $\frac{1}{4}$ -  $\frac{1}{3}$ = $\frac{12 - 3 - 4}{12} = \frac{5}{12}$ ------(i) 

Let the total number of balls in the jar be x . 

As the jar contains 10 orange balls , 

  P(orange ball) =  = $\frac{10}{x} = \frac{5}{12}$    ..........(using(i))

$\Rightarrow$  x = 24 , 

Hence, the total number of balls in the jar = 24

There are four complete weeks in the month of March . There  are three more (extra) days in the month . If the month starts on Sunday or Monday or Tuesday, then there are 5 Tuesdays in the month of March, so probability is  $\frac{3}{7}$

Let Event A = getting a queen of club  , 

Event B = getting a king of spade , 

P(A$\cup$B) is to be calculated , 

P(A$\cup$B) = P(A) + P(B) - P(A$\cap$B) ,

 n(A$\cap$B) = $\varnothing$  , 

P(A$\cap$B) = 0 (Mutually Exclusive events)  

P(A$\cup$B) = P(A) + P(B),

=  $\frac{1}{52}$ + $\frac{1}{52}$  =   $\frac{1}{26}$ , 

Note : There is a single unit queen of club and king of spade in a card pack , So P(A) = $\frac{1}{52}$ and P(B) = $\frac{1}{52}$

As per question , 

P(A') = $\frac{8}{13}$ (A can't solve the problem), 

P(A) = $\frac{5}{13}$ (A can solve the problem) , 

P(B) = $\frac{9}{13}$  (B can solve the problem) ,  

P(A$\cup$B) = P(A) + P(B) - P(A$\cap$B) =  $\frac{5}{13}$ + $\frac{9}{13}$ - $\frac{45}{169}$  , 

 =  $\frac{14}{13}$ -  $\frac{45}{169}$ =   $\frac{143 - 45}{169}$  ,  

 =  $\frac{98}{169}$

Since A and B are independent events . 

(i)  $\therefore$ $P\left(\frac{A}{B}\right) = P\left(A\right) = \frac{1}{2}$

(ii)   $P\left(\frac{A\cap B}{A\text{'}\cup B\text{'}}\right) = 0$

n{(A$\cap$B) $\cap$(A$\cap$B)'} = 0

It is the case of without replacement . 

P(A) = $\frac{4}{52}$ =  $\frac{1}{13}$  , 

Since the second trial is made without replacement, only 3 aces and 51 cards are left after the first trial . 

So, P(A) = $\frac{3}{51}$ = $\frac{1}{17}$

Total number of balls in the bag = 5 + 4 + 6 = 15 , 

Let S be the sample space , then the number of ways in which 3 balls can be drawn out of 15 balls = n(S) = ¹⁵C_{3 ,}

 Let A be the event of drawing one red, one white and one green ball, then the required number of ways . 

  P(A) =$\frac{n\left(A\right)}{n\left(S\right)} = \frac{{}^{5}C_{1 }x {}^{4}C_{1}x {}^{6}C_1}{{}^{15}C_3} = \frac{24}{91}$ _, 

Total way = ⁵²C₃ = 22100 ,

There are 4 suits in a pack of card , so 3 suits can be selected in ⁴C₃

way and one card each from different suits can be selected in 

¹³C₁ x  ¹³C₁​ x  ¹³C₁​ , 

way , 

So, favourable ways =  ⁴C₃ x  ¹³C₁​  x  ¹³C₁​  x  ¹³C₁​  = 8788 , 

Required probability = $\frac{8788}{22100} = \frac{169}{425}$

Given , P(A) = 0.6, P(B) = 0.7 , 

Here, A and B are independent events. 

$\therefore$P(A$\cap$B) = P(A) x  P(B)  

  =  0.6 x 0.7 = 0.42 , 

P$(\overline{A} \cap \overline{B})$ = P$\left(\overline{A}\right)$ x P$\left(\overline{B}\right)$ , 

 = 0.4 x 0.3 = 0.12  , 

Probability that A and B will say same thing = Probability that both speak truth or false , 

=  P(A$\cap$B) + P $(\overline{A} \cap \overline{B})$  , 

= 0.42 + 0.12 = 0.54 

As per question,

$$\begin{gathered} P\left(A\right) = \frac{1}{2 } , P\left(\overline{A}\right) = \left(\frac{1}{2}\right) , \\ P\left(B\right) = \frac{1}{3} , P\left(\overline{B}\right) = \frac{2}{3} , \\ P\left(C\right) = \frac{1}{4} , P\left(\overline{C}\right) = \frac{3}{4} , \end{gathered}$$

Problem will be solved , if atleast one of them solve it, so first we calculate probability that it is not solved  

$$\begin{gathered} P\left(\overline{A}\right) . P\left(\overline{B}\right) . P\left(\overline{C}\right) = \frac{1}{2} x \frac{2}{3} x \frac{3}{4} = \frac{1}{4} , \\ \therefore \end{gathered}$$

Required probability = 1 - $\frac{1}{4} = \frac{3}{4}$

As given multiple choice questions are there and each question has four option, in which one option is right.

so, probability of getting a correct option of question is = $\frac{1}{4}$

P(All three questions are correct) = $\frac{1}{4} . \frac{1}{4} . \frac{1}{4} = \frac{1}{64} ,$

P(All three questions will not correct) = 1 - $\frac{1}{64} = \frac{63}{64}$

As per conditional probability  , 

$P\left(\frac{A}{A\cup B}\right)$ = $\frac{P\{A\cap (A\cup B\left)\right\}}{P(A\cup B)}$ = $\frac{P\left(A\right)}{P(A\cup B)}$, 

P(A$\cup$B) = P(A) + P(B) - P(A) . P(B) , 

=  $\frac{1}{3}$ +  $\frac{2}{5}$ - $\frac{2}{15}$ =  $\frac{11}{15}$ -  $\frac{2}{15}$  = $\frac{9}{15} = \frac{3}{5}$ , 

$P\left(\frac{A}{A\cup B}\right)$  =  $\frac{{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{3}{5}}} = \frac{5}{9}$

Four coins are tossed = 2⁴ = 16 (outcome) , 

Favourable outcomes are = $\frac{4!}{2! x 2!} = \frac{4 x 3}{2!}$ = 6 , 

So probability of getting exactly 2 tails = $\frac{6}{16} = \frac{3}{8}$

All possible outcomes = 2ⁿ  = 2⁵ = 32 ,

Tails appears in odd number of times it mean (1, 3 or 5 tail appear)

(i) . 1 tail appear (4 head appear) (apply the concept of permutations of similar objects),   

$\frac{5!}{4!}$ =  5 , 

(ii)  3 tail appear (2 head appear) $\frac{5!}{3! x 2!}$ = 10 , 

(iii)  5 tail appear  

$\frac{5!}{5!}$ = 1 , 

All possible outcomes = 16  , 

Now, probability of getting  of odd number of tail appear =  $\frac{16}{32} = \frac{1}{2}$

The sample space of the given experiment is , 

{T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66} , 

The total number of possible outcomes in the sample space = 22. 

If a pair of dice is tossed , then the number of elements in its sample space =  6 x 6 = 36. When a coin  is tossed, the number of elements  in its sample space = 2 . 

Therefore , if a pair of dice and a coin are tossed simultaneously, then the number of elements in its sample space S = 36 x 2 = 72 , 

As every subset of S is an event and number of subsets of S = 2⁷², therefore , the number of possible events =   2⁷²​ . 

If we toss a coin once . 

we get S = {1, 2, 3, 4, 5, 6}, n(S) = 6 ,

event E = multiple of 3 , 

E = {3, 6} , 

n(E) = 2 , 

P(E) = $\frac{2}{6} = \frac{1}{3}$ , 

odd in favour of event E = 1:2 , 

Note : If the probability of event E is given to be P.  

Odds in favour of event (E) = $\frac{P}{1 - p}$ , 

and odds against event (E) = $\frac{1 - P}{P}$ , 

In this case P = $\frac{1}{3}$ ,  1 - P = $\frac{2}{3}$ , 

odds in favour of Event (E) = $\frac{P}{1 - p}$ , 

=  $\frac{{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{2}{3}}} = \frac{1}{2} = 1:2$

The man can arrive anytime between 3 to 5 pm . 

So total duration = 2 hours = 120 minutes . 

Now he sees the goal only if he arrives with in first 20 minutes. 

So, P(he sees the goal) = $\frac{favourable duration}{total duration} = \frac{20 minutes}{120 miuutes} = \frac{1}{6}$ , 

Hence, P(he misses the goal) = 1 - $\frac{1}{6}$ = $\frac{5}{6}$ .

n(S) = All possible orders ⁶P₄= 360 ,  

A be the event that A, B, C and D are first four in any order 4! = 24 , 

n(A) = 24 , 

P(A) = $\frac{n\left(A\right)}{n\left(S\right)} = \frac{24}{360} = \frac{1}{15}$

Let A is a event showing correct arrangement of the word "SPORT" ,

n(A) = 1 (there is a only one option)  , 

n(S) = all possible arrangements = 5! , 

 = 120 , 

P(A) =     $\frac{n\left(A\right)}{n\left(S\right)}$      = $\frac{1}{120}$

A = is a event getting a letter with right address (that is only one way)

n(A) = 1 , 

n(S) = all possible arrangements of despatches = 4! =  24 , 

P(A) = $\frac{n\left(A\right)}{n\left(S\right)} = \frac{1}{24}$

Let A = A is getting 1, 2, 5 or 6 , 

P(A) = $\frac{4}{6} = \frac{2}{3}$ , 

B = is getting 3, 4 or 6  , 

P(B) = $\frac{3}{6} = \frac{1}{2}$  , 

A and B both are independent event , 

P(A$\cap$B) = P(A) . P(B) , 

 = $$\begin{gathered} \frac{2}{3} = \frac{1}{2} \\ = \frac{1}{3} \end{gathered}$$

As per question P(A) = 80% = $\frac{4}{5}$ , 

P(B) = 90% = $\frac{9}{10}$ , 

P(A') =  $\frac{1}{5}$       , P(B') = $\frac{1}{10}$  , 

= P(A'$\cap$B) + P(A$\cap$B') {Contradict, mean one judge will be in favour then another judge will be in against.}, 

= P(A') . P(B) + P(A) . P(B') , 

$= \frac{1}{5} x \frac{9}{10} + \frac{4}{5} x \frac{1}{10}$ , 

$$\begin{gathered} = \frac{9}{50} + \frac{4}{50} , \\ = \frac{13}{50} , \\ = \frac{13}{50} x 100 , \end{gathered}$$

 = 26% .

Let A starts first and its chance  to get sum of 5 , 

P(A) = $\frac{4}{36} = \frac{1}{9}$ , 

 = P(A) + $P\left(\overline{A}\right) P\left(\overline{B}\right)$P(A) + P(A') P(B') P(A') P(B')   P(A)  + --------$\infty$ ,  

$= \frac{1}{9} + \left(\frac{8}{9}\right) \left(\frac{8}{9}\right) \left(\frac{1}{9}\right) + {\left(\frac{8}{9}\right)}^4. \frac{1}{9}+ -----\infty$ , 

$= \frac{1}{9}\left[1 + {\left(\frac{8}{9}\right)}^2 + {\left(\frac{8}{9}\right)}^4 + ------\infty \right]$ , 

$= \frac{1}{9}\left[\frac{1}{1 - \left({\displaystyle \frac{64}{81}}\right)}\right] = \frac{1}{9} \left[\frac{81}{81 - 64}\right] ,$

$= \frac{1}{9}\left[\frac{81}{17}\right] = \frac{9}{17}$

All Possible arrangements of students = 10! = n(S) , 

Now A is a event which shows all possible arrangements in which two persons will sit together . 

n(A) = 9! x 2!, 

P(A) = $\frac{n\left(A\right)}{n\left(S\right)} = \frac{9! x 2!}{10!} = \frac{1}{5}$

20 delegates can sit at a round table in (20 - 1)! i.e. 19! ways . So, the total number of possible outcomes are  = 19! , 

Considering two particular delegates as one delegates(because they are sitting together), there are 19 delegates which can be seated on a round table in (19 - 1)!  i.e. 18! ways . But these two particular delegates delegates can be arranged among themselves in 2! ways. Therefore, the number of arrangements in which two particular delegates sit together = 18! x 2! . 

So, the number of favourable outcomes = 18! 2! , 

$\therefore The required probability = \frac{18! x 2!}{19!} = \frac{2}{19}$

As per question P(A) =  $\frac{1}{2}$, P(B) =   $\frac{1}{3}$. 

As only one player can win the game , So A and B are mutually exclusive (events). 

P(Neither of them winning the game) =

P(A'$\cap$B')   =  P(A$\cup$B)' , 

=  1 -   P(A$\cup$B), 

= 1 - {P(A) + P(B)} , 

= 1 - { $\frac{1}{2}$ + $\frac{1}{3}$ } , 

 =  1 -  $\left\{\frac{5}{6}\right\}$

=  $\frac{1}{6}$

(i) . P(B - A) = P(B) - P(A) , 

(ii) P(A) $\le$ P(B) . 

When two dice are rolled once, then the sample space has 36 equally likely outcomes. 

A = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}, 

and   B = {(2,6), (3,5), (4,4), (5,3), (6,2)} . 

Now that A $\cap$B = {(3,5), (5,3)} . 

Here n(A) = 18, n(B) = 5 and n(A $\cap$B ) = 2 . 

$\therefore$ P(A)  =  $\frac{18}{36}$ , P(B) = $\frac{5}{36}$ and P(A $\cap$B )   =   $\frac{2}{36}$  ,

$\therefore$ Required probability = P(an odd number on first die or a total of 8), 

= P(A$\cup$B) , 

=  P(A) + P(B) -   P(A $\cap$B ) , (addition theorem)

=  $\frac{18}{36}$ +   $\frac{5}{36}$ -   $\frac{2}{36}$ =  $\frac{21}{36}$ =   $\frac{7}{12}$

Let event A = Person will hit balloon in first trial .

 event B = Person will hit balloon in second trial .

 event C = Person will hit balloon in third trial .

 event D = Person will hit balloon in fourth trial .

P(A) = 0.5,  P(B) = 0.2,  P(C) = 0.1 ,  P(D) = 0.8 , 

person can't shoot the balloon . 

in any trial = P(A'$\cap$B'$\cap$C'$\cap$D') , 

Now, he can shoot in any trial , 

P(A$\cup$B$\cup$C$\cup$D) = 1 -   P(A'$\cap$B'$\cap$C'$\cap$D') , 

=  1 - (0.5)(0.8)(0.9)(0.2) , 

=  1 - 0.072 , 

=  0.928

Let A , B, C be the events that  A hits the target , B hits the target, 

C hits the target respectively , 

Then  P(A) =   $\frac{4}{5}$ ,   P(B) = $\frac{3}{4}$ ,  P(C) = $\frac{2}{3}$ ,  

$P\left(\overline{A}\right) = \frac{1}{5} , P\left(\overline{B}\right) =\frac{1}{4}, P\left(\overline{C}\right) = \frac{1}{3}$ , 

CASE   I :  P(A, B and C all hit the target) = P(A$\cap$B$\cap$C)  , 

= P(A)P(B)P(C) =  $\frac{4}{5}$ x  $\frac{3}{4}$ x   $\frac{2}{3}$  = $\frac{2}{5}$ , 

CASE   II :  P(A and B hit but not C)  = $P(A\cap B\cap \overline{C})$ , 

=  P(A) P(B) $P\left(\overline{C}\right)$ , 

=    $\frac{4}{5}$ x  $\frac{3}{4}$ x  $\frac{1}{3} = \frac{1}{5}$ , 

CASE III  :  P(A and C hit but not B) = $P(A\cap C\cap \overline{B})$  , 

 = P(A) P(C) $P\left(\overline{B}\right)$  , 

 =  $\frac{4}{3} x \frac{2}{3} x \frac{1}{4} = \frac{2}{15}$ , 

CASE IV  :  P(B and C  hit but not A)  =  $\left(B\cap C\cap \overline{A}\right)$  , 

 =   P(B)P(C)$P\left(\overline{A}\right)$  , 

 =   $\frac{3}{4} x \frac{2}{3} x \frac{1}{5} = \frac{1}{10}$ , 

All the above cases being mutually exclusive , we have the required probability . 

$\frac{2}{5} + \frac{1}{5} + \frac{2}{15} + \frac{1}{10} = \frac{5}{6}$

Let E₁ , E₂  and A be the events defined as follows , 

E₁  = five occurs , E₂= five does not occurs , 

and A = the man reports that it is a five  . 

We have , P(E₁) = $\frac{1}{6}$ , P(E₂) = $\frac{5}{6}$ , 

Now $P\left(\frac{A}{E_1}\right)$ = probability that the man reports that their is a five on the  die given , that five has occured on the die  , 

= probability that the man speak truth = $\frac{3}{4}$ , 

and $P\left(\frac{A}{E_2}\right)$ = probability that the man reports that their is a five on the  die given , that five has not occured on the die  , 

 = probability that the man does not speak truth = 1 -  $\frac{3}{4}$ =  $\frac{1}{4}$   ,

We have to find $P\left(\frac{E_1}{A}\right)$ , 

By baye's rules , we have 

$$\begin{gathered} P\left(\frac{E_1}{A}\right) = \frac{P\left(E_1\right) . P\left({\displaystyle \frac{A}{E_1}}\right)}{P\left(E_1\right) . P\left(\frac{A}{E_1}\right) + P\left(E_2\right) . P\left(\frac{A}{E_2}\right)} , \\ = \frac{{\displaystyle \frac{1}{6}} x {\displaystyle \frac{3}{4}}}{{\displaystyle \frac{1}{6}} x {\displaystyle \frac{3}{4}} + {\displaystyle \frac{5}{6}} x {\displaystyle \frac{1}{4}}} = \frac{3}{8} \end{gathered}$$

Let A : Two drawn balls are white , 

E₁ :  All the balls are white , 

E₂ : Three balls are white , 

E₃ : Two balls are white ,  

Since E₁ , E₂ , and  E₃ and mutually exclusive and exhaustive exams  .(having equal chance to occur)

P( E₁) = P( E₂) = P(E₃) = $\frac{1}{3}$ , 

Now, $$\begin{gathered} P\left(\frac{A}{E_1}\right) = \frac{{}^{4}C_2}{{}^{4}C_2} = 1 , \\ P\left(\frac{A}{E_2}\right) = \frac{{}^{3}C_2}{{}^{4}C_2} = \frac{3}{6} = \frac{1}{2} , \\ P\left(\frac{A}{E_3}\right) = \frac{{}^{2}C_2}{{}^{4}C_2} = \frac{1}{6} , \end{gathered}$$  

 $\therefore$  $P\left(\frac{E_1}{A}\right) = \frac{P\left(E_1\right) .+ P\left({\displaystyle \frac{A}{E_1}}\right)}{P\left(E_1\right) . P\left(\frac{A}{E_1}\right) + P\left(E_2\right) . P\left(\frac{A}{E_2}\right) + P\left(E_3\right) . P\left(\frac{A}{E_3}\right)}$

$$\begin{gathered} = \frac{{\displaystyle \frac{1}{3}} x 1}{{\displaystyle \frac{1}{3}} x 1 + {\displaystyle \frac{1}{3}} x {\displaystyle \frac{1}{2}} + {\displaystyle \frac{1}{3}} x {\displaystyle \frac{1}{6}}} \\ = \frac{1}{1 + {\displaystyle \frac{1}{2}} + {\displaystyle \frac{1}{6}}} = \frac{6}{10} = 0.6 \end{gathered}$$$$\begin{gathered} = \frac{{\displaystyle \frac{1}{3}} x 1}{{\displaystyle \frac{1}{3}} x 1 + {\displaystyle \frac{1}{3}} x {\displaystyle \frac{1}{2}} + {\displaystyle \frac{1}{3}} x {\displaystyle \frac{1}{6}}} \\ = \frac{1}{1 + {\displaystyle \frac{1}{2}} + {\displaystyle \frac{1}{6}}} = \frac{6}{10} = 0.6 \end{gathered}$$

Let E₁ : First ball is red , 

E₂ : First ball is black, 

A : second ball is red , 

Total number of balls is 10. 

Then P(E₁) = $\frac{3}{10}$ , P(E₂) = $\frac{7}{10}$ , 

$\therefore P\left(\frac{A}{E_1}\right) = \frac{2}{9} and P\left(\frac{A}{E_2}\right) = \frac{3}{9}$ , 

Then by Baye's theorem, probability of second selected ball is red when first selected ball is also red is given by 

$$\begin{gathered} P\left(\frac{E_1}{A}\right) = \frac{P\left(E_1\right) . P\left({\displaystyle \frac{A}{E_1}}\right)}{P\left(E_1\right) . P\left(\frac{A}{E_1}\right) + P\left(E_2\right) . P\left(\frac{A}{E_2}\right)} \\ = \frac{{\displaystyle \frac{3}{10}} x {\displaystyle \frac{2}{9}}}{\frac{3}{10} x \frac{2}{9} + {\displaystyle \frac{7}{10}} x {\displaystyle \frac{3}{9}}} = \frac{6}{6 + 21} , \\ = \frac{6}{27} = \frac{2}{9} \end{gathered}$$  

Hence , the probability that the first  selected ball is red , is  $\frac{2}{9}$

ER