TS ICET 2019 Question Paper Shift-1 (24th May)

Arithmatic mean=$$\ \frac{sumof observations }{ total no. Of observations}$$ 

sum of first n natural no.=$$\ \frac{n(n+1)}{ 2}$$

sum of first 100 natural no.=$$\ \frac{100*101 }{2 }$$=5050

sum of first 55 natural no.=$$\ \frac{55*56}{2 }$$=1540

sum of 56,57,...…,100=5050$$\ \ \ -\ $$1540

=3510

Arithmatic mean=$$\ \frac{3510 }{45 }$$=78

.     

 Arithmetic mean=26.5

Median=25.4

Using empirical formula,

mode=3$$\times\ $$median$$\ -\ $$2$$\times\ $$mean

=3$$\times\ $$25.4$$\ -\ $$2$$\times\ $$26.5

=76.2$$\ -\ $$53

=23.2

Mean($$\overline{x}$$)=$$\frac{1}{n}$$ $$\Sigma_{i=1}^{10}x_i\ $$=$$\ \frac{12 }{10 }$$=1.2

Standard deviation= $$\sqrt{\frac{\Sigma_{i=1}^{10}x_i^{2}{n}}}\$$

here we have 

number of subject n = 7 

square of sum of deviation are $$ \sum_di^2$$= 21 

now we know the fourmula of the coefficient of rank correlation  is rR = 1- $$\times {n-1}\sum_di^2\frac\times{n}{n^2-1}$$ 

       the coefficient of rank correlation is rR = 1- $$\times {7-1}\sum_21 \frac\times{7}{7^2-1}$$

                                                               rR = 1- $$\frac{3}{8}$$

                                                               rR=  $$\frac {5}{8}$$ answer                                                                 

probility of  solving problems by P, is= $$\frac{1}{3}$$

probility of not solving the problem by P , is= $$\frac{2}{3}$$

probility of solving problems by q, is= $$\frac{1}{4}$$ 

probility of not solving the problem by Q , is= $$\frac{3}{4}$$

probility of solving problems by r, is= $$\frac{1}{2}$$

probility of not solving the problem by R , is= $$\frac{1}{2}$$

so the probility of solving the problem are =problem is solved by p * problem is not solved by q and r + problem is solved by q * problem is not solved by p and r + problem is solved by r * problem is not solved by p and q + problem is solved by p and q and not solved by r + problem is solved by q and r and not solved by p + problem is solved by p and r and not solved by q + problem is solved by all three of them 

P{solved}= $$\frac{1}{3}$$* $$\frac{3}{4}$$*$$\frac{1}{2}$$ + $$\frac{1}{4}$$* $$\frac{2}{3}$$*$$\frac{1}{2}$$ + $$\frac{1}{2}$$* $$\frac{3}{4}$$*$$\frac{2}{3}$$ + $$\frac{1}{3}$$* $$\frac{1}{4}$$*$$\frac{1}{2}$$ + $$\frac{1}{4}$$* $$\frac{1}{2}$$*$$\frac{2}{3}$$ + $$\frac{1}{3}$$* $$\frac{1}{2}$$*$$\frac{3}{4}$$ + $$\frac{1}{3}$$* $$\frac{1}{4}$$*$$\frac{1}{2}$$ 

p{soleed}= $$\frac{18}{24}$$

p{solved}=$$\frac{3}{4}$$ answer 

or 

probility of not solving problems by P,Q and R are = $$\frac{2}{3}, \frac{3}{4}$$ and $$\frac{1}{2}$$ respectively 

so the probility of problem is solved is = 1-probility of not solving problems by P,Q and R are = $$\frac{2}{3}, \frac{3}{4}$$ and $$\frac{1}{2}$$

                                                         =1- $$\frac{2}{3}* \frac{3}{4}$$* $$\frac{1}{2}$$

                                                        =1-$$\frac{6}{24}$$

so the probility of problem is solved is = $$\frac{3}{4}$$ 

when 4 dices are thrown the number of total event occurs = 6*6*6*6= 1296

showing same number event are ={1,1,1,1] {2,2,2,2],{3,3,3,3],{4.4.4.4},{5,5,5,5},{6,6,6,6}

total number of favorable events are =6 

Probility of getting same number on four dices are = $$\frac{number of favorable cases}{total number of events occurs}$$

P{e4}= $$\frac{6}{6^4}$$

P{e4}=$$\frac{1}{6^3}$$ answer

total number of event occurs when 2 card are drawn from pack of 52 cards = 52*51= 2652 

number of suits are there in packs of cards = 4

in a suit to total number of cards are there = 13 

total number of  favorable events when two cards are drawn from suits = 13*12*4 =156*4 

 now the probility of getting both cards are from  same  suits are =$$ \frac{total number of favorable events}{total  number of event occurs when 2 card are drawn from pack of 52 cards}$$

$$p{e2}=\frac{156*4}{2652}$$

$$p{e2}=\frac{4}{17}$$ answer  

now the total number of event occurs when two unbiased dice are thrown =36 

we have to choose prime number which are 2,3 and 5 

so the favorable outcomes pair  are $${{2, 2}}$$, pair  $${{3, 3}}$$,pair  $${{5, 5}}$$ 

then the probility of getting same prime number are 

$$P{ {same- prime}}$$=$$\frac{number of favorable outcomes}{total number of possible outcomes}$$

$$p{ {same- prime}}$$=$$\frac{3}{36}$$

$$p {{same- prime}}$$=$$\frac{1}{12}$$ answer