Question

Q-1) If (x,y )=(y,x),find the value of 9 (y-x)

Q-2)The mean of 5 numbers is 30 .If one number is excluded their mean becomes 28 .Find the excluded number.

Q-3) If the probability of winning a race is 1/6 less than the twice the probability of losing the race .Find the probability of winning the race .

# 20 pts .

Answer 1

1)

( x,y) = (y,x)

hence the ordinate and abssica both is zero


therefore , 9 (y- x )

=> 9 (0-0) .

=> 0 .

------------------------------------------------
2) mean of five number is =>

let the number be => x , x2 , x3 , x4 ,x5

sum of observation / total number of observation .

=> 30 = x + x2 + x3 + x4 + x5 / 5


=> x + x2 + x3 +x4 + x5 = 150-------(1)

now , one number is exclude

let , x is exclude

therefore =>

x2 + x3 + x4 + x5/ 4 = 28 -------(2)



x2 + x3 + x4 + x5 = 112 -----(3)

now on putting the value of eqation ( 3) in equation (1) we get


x + 112 =>150



therefore excluded number is => 38 .



---------------------------------------------------------



3)
according to question =>

probablblity of winning a race = >

P(E)= 2 x probability of losing game - 1/ 6 -------(1)





P(E) = 2* P( L ) - 1/6


now adding the probability of losing the game on both side .


we get = >


p(E) + P (L ) = > 3 P(L ) -1/6



=> 1 = { 18 p(L ) -1}. / 6


p ( L ) = 7 / 18




now .

probably of winning the race =>


P(E ) + P ( L ) = 1



P( E) + 7/18 = 1



hence , the probability of winning the game is


=> 1- 7/18

=> 11/18

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hope it helps!!!


$thanks$

Answer 2

HELLO DEAR,

------------------------(1)-------------------------

given that:-

(x,y) = (y,x)

we get two condition here,

----(1)

x=y

we get,
9(y-x) =9(x-x) =9×0 =0

-----(1)

y=x

9(y-x)=9(x-x)=9×0 =0

------------------------(2)------------------------

let the excluded no. be x

mean of excluded one no.
_______________________ =28
5

mean of excluded one no. = 140

the mean of five no. = 30

sum of five no.
________________ = 30
5

=> sum of five no. = 150

=> sum of 4 no. + x = 150

=> 140 + x =150

=> x =150 - 140

=> x =10

so the excluded no. = 10

-------------------------(3)----------------------

let the probability of loosing race =x

then the probability of winning the race = 2x-1/6

now.

p(w) + p(L) =1

=> 2x -1/6 +x =1

=> 3x = 1+1/6

=> 3x =7/6

=> x =7/18


now.

p(w) = 2x -1/6

=> 2(7/18) -16

=>7/9 - 1/6

=> (42-9)/54

=> 33/54

=> p(w) = 11/18


I HOPE ITS HELP YOU DEAR,
THANKS