Probability

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New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

n = 33, p = success, q = failure

3P (x = 0) = P (x = 1)

3 3 C 0 p 0 q 3 3 = 3 3 C 1 p q 3 2            

p = 1 1 2 , q = 1 1 1 2 q p = 1 1          

………. (i)

Subtracting, (ii) – (i), we get 1320

 

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

  6 1 1 = Required probability         

 

After solving, we get n = 4

New answer posted

2 months ago

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P
Payal Gupta

Contributor-Level 10

Since student guesses only two wrong. So there are three possibilities

(i) both wrong in section A

(ii) both wrong in section B

(iii) one wrong in each section A and B.

 Required possibilities =

=4C4*6C4(34)4*(14)4(34)2+4C3*6C5(34)3(14)5*14*34 +4C2*6C6*(34)2(14)2*(14)6

=27410[15*27+24*3+2]=27*479410

New answer posted

2 months ago

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P
Payal Gupta

Contributor-Level 10

 ?  x is a random variable.

k+2k+4k+6k+8k=1k=121

P ( (1<x<4)|x2)=4k7k=47

New answer posted

2 months ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

p                          q                          r                           s

 F                           T  &nb

...more

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

 

a = | 4 . 2 3 ( 1 ) 2 1 5 |

= | 8 + 3 2 1 5 | = 2

L = 4 a = 8

New answer posted

2 months ago

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P
Payal Gupta

Contributor-Level 10

Consider the equation of plane,

P: (2x+3y+z+20)+λ (x3y+5z8)=0

?  Plane P is perpendicular to 2x + 3y + z + 20 = 0

So,  4+2λ+99λ+1+5λ=0

λ=7

P : 9x – 18y + 36z – 36 = 0

Or P : x – 2y + 4z = 4

If image of

(2, 12, 2)

In plane P is (a, b, c) then

a21=b+122=c24

and  (a+22)2 (b122)+4 (c+22)=4

clearly

a=43, b=56andc=23

So, a : b : c = 8 : 5 : 4

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Total number of numbers from given condition = n (s) = 26

Every required number is of the form

A = 7 . ( 1 0 a 1 + 1 0 a 2 + 1 0 a 3 + . . . . )  + 111111

Here 111111 is always divisible by 21.

  Required probability = 2 2 2 5 = p   1 1 3 2 = p  96p = 33

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

The given two lines are coplanar

| 0 3 1 2 0 3 1 α 0 1 | = 0 α = 5 3

Now, n = | i ^ j ^ k ^ 0 3 1 2 0 3 | = i ^ ( 9 ) j ^ ( 2 ) + k ^ ( 6 ) = ( 9 , 2 , 6 )

Equation of plane :

= | ( 9 . 5 3 + 0 + 0 1 3 ) 8 1 + 3 6 + 4 | = 2 1 2 1 = 2 1 1

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

We have,  1-  (probability of all shots result in failure)  > 1 4

1 - 9 10 n > 1 4 3 4 > 9 10 n n 3

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