JEE 2012 :: General Questions :: Mathematics

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Let X and Y be two events such that  $P(X|Y)=\frac{1}{2},P(Y|X)=\frac{1}{3}$ and  $P(X \cap Y) \frac{1}{6}$ , which of the following is/are correct?

Answer:

Explanation:

 Concept involved 

(i) Conditional probability i.e

$P(A/B)=\frac{P(A\cap B)}{P(B)}$

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

(iii) independent event, then

 $P(A \cap B)=P(A).P(B)$

  sol. Here, $P(X/Y)=\frac{1}{2},P\left(\frac{Y}{X}\right)=\frac{1}{3}$

 and   $P(X \cap Y)=6$

$\therefore$  $P\left(\frac{X}{Y}\right)=\frac{P(X \cap Y)}{P(Y)}$

 $\Rightarrow$  $\frac{1}{2}= \frac{1/6}{P(Y)}$

 $\Rightarrow$  $P(Y)=\frac{1}{3}$............(i)

    $P(\frac{Y}{X})=\frac{1}{3}$

  $\Rightarrow$   $\frac{ P(X\cap Y)}{P(X)}=\frac{1}{3}$

$\Rightarrow$  $\frac{1}{6}=\frac{1}{3}P(X)$

$\therefore$  $P(X)=\frac{1}{2}$...........(ii)

 $p(X \cup Y)=P(X)+P(Y)-P(X \cap Y)$

 =$\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3}$..........(iii)

   $P(X \cap Y)=\frac{1}{6}$

 and    $P(X) P(Y)=\frac{1}{2} \frac{1}{3}=\frac{1}{6}$

 $\Rightarrow$  $P(X \cap Y)=P(X) .P(Y)$

 $\Rightarrow$ Independent events ...........(iv)

   $P(X^{c} \cap Y)=P(Y)-P(X \cap Y)$

  =$\frac{1}{3}-\frac{1}{6}=\frac{1}{6}$ ..........(iv)

A tangent PT is drawn to the circle $x^{2}+y^{2}=4$ at the point $P(\sqrt{3},1)$ . a straight line L perpendicular to PT is a tangent to thye circle $(x-3)^{2}+y^{2}=1$

A possible equation of L is 

Answer:

Explanation:

(i) Equation of a tangent to at (x₁,y₁) is 

x²+y²=r²

at (x₁,y₁) is 

xx₁+yy₁=r²

(ii)If ax+by+c =0 is tangent to (x-h)²+(y-k)²=r²

|cp|=r

Here tangent to $x^{2}+y^{2}=4$ at the point

$P(\sqrt{3},1)$  is $\sqrt{3} x+y=4$.............(i)

 As. L is perpendicular to $\sqrt{3} x+y=4$

 $\Rightarrow$  $x-\sqrt{3}y= \lambda$

 which is tangnet to $(x-3)^{2}+y^{2}=1$

 $\Rightarrow$   $\frac{|3-0-\lambda|}{\sqrt{1+3}}=1$

$\Rightarrow$  $|3-\lambda|=2$

$\Rightarrow$  $3-\lambda=2,-2$

  $\therefore$  $\lambda=1,5$

 $\Rightarrow$  $L:x- \sqrt{3} y=1, x-\sqrt{3}y=5$

A tangent PT is drawn to the circle $x^{2}+y^{2}=4$ at the point $P(\sqrt{3},1)$ . a straight line L perpendicular to PT is a tangent to the circle $(x-3)^{2}+y^{2}=1$

A common tangent of the two circles is 

Answer:

Explanation:

(i) Equation of a tangent to at (x₁,y₁) is 

x²+y²=r²

at (x₁,y₁) is 

xx₁+yy₁=r²

(ii)If ax+by+c =0 is tangent to (x-h)²+(y-k)²=r²

|cp|=r

Here, equation of common tangnet be 

 $y=mx \pm 2\sqrt{1+m^{2}}$

 which is also the tangent to

  $(x-3)^{2}+y^{2}=1$

 $\Rightarrow$   $\frac{|3m-0+2\sqrt{1+m^{2}}|}{\sqrt{m^{2}+1}}=1$

 $\Rightarrow$   $3m+2\sqrt{1+m^{2}}=\pm \sqrt{1+m^{2}}$

 $\Rightarrow$  $3m=-3 \sqrt{1+m^{2}}$

or   $3m=-\sqrt{1+M^{2}}$

 $\Rightarrow$  $m^{2}=1+m^{2}$

or     $9m^{2}=1+m^{2}$

 $\Rightarrow$   $m \epsilon \phi$   or  $m=\pm \frac{1}{2 \sqrt{2}}$

$\therefore$    $y=\pm \frac{1}{2 \sqrt{2}} x \pm 2\sqrt{1+\frac{1}{8}}$

 $\Rightarrow$   $y=\pm \frac{x}{2 \sqrt{2}}\pm \frac{6}{2\sqrt{2}}$

 $\Rightarrow$  $2 \sqrt{2} y= \pm (x+6)$

Let $a_{n}$ denote the number of all n-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits, in them are 0. Let $b_{n}$ = The number of such n-digit integers ending with digit 1 and $c_{n}$ = The number of such n-digit integers ending with digit 0.

The value of $b_{6}$ is 

Answer:

Explanation:

Since $a_{n}$ be the n-digit positive integer formed by the digits 0,1 or both such that no two consecutive digits are zero.

 $b_{n}$=numbers which are ending with 1

$C_{n}$= number which are ending with 0

$\therefore$  $a_{n}=b_{n}+c_{n}$

 i.e, $a_{n}$=1(0 or 1)...............(0 or 1)

$b_{n}=$1 (0 or 1) ........1   

                                            nth place

 $c_{n}$=1(0 or 1) ......0

$b_{6}$= Six digit number ending with 1

Now, the four places are to be filled  

i,e,

 For 3 places, all 1's are used =1 way

 one zero is used= $^{3}C_{1}$=3

 two zeros are used=1 way

 0  1  0  1  ............

 Total=5 ways

  for 3 places,

   all 1's are used =1ways

  one zero is used = $^{2}C_{1}$=2 ways

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

                                               =3 ways

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

 Thus, $b_{6}=5+3=8$

Let $a_{n}$ denote the number of all n-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits, in them are 0. Let $b_{n}$ = The number of such n-digit integers ending with digit 1 and $c_{n}$ = The number of such n-digit integers ending with digit 0.

Which of the following is correct?

Answer:

Explanation:

 Since $a_{n}$ be the n-digit positive integer formed by the digits 0,1 or both such that no two  consecutive digits are zero.

 $b_{n}$=numbers which are ending with 1

$C_{n}$= number which are ending with 0

$\therefore$  $a_{n}=b_{n}+c_{n}$

 i.e, $a_{n}$=1(0 or 1)...............(0 or 1)

$b_{n}=$1 (0 or 1) ........1   

                                            nth place

 $c_{n}$=1(0 or 1) ......0

As   $a_{n}=b_{n}+c_{n}$

 i.e,   $a_{n}=1$.................(1 or 0).

$\Rightarrow$  $a_{n}=a_{n-1}+a_{n-2}$

 $\therefore$    $a_{17}=a_{16}+a_{15}$

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