IIT JEE Advanced 2006 Paper 1

Note: Common initial solution for Questions 107–109:
• We have n urns and (n + 1) balls.
• ith urn contains i white balls and (n + 1 – i) red balls.
• w = event getting white ball.
• ui = selecting of ith urn.
Now, we have
P(ui) α i
P(ui) = ki
P(u1) = k , P(U2) = 2k , ... , P(un) = nk
Therefore,
P(u1) + P(U2) + , ... + P(un) = 1
k( 1 + 2 + ... + n) = 1
Now, k

n(n+1)

2

= 1 ⇒ k =

2

n(n+1)

P(u1) =

2

n(n+1)

; P(u2) =

2.2

n(n+1)

, ... , P(un) =

2n

n(n+1)

P(W) = P(u)1P(

W

u1

) + P(u2)P(

W

u2

) + ... + P(un)P(

Wn

un

)
= P(

W

u1

)
which is the probability of white ball from ith urn.
1st urn : 1 W nr
2nd urn : 2 W (n - 1) R
3rd urn : 3 W (n - 2) R
P (W) = (

2

n(n+1)

)(

1

(n+1)

) + (

4

n(n+1)

)(

2

n+1

) + (

6

n(n+1)

)(

3

n+1

) + ... + (

2n

n(n+1)

)(

n

n+1

)
=

2

n(n+1)2

(1 + 2+32 + ... + n2)
=

2

n(n+1)2

n(n+1)(2n+1)

6

=

1

3

2n+1

(n+1)

Therefore,

lim

n→∞

P (W) =

2

3