- 1.A box has W white marbles and B black marbles ...show thatprobability of picking a white marble in the first try =probability of picking a white marble in second try with first marble already pickedor in other words, Prove thatP(W1) = P(W2) = W/(W+B)
- hi,good to see some puzzle here.....keep going on posting the puzzles......P(W1) = W/(W+B).......this is obviousnow,for P(W2) = P1 + P2

where P1 = probability that the first ball is white and second ball is also whiteP2 = probability that the first ball is black and second ball is whiteP1 = W/(W + B) x ( W - 1)/(W + B - 1)P2 = B/(W + B) x W/(W + B - 1)adding, P1 + P2 = (W^2 - W + WB)/(W + B)x(W + B -1) = W/(W + B)rajwrote:*"Vijaykumar Nayak (RBIN/EDS1) *" <Vijaykumar.Nayak@...>*1.A box has W white marbles and B black marbles ...show thatprobability of picking a white marble in the first try =probability of picking a white marble in second try with first marble already pickedor in other words, Prove thatP(W1) = P(W2) = W/(W+B)`To unsubscribe from this list, go to the ONElist web site, at`

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Protect your identity with Yahoo! Mail AddressGuard - P(W1) = W/(W+B)

P(W2) = [W/(W+B) * (W-1)/(W+B-1)] + [B/(W+B) * W/(W+B-1)]

= [W(W-1) + B*W] / [(W+B)*(W+B-1)]

= [W(W+B-1)] / [(W+B)*(W+B-1)]

= W/(W+B)

> 1.A box has W white marbles and B black marbles ... show that

> probability of picking a white marble in the first try =probability of

> picking a white marble in second try with first marble already picked

> or in other words, Prove that

> P(W1) = P(W2) = W/(W+B) - P(w1) is obviously W/(W+B), since he has to choose

from W+B marbles out of which W are white..

Now P(w2) = probability of getting a white marble in

the second turn with a white one having been chosen in

the first + probability of a white marble being chosen

with a black marble in the first chance

= (W/(W+B))*((W-1)/(W+B-1)) + (B/(W+B))*((W)/(W+B-1))

= ((W(W-1) + BW))/((W+B-1)(W+B))

= W(W+B-1)/((W+B-1)(W+B))

= W/(W+B)

Hope this satisfies you..

Bye,

Atul.

--- "Vijaykumar Nayak (RBIN/EDS1) *"

<Vijaykumar.Nayak@...> wrote: > 1.A box has W

white marbles and B black marbles ...>

________________________________________________________________________

> show that

> probability of picking a white marble in the first

> try =probability of picking a white marble in second

> try with first marble already picked

>

> or in other words, Prove that

>

> P(W1) = P(W2) = W/(W+B)

>

>

>

>

Yahoo! India Mobile: Download the latest polyphonic ringtones.

Go to http://in.mobile.yahoo.com Probability of picking a white ball in first try is W/(W+B).

Now comming to picking a white ball in second try..here two cases

picking a while ball in first try or picking a black ball in first try.

case 1:

Probabilityy of picking a white ball in first try and white ball in second try is

W/(W+B) X (W-1)/(W+B-1) = W(W-1)/(W+B-1)(W+B)

Case 2:

Probability of picking a black ball in first try and white ball in second try is

B/(W+B) X (W)/(W+B-1) = BW/(W+B-1)(W+B)

So total probability is addition of above two

which is (BW + W(W-1))/(W+B-1)(W+B) = W(W+B-1)/(W+B-1)(W+B) = W/(W+B)

Hence proved

-kranthi

On Tue, 04 Nov 2003 Vijaykumar Nayak (RBIN/EDS1) * wrote :>1.A box has W white marbles and B black marbles ...

>

>show that

>probability of picking a white marble in the first try =probability of picking a white marble in second try with first marble already picked

>

>or in other words, Prove that

>

>P(W1) = P(W2) = W/(W+B)

>

>

>