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• 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
Message 1 of 5 , Nov 4, 2003
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)

• hi, good to see some puzzle here.....keep going on posting the puzzles...... P(W1) = W/(W+B).......this is obvious now,for P(W2) = P1 + P2 where P1 =
Message 2 of 5 , Nov 4, 2003
hi,

good to see some puzzle here.....keep going on posting the puzzles......

P(W1) = W/(W+B).......this is obvious

now,for P(W2) = P1 + P2
where P1 = probability that the first ball is white and second ball is also white
P2 = probability that the first ball is black and second ball is white

P1 =  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)

raj
"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)

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• 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)
Message 3 of 5 , Nov 4, 2003
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
Message 4 of 5 , Nov 6, 2003
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)
>
>
>
>

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• 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
Message 5 of 5 , Nov 10, 2003

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)
>
>
>

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