There were initially no black marbles in a jar. Subsequently

Kindly correct me where am I wrong ?


For me it should be D

1. No. of ways of selecting 1st & 2nd black marble : (3 black & 3 other) 3/6 * 2/5 ----- (45 black & 45 other) 45/90 * 44/89 ----- (1755 black & 1755 other) 1755/3510 * 1754/3509 ----

In all cases (P) = 0.5 * 0.4(approx. to 1st place decimal) ... suff.


2. No. of ways of selecting 1st & 2nd black marble : 8/10 * 7/9 = 28/45 ----- its very clearly specified that total marbles are 10 & 8 out of them are black ... suff.

In statement 1, your calculation in getting 0.4 is incorrect. Infact in your example 1754/3509 would be very very close to 0.5 itself. Basically this value could range from 0 to 0.5.
In statement 2, total is not 10. 10 is only number of total marbles added. You still dont know how many were already present.

As shown by others earlier, choice C is the correct one.

Thanks for your reply Vips , I concede that B and A( Especially when total balls have lower values 2,4 --- the values do diverge beyond 0.4) are not sufficient..... but would you hold A not sufficient even if after the new balls are added , the total balls becomes 52 or greater ???

Thats right. Even if number of balls is 500 or 600, the probability of such event would be different (though very close to each other). As long as you are getting multiple values instead of one unique solution (for the question asked - it could be a range asked in question) from a statement, just mark the option as insufficient.

You could consider easier cases for (1).

If after the new marbles are added, there is 1 black marble and 1 non-black marble in the jar (so if initially there was only 1 non-black marble and 1 black marble was added), then the probability of selecting 2 black marbles in a row will be 0 (since we have only 1 black marble).

But if after the new marbles are added, there are 2 black and 2 non-black marbles in the jar, then the probability of selecting 2 black marbles in a row will obviously be more than 0.

Hope it's clear.

P ( 1 & 2 Black ) = P ( first black ) * P ( second black ) = ?

model of jar > Black + Others = Total
P ( first b ) = b/t
after selecting one marble and throwing it away, the model will change and total num of marbles will be reduced by one, so ...
P (second black) = (b-1)/(t-1)

stmt 1. B = 50% * T

P( first black ) = b/t = 0.5
but we dont know t and b to find P ( sec b )
insuf.

stmt 2. new barles = 10 , 8 are black

no info about Total
insuf.

stmat 1+2 :

b = 0.5 t = 8
t = 16
P = 0.5 * [(8-1)/(16-1)]

ans : C

Can anyone explain more on this question, I am still NOT clear

According to me the correct option should be D. Why C?

You should be more specific. Why do you think the answer is D? What is confusing in the solutions above? Also, have you read this: there-were-initially-no-black-marbles-in-a-jar-subsequently-59700.html#p1148705

(1) After the new marbles are added, 50% of all marbles are black.
If only one draw was asked , then this info was sufficient, but here we have to draw two marbles one after the another without replacement
First draw :- half marbles are black so probablity of drawing black marble is half marble (50% probablity)
Second Draw :- Now we since we don't know exactly how many marbles are left in the jar we cannot find out the probablity of second draw.
INSUFFICIENT
(2) Among the 10 added marbles, 8 are black
Since we do not know how many marbles are already present in the jar we do not know how adding 8 black and 2 other colour marbles will change the number of total marbles
INSUFFICIENT

MERGE:- Since there were no marbles earlier therefore adding 8 marbles will result in a total of 8 marbles in the jar and these 8 marbles are half of the total marbles. therefore the other marbles are 8 in number.
Now we know
Black marbles = 8
Other marbles = 8
Total marbles = 16
Probability of picking two black marbles without replacement =\(\frac{8}{16}*\frac{7}{15}\)

SUFFICIENT

ANSWER IS C

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