chetan2u wrote:
The probability of picking 2 red balls one after another without replacement from a bag containing only red and white balls is 14/33. What will be the probability of picking 2 white balls one after another w/o replacement from that bag, given that it can hold at the most 20 balls?
A. 1/33
B. 3/33
C. 5/33
D. 9/33
E. 19/33
Hi,
Lets see what info is being provided by the Q...1) The bag contains only red and white balls.
2) Prob of picking two red balls
w/o replacement =14/33
3) A critical info
it can hold at the most 20 balls...
what does " Prob of picking two red balls
w/o replacement =14/33" mean if there are R red balls and W white balls..
\(\frac{R}{{R+W}} *\frac{{R-1}}{{R+W-1}}=\frac{14}{33}\)..
or \(\frac{R{(R-1)}}{{(R+W)}{(R+W-1)}}= \frac{14}{33}\).....
WHAT DOES RHS MEAN14/33..
this means the actual numbers can be taken as 14x/33x, where x is the common termWHAT DOES LHS MEAN\(\frac{R{(R-1)}}{{(R+W)}{(R+W-1)}}\)..
it means both nemerator and denominator are multiple of consecutive numbers..
from above two points we can say ..
14x is the product of two consecutive integers..
2*7*x..
33x is also the product of two consecutive integers..
3*11*x
as we see, x can be fitted as 4 to get numerator as 7*8 and denominator as 11*12...
so \(\frac{R{(R-1)}}{{(R+W)}{(R+W-1)}}\)..= \(\frac{{(8*7)}}{{(12*11)}}\)..
.... or \(\frac{8{(8-1)}}{{(8+4)}{(8+4-1)}}\)..
thus R = 8 and w=4..
thus prob of picking 2 white balls one after another w/o replacement from that bag=4/12*3/11=1/11=3/33..
ans B
...
You mention this as a critical info yet I fail to see how exactly this critical info was used in the process of determining the answer. Could you please elaborate?