SOLUTION:
6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9
This is with replacement case (and was solved incorrectly by some of you):
Quote:
P=2*3/9*2/9=4/27
We are multiplying by 2 as there are two possible wining scenarios RW and WR.
Answer: D.[/quote]
The probability of drawing a red and a white ball has to be( 2/9)*(3/9) and then the solution follows. How come it is (2*3/9)
/ (2/9), i dont get it again .
Will continue again with rest of the questions.
P.S. I have never liked questions who have just been tough but no brainerbut the one in the list here are some of the best questions to train on many tricks and fundamentals I believe. More I see the stuff on quant , more I admire the work of the people engaged with the site. Bravo.. and cheers from cold Paris but grâce à GMAT for keeping me hot
[/quote]
We want out of two balls one to be red (R) and another white (W).
There are two ways this could happen: we can draw FIRST ball red and SECOND white OR FIRST ball white and SECOND red.
Probability of each case is: \(\frac{3}{9}*\frac{2}{9}\), so \(P=\frac{3}{9}*\frac{2}{9}+\frac{2}{9}*\frac{3}{9}=2*\frac{3}{9}*\frac{2}{9}\).
One more thing worth of mentioning: \(\frac{3}{9}*\frac{2}{9}\) would be correct answer if we were asked to determine the probability of FIRST ball being red and SECOND white, OR FIRST white and SECOND red. These are the cases in which order of drawing matters.
Hope it's clear
Perfect. That is how I solved it.
Next attack : Inequality : Will be going through your links to flex "inequality muscles" . Can't thank enough for the efforts you guys have put.