Basic probability question - please help

Question

The question: What is the probability of randomly taking out a non-blue ball for the first time, and then taking out a non-red ball from a bucket with 4 white, 2 red, and 4 blue balls?

Instead of stacking the probabilities of the other colors, I tried solving it like this:
The probability of taking out a non-blue ball for the first time is 1-(4/10).
Then, the probability of taking out a non-red ball for the first time is either 1-(2/9) or 1-(1/9), depending on if a red ball was already chosen.
I am unsure how to express the connection between the last two probabilities.
The answer to the question is 44/90.
Would appreciate some help

AndrewN · Answer

Hello,  HaSmamit . I think the question could use a little rephrasing, such as "non-blue ball on the first selection," as well as some comment on how the first ball selected would not be put back before the second selection was made. Anyway, rather than look for shortcuts to these types of questions, I typically employ logic to arrive at an accurate conclusion.

1) There are only four ways in which desirable selections may be made: WW, WB, RW, RB.

2) We can calculate the probabilities of each sequence independently and add at the end:

WW:  \frac{4}{10} * \frac{3}{9} = \frac{12}{90}

WB:  \frac{4}{10} * \frac{4}{9} = \frac{16}{90}

RW:  \frac{2}{10} * \frac{4}{9} = \frac{8}{90}

RB:  \frac{2}{10} * \frac{4}{9} = \frac{8}{90}

\frac{12}{90} + \frac{16}{90} + \frac{8}{90} + \frac{8}{90} = \frac{44}{90}

This may not be the fastest way to solve the question, but if you forget a certain formula or cannot decide which one to use, such a method can certainly help you. I would recommend checking out the pertinent sections of the  Ultimate GMAT Quantitative Megathread  and  ALL YOU NEED FOR QUANT ! ! !  if you need to brush up on theory.

Good luck with your studies.

- Andrew

ThatDudeKnows · Answer

HaSmamit

I agree with everything that AndrewN said. Here's a little tweak just to help you keep evolving how you think about probability questions. You came at it from the "1 minus" angle (but got stuck), and Andrew came at it from the add up the ways to win. On some questions, it makes sense to put both approaches to work at the same time.

To win, we need either
Red and notRed = 2/10 * 8/9 = 16/90
White and notRed = 4/10 * 7/9 = 28/90
16/90 + 28/90 = 44/90

Kinshook · Answer

Asked: What is the probability of randomly taking out a non-blue ball for the first time, and then taking out a non-red ball from a bucket with 4 white, 2 red, and 4 blue balls?

There are followings possibilities = {WW, WB, RW, RB}

WW = \frac{4}{10}*\frac{3}{9 } = \frac{12}{90} 
 WB = \frac{4}{10}*\frac{4}{9} = \frac{16}{90} 
 RW = \frac{2}{10}*\frac{4}{9} = \frac{8}{90} 
 RB = \frac{2}{10} *\frac{4}{9} = \frac{8}{90}

Adding all the probabilities, we get =  \frac{(12+16+8+8)}{90} = \frac{44}{90}

GmatKnightTutor · Answer

Might find this a helpful type of question to go through, too:

How to get better at GMAT Quant. Recognize how ratios with 3 elements work]

Basic probability question - please help

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