HaSmamit
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
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