1
$\begingroup$

A bag contains 9 red marbles, 5 blue marbles and 7 green marbles. Need to select one marble at a time without replacement. What is the probability that the first marble is blue, the second is blue or green an the third is red.

My Solution:

$= P(\text{blue}) * P(\text{blue} \cup \text{green}) * P(\text{red})$

$\displaystyle = {5\over 21} * ({4\over 20} + {7\over 20}) * ({9\over 19})$

Does that sound that right?

$\endgroup$
  • 1
    $\begingroup$ It's fine. If the second choice was blue or red, one would have to change the analysis. $\endgroup$ – André Nicolas May 25 '14 at 2:40
1
$\begingroup$

That looks like the right answer. These are conditional probabilities, so

\begin{align*} & \mathrm{Pr}(\text{marble 1 is blue} \cap \text{marble 2 is blue or green} \cap \text{marble 3 is red}) \\ &= \mathrm{Pr}(\text{marble 1 is blue}) \times \mathrm{Pr}(\text{marble 2 is blue or green} \mid \text{marble 1 is blue}) \times \mathrm{Pr}(\text{marble 3 is red} \mid \text{marble 1 is blue} \cap \text{marble 2 is blue or green}) \\ &= \frac{5}{9+5+7} \times \frac{(5+7)-1}{(9+5+7)-1} \times \frac{9}{(9+5+7)-2}\\ &= \frac{33}{532}. \end{align*}


Alternatively, we can use a counting argument. There are $3! \binom{9+5+7}{3}=7980$ ordered $3$-tuples of $\{r_1,\ldots,r_9,b_1,\ldots,b_5,g_1,\ldots,g_7\}$. Of these, precisely $5 \times (5+7-1) \times 9=495$ satisfies the blue-blue/green-red pattern. We have $$\frac{5 \times (5+7-1) \times 9}{3! \binom{9+5+7}{3}}=\frac{33}{532}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.