2
$\begingroup$

I was stuck on this question in a school assignment. The question is as follows: In a box are nine identically shaped gemstones. Four of these are diamonds and five are not.What is the probability that at least one diamond is selected, if a person took out two gems blindfolded?

I tried doing $\frac{(4)(8)}{{9}\choose{2}}\frac{1}{2!} = \frac{16}{36}$, where the $4$ is the first selected diamond, and the $8$ represents the remaining gems after the first is selected. The fraction $\frac{1}{2!}$ is for correcting the overcounting. However when I tried drawing an extended tree diagram, I found that the answer is $\frac{26}{36}$. Where did I go wrong?

$\endgroup$
0

2 Answers 2

Reset to default
3
$\begingroup$

It's not necessarily the first gem, or just one gem, that's the diamond. $4 \cdot 8$ counts all ordered pairs of gems in which the first gem is the diamond, and the second gem is anything. A set of two diamonds is counted twice, but set of two gems with only one diamond is only counted once, so you don't correct the overcount when you divide by $2$.

Instead, we can count $\binom 42$ cases with two diamonds and $\binom 41 \binom 51$ cases with one diamond, for a total of $26$. Divide this by the $\binom 92$ total cases, and we get the answer your tree diagram gave you.

$\endgroup$
2
$\begingroup$

Although you can proceed by adding Pr($1$ diamond) + Pr($2$ diamond), Pr questions of the "at least one" type are best tackled by finding the complement, viz. $Pr = 1 - \frac{\dbinom5 2}{\dbinom8 2}$

Or, using probabilities directly, $1 -\left(\frac{5}{9}\cdot\frac{4}{8}\right)$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .