2
$\begingroup$

There are 7 empty boxes numbered sequentially from 1 to 7. A sack of gold is placed into one of the boxes at random, and, independently at random, a snake is put into one of the seven boxes. Let $E_1$ be the event that the snake and the sack of gold are in different boxes, and let $E_2$ be the event that the sack of gold is not in boxes 1,2,3. What is the conditional probability that the snake is in one of the three boxes 3,4 or 5, given the event $E_1\cap E_2$?

Warning: the symmetry principle should be applied with caution in this problem, it may be safer to completely avoid using this principle here

given

the GOLD is in either box 4, 5, 6, 7

and the SNAKE is NOT in the same box as the GOLD

what is the probability the SNAKE is is either box 3, 4, 5

so I'm thinking that (4 choose 1) for choosing which box the gold is in...

and then there is 6 more boxes to choose for the snake because it'snot in the box with the gold so (6 choose 1)

these two facts combined might make up the denominator?

and for the numerator, it would be (4 choose 1) for where gold goes, and then (3 choose 1) for where the snake goes (either 3, 4, 5)...

but this does not seem right because snake and gold could be in the same box using my technique for the numerator

$\endgroup$
  • $\begingroup$ There are two combinations using your method in which the snake and the gold are in the same box, if you take that off the numerator and see if that looks more sensible $\endgroup$ – MRobinson Aug 31 '18 at 7:12
  • $\begingroup$ Please avail yourself of this tutorial and reference on how to typeset math on this site. This being your eighth question, I think it would be appropriate that you learn how to typeset them properly. $\endgroup$ – joriki Aug 31 '18 at 7:14
0
$\begingroup$

You can get quite a long way by thinking about the symmetry of the problem. If $E_1\cap E_2$ holds then no matter which box the gold is in, there are $6$ possibilities for the snake, which include $1,2,3$. So (given $E_1\cap E_2$) the snake is in $\{1,2,3\}$ with probability $1/2$.

Since the condition $E_1\cap E_2$ is unchanged by reordering boxes $1,2,3$, each of these boxes has the same conditional probability of having the snake. Similarly, each of $4,5,6,7$ has an equal conditional probability of having the snake. You can now work out the individual conditional probabilities of being in $3$, in $4$ and in $5$ and add them up.

$\endgroup$
0
$\begingroup$

Draw a $7\times7$ grid of squares. Devote its $7$ rows to the $7$ places where the gold can be, and the $7$ columns to the $7$ places where the snake can be. Paint the squares corresponding to $E_1\cap E_2$ grey, and count them; they're all equiprobable a priori. Then put a $\checkmark$ on all grey squares that have the snake in one of the boxes $3$–$5$, and count the checkmarks as well. Now you are one step away from the result.

$\endgroup$

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.