# Help two events

• Oct 20th 2007, 11:51 AM
Infiniti
Help two events

"The Heavenrich family of five people randomly picks a name of a family member from a hat to decide for whom to buy a present. What is the probability that nobody picks their own name? If someone picks their own name, all the names are returned and everyone picks again. What is the probability that it took exactly two tries for everyone to pick a name of another family member?
• Oct 20th 2007, 03:28 PM
Truthbetold
Quote:

"The Heavenrich family of five people randomly picks a name of a family member from a hat to decide for whom to buy a present. What is the probability that nobody picks their own name? If someone picks their own name, all the names are returned and everyone picks again. What is the probability that it took exactly two tries for everyone to pick a name of another family member?
Ah yes. Probability. I think I know how to do the first part.

Let's give names to the people just for easier reference: Dad, Mom, Kid 1, kid 2, kid 3.

There is a 4/5 chance he will NOT pick his name.

Next mom picks.
There is a 3/5, because one name that is NOT dad's is gone, she won't pick name.

Then kid 1 and kid 2 pick.
There probability is 2/5 and 1/5, respectively.

NOw Kid 3 picks.
There is only one name left, here is where I'm not sure about this anymore, that is not his. So 1/5.

Finally,
multiply them together because they are dependent events.
• Oct 20th 2007, 03:41 PM
Plato
Quote:

Originally Posted by Truthbetold
Let's give names to the people just for easier reference: Dad, Mom, Kid 1, kid 2, kid 3. Let's say Dad picks first.
There is a 4/5 chance he will NOT pick his name.
Next mom picks.
There is a 3/5, because one name that is NOT dad's is gone, she won't pick name.
Then kid 1 and kid 2 pick.
There probability is 2/5 and 1/5, respectively.
NOw Kid 3 picks.
There is only one name left, here is where I'm not sure about this anymore, that is not his. So 1/5.
Finally,
multiply them together because they are dependent events.

Hi there Truthbetold
The correct answer is $\displaystyle \frac{{\left( {76} \right)\left( {44} \right)}}{{(120)^2 }}$.
Does that agree with you approach?