Gravitational Attraction
What would happen if two people out in space a few meters apart, abandoned by their spacecraft, decided to wait until gravity pulled them together? My initial thought was that …
(Note: this problem has an error...can you find it? I will post the correction ~~soon~~ here, which itself has some interesting properties)
We often make the important distinction between drawing cards with or without replacement when determining the results of card games. Clearly, if you draw a Jack, then the probability of drawing another Jack is different whether you replace and shuffle, or you leave the drawn Jack out. In an extreme case, imagine a deck with only one Jack.
After going through an example in class, and applying replacement, a student asked about the same example without replacement. I didn't have time to go over the example in class, but I sketched how the calculation would be different, and would yield a different answer. After class, going through the calculation, I was in for a surprise.
You draw two cards from a deck, and ask what is the probability that the first is a black card, and the second is a jack. In math notation, we want:
In replacement, we replace the first card after drawing it, reshuffle, and then draw the second. Thus the two events are independent.
Without replacement is a little trickier to set up, because the second draw depends on the first.
the same answer!!
(Note: this problem has an error...can you find it? I will post the correction ~~soon~~ here, which itself has some interesting properties)
The only conclusions are
Cool!
(Note: this problem has an error...can you find it? I will post the correction ~~soon~~ here, which itself has some interesting properties)