09 Apr 08
Originally posted by deriver69I hope not. The whole point of my thread was that these problems are simple to do, but the results are often surprising. I was hoping more for "oh, that's neat!" than "you're wrong, and you smell funny too." 😕
I wonder if this will generate another few hundred replies. I was impressed with the long debate over the last problem.
09 Apr 08
Originally posted by PBE6Simple enough problem.
Bob rolls two 6-sided dice, one red and one blue. Given that he rolls an 8, what is the probability that he rolled a 6 with the blue die?
Before the extra information (the pip count of the roll), The odds are clearly 1 in 6.
After the extra information, we can eliminate some of the possibilities, and then examine the ones which remain and recalculate the chances.
In this example, is it much easier to see that the pip count of the red and the pip count of the blue are distinct.
It was less intuitive to think of visually identical sides of the silver/silver card as being distinct, even though from a probability perspective, they are.
09 Apr 08
2 edits
Originally posted by eldragonflyCorrect!
this is easy.
there are only 5 ways to roll and 8 on two dice.
1/5
Back to the bag of quarters. You are given a bag of "n" quarters, one of which is double-headed. Your friend picks a quarter from the bag randomly, and proceeds to flip it. On the first "k" tosses, the quarter lands heads up all "k" times. Given this information, what is the probability that your friend picked the double-headed quarter?
09 Apr 08
1 edit
Originally posted by PBE6My initial table, dividing possibilities by regular coin or not, and by whether or not a tails comes up. (N coins, K tosses)
Correct!
Back to the bag of quarters. You are given a bag of "n" quarters, one of which is double-headed. Your friend picks a quarter from the bag randomly, and proceeds to flip it. The quarter lands heads up "k" times. Given this information, what is the probability that your friend picked the double-headed quarter?
Special Coin, Tails appears - 0
Special Coin, All Heads - 1/N = (2^K)/(N*2^K)
Regular Coin, All Heads - (N-1)/N * 1/(2^K) = (N-1)/(N*2^K)
Regular Coin, Tails appears - (N-1)/N * (2^K-1)/(2^K) = ((N-1)(2^K-1))/(N*2^K)
Eliminate the cases where tails appears. Since I made all the denominators the same, I can get rid of it by multiplying both numbers by it.
So I have the following numbers based on the coin shoing heads for K tosses.
Special - 2^K
Normal - N-1
So the odds the coin is the double-headed is (2^K)/(2^K+N-1) or 1-(N-1)/(2^K+N-1)
09 Apr 08
Originally posted by kbaumenLol. 🙂
I got the same answer and tried to find a flaw because it was also eldragonfly's answer. But I couldn't. I guess I shouldn't be that stereotypical.
OK, one more. The rate of infection from a particular disease is 1 in 1,000,000 in the general population. A hospital wants to administer a test that is 99% accurate (i.e. if 100 people who have the disease get tested, 99 will test "positive" and 1 will test "negative" falsely) and 95% specific (i.e. if 100 who don't have the disease get tested, 95 will test "negative" and 5 will test "positive" falsely).
Given that a person gets a positive result on the test, what is the chance that they really have the disease?
09 Apr 08
Originally posted by geepamoogleI know you got it right, I just wanted to give eldragonfly a chance to solve it too.
My initial table, dividing possibilities by regular coin or not, and by whether or not a tails comes up. (N coins, K tosses)
Special Coin, Tails appears - 0
Special Coin, All Heads - 1/N = (2^K)/(N*2^K)
Regular Coin, All Heads - (N-1)/N * 1/(2^K) = (N-1)/(N*2^K)
Regular Coin, Tails appears - (N-1)/N * (2^K-1)/(2^K) = ((N-1)(2^K-1))/(N*2^K)
Elim ...[text shortened]... ecial - 2^K
Normal - N-1
So the odds the coin is the double-headed is [b](2^K)/(2^K+N-1)[/b]
09 Apr 08
I'd like to post a problem also. I know that's not probability and is quite easy, but still I've heard some people answering wrongly.
Consider a bacteria in a sterile glass. Only the bacteria and the glass. Once in a minute, the bacteria reproduces and so the number of bacterias in the glass is doubled then. In an hour, the glass is full. Now two of these kind of bacterias are placed in the glass. How long would it now take to have the glass full?
09 Apr 08
1 edit
Originally posted by PBE6Sorry, decided to PM my solution, to let others have fun.
Lol. 🙂
OK, one more. The rate of infection from a particular disease is 1 in 1,000,000 in the general population. A hospital wants to administer a test that is 99% accurate (i.e. if 100 people who have the disease get tested, 99 will test "positive" and 1 will test "negative" falsely) and 95% specific (i.e. if 100 who don't have the disease get tested, 95 ts a positive result on the test, what is the chance that they really have the disease?
09 Apr 08
Originally posted by kbaumenI'll PM you my answer too.
I'd like to post a problem also. I know that's not probability and is quite easy, but still I've heard some people answering wrongly.
Consider a bacteria in a sterile glass. Only the bacteria and the glass. Once in a minute, the bacteria reproduces and so the number of bacterias in the glass is doubled then. In an hour, the glass is full. Now two of these ...[text shortened]... kind of bacterias are placed in the glass. How long would it now take to have the glass full?