I toss 2 coins, one lands heads up. What is the probability the other coin is heads up also?
I toss 2 coins, one lands heads up. What is the probability the other coin is heads up also?
> I toss 2 coins, one lands heads up. What is the
> probability the other coin is heads up also?
You have to be a little clearer with your statement. You probably don't mean, "I toss the first one and it's heads; what is the chance that the second one will also be heads," which, of course, is one-half.
I sense that you mean, "I tossed two coins and at least one of them was a head. What is the chance that the other one was also a head?" In that case, the answer is one-third.
There are four ways two coins can land, but when you announce "one lands heads up," you eliminate the tails-tails possibility. So three equiprobable results remain: H-H, H-T, and T-H. Only one of them makes "the other coin heads up also," so the answer is one chance out of three.
Don
OK, I'll come up with a more challenging question.
> You have to be a little clearer with your statement.
> You probably don't mean, "I toss the first one
> and it's heads; what is the chance that the second one
> will also be heads," which, of course, is
> one-half.
> I sense that you mean, "I tossed two coins and
> at least one of them was a head. What is the chance
> that the other one was also a head?" In that
> case, the answer is one-third.
> There are four ways two coins can land, but when you
> announce "one lands heads up," you eliminate
> the tails-tails possibility. So three equiprobable
> results remain: H-H, H-T, and T-H. Only one of them
> makes "the other coin heads up also," so the
> answer is one chance out of three.
> Don
> I toss 2 coins, one lands heads up. What is the
> probability the other coin is heads up also?
do they have that game in a casino?
I never thought I'd see War in a casino! Hell, it was one of the JOKE GAMES in the movie Vegas Vacation!
Bettie
Soon they'll have you simply picking a number between one and ten (I loved that part.)
My all-time favorite casino game to date .. playing tic-tac-toe against a live chicken.
what's the probability of a patient having cancer if the mammogram is negative, where, if the patient has cancer the mammogram will detect it 80% of the time and if the patient does not have cancer the mammogram will issue a false positive 10% of the time? The probability of cancer is 1%.
> Soon they'll have you simply picking a number between
> one and ten (I loved that part.)
> My all-time favorite casino game to date .. playing
> tic-tac-toe against a live chicken.
Hi,
Last time I checked this was a blackjack forum and not a homework forum
Good luck though,
MGP
> what's the probability of a patient having cancer if
> the mammogram is negative, where, if the patient has
> cancer the mammogram will detect it 80% of the time
> and if the patient does not have cancer the mammogram
> will issue a false positive 10% of the time? The
> probability of cancer is 1%.
Contrary to opinions posted on some websites, I am loath to delete posts without a compelling reason.
However, I am scratching my head trying to find the blackjack content in this thread.
Grosjean also gives a couple examples of Bayes' Theorem in Beyond Counting. I thought some people might want to expand their knowledge beyond learning systems that gain percentages of mediocre systems to begin with or cracking jokes about Hooters.
> Contrary to opinions posted on some websites, I am
> loath to delete posts without a compelling reason.
> However, I am scratching my head trying to find the
> blackjack content in this thread.
> Grosjean also gives a couple examples of Bayes'
> Theorem in Beyond Counting. I thought some people
> might want to expand their knowledge beyond learning
> systems that gain percentages of mediocre systems to
> begin with or cracking jokes about Hooters.
whatever happened to that guy who came in here a few weeks back with the intent to "wrestle control away" from the math studs? we could sure use his help about now.
p(cancer|neg) = p(neg|cancer)/[p(neg|cancer)p(cancer) + p(negative|benign)p(benign)]
= (.2)(.01)/[(.2)(.01)+(.9)(.99)] = .0022.
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