Explore what probability means and why it's useful.
Log in Wendy Sugimura 9 years agoPosted 9 years ago. Direct link to Wendy Sugimura's post “If two standard dice are ...” If two standard dice are rolled. What is the probability that the total of two dice is less than 6? • (26 votes) Rhyss a year agoPosted a year ago. Direct link to Rhyss's post “less than 6 would not inc...” less than 6 would not include 6 so Length=10 ⁂ p()=10/36 (12 votes) bgljade 7 years agoPosted 7 years ago. Direct link to bgljade's post “A card is drawn from a st...” A card is drawn from a standard deck of 52 cards. Find the probability that is • (12 votes) Avinash Athota 7 years agoPosted 7 years ago. Direct link to Avinash Athota's post “I am just warning you, I ...” I am just warning you, I don't know much about cards that much, so my numbers may be off. Jim 3 years agoPosted 3 years ago. Direct link to Jim's post “Can't you multiply the po...” Can't you multiply the possibility(fraction) with the the same numerator or denominator to get a different but equivalent answer? • (4 votes) Andrew H. 3 years agoPosted 3 years ago. Direct link to Andrew H.'s post “Yes you can multiply prob...” Yes you can multiply probabilities with fractions that are equal to one. We usually want the fraction in the simpliest form though. (5 votes) Isaac 🤠 a year agoPosted a year ago. Direct link to Isaac 🤠's post “im hungry 🍞” im hungry 🍞 • (5 votes) LennyH a year agoPosted a year ago. Direct link to LennyH's post “me too😁” me too😁 (2 votes) Trin 3 years agoPosted 3 years ago. Direct link to Trin's post “does probability always h...” does probability always have to be written like a fraction? How do you know when to write it as a percentage? • (2 votes) green_ninja 3 years agoPosted 3 years ago. Direct link to green_ninja's post “Usually, the question con...” Usually, the question concerning probability should specify if they want either fractions or percentages. Here on KA, you can tell if they're asking for a percentage if you see a % sign by the answer box, while for fractions / decimals a small dialogue box will pop up after you click on the answer box telling you which form to put it in. (I've also seen them state which form to use in italics right after the question.) Hope this helps!😀 (7 votes) Nethra 2 years agoPosted 2 years ago. Direct link to Nethra's post “Um...there would be 7 dog...” Um...there would be 7 dogs instead of 9. And there would only be 2 brown dogs now. Which is equal to the number of white dogs. Or is there a more complex reason to this? I don't know. Anyway I hope this helps. (4 votes) moses.muramira22a 8 months agoPosted 8 months ago. Direct link to moses.muramira22a's post “what is the formula of pr...” what is the formula of probability • (2 votes) daniella 4 months agoPosted 4 months ago. Direct link to daniella's post “Classical Probability (Eq...” Classical Probability (Equally Likely Outcomes): Probability of an Event Not Occurring: Probability of Independent Events: Probability of Mutually Exclusive Events: Conditional Probability: (3 votes) beatlemaniac 9 months agoPosted 9 months ago. Direct link to beatlemaniac's post “Ok, I think I get it. So,...” Ok, I think I get it. So, would the probability of picking a yellow marble be 37.5%? I got 37.5% by turning 3/8 into a percentage. If I'm correct, this is a lot easier than I thought. • (2 votes) Joanne Cai 9 months agoPosted 9 months ago. Direct link to Joanne Cai's post “Yes that would be indeed ...” Yes that would be indeed correct! (2 votes) Jan Register 3 years agoPosted 3 years ago. Direct link to Jan Register's post “3 red marbles and 3 blue ...” 3 red marbles and 3 blue marbles. do not replace first marble in bag before picking again. probability that both marbles are blue • (1 vote) Jerry Nilsson 3 years agoPosted 3 years ago. Direct link to Jerry Nilsson's post “There are 6 marbles in to...” There are 6 marbles in total, and 3 of them are blue, so the probability that the first marble is blue is 3∕6 = 1∕2 Given that the first marble was blue, there are now 5 marbles left in the bag and 2 of them are blue, and the probability that the second marble is blue as well is 2∕5 So, the probability that both marbles are blue is 1∕2 ∙ 2∕5 = 1∕5 (3 votes) Jordania213 5 years agoPosted 5 years ago. Direct link to Jordania213's post “The mall has a merry-go-r...” The mall has a merry-go-round with 12 horses on the outside ring. If 12 people randomly choose those horses, what is the probability they are seated in alphabetical order? I've been stuck on this problem for so long and I have no clue to what is the right way to solve this problem? • (2 votes)Want to join the conversation?
[ ]
| 1-1 2-1 3-1 4-1 |
| 1-2 2-2 3-2 |
| 1-3 2-3 |
| 1-4 |
[ ]
total =6x6
a.) a heart or a face card.
b.) a jack or an ace card
c.) a 10 or a spade.
a. there are 13 heart cards and 12 face cards (aces aren't faces, right?), of which 3 are repeated, so 13+12-3 = 22/52 = 11/26
b. there are 4 jacks and 4 aces, so 4+4 = 8/52 = 4/26 = 2/13
c. there are 4 tens and 13 spades, and one 10 is repeated, so 4+13-1 = 16/52 = 8/26 = 4/13
I hope that helps!
Example: 3/4 chance times 3/3(numerator) equals 9/12. At my school, they say you can multiply fractions with the same numerator/denominator, but I haven't taken probability yet in my grade.
To find the probability of an event happening, you divide the number of ways the event can happen by the total number of possible outcomes.
If you want to find the probability of an event not happening, you subtract the probability of the event happening from 1.
If two events don't affect each other (like flipping a coin twice), you multiply their individual probabilities to find the probability of both events happening.
If two events can't happen at the same time (like rolling a die and getting a 1 or a 2), you add their individual probabilities to find the probability of either event happening.
If you want to find the probability of one event happening given that another event has already happened, you divide the probability of both events happening by the probability of the second event happening.
