why isn't the prob of rolling two doubles 1/36? prob of rolling any number on 1 dice is 1/6 shouldn't you multiply the prob of both dice like in the first coin flip video? I understand the explanation given, but I'm trying to figure out why the same coin logic doesn't work.
You need to consider how many ways you can roll two doubles, you can get 1,1 2,2 3,3 4,4 5,5 and 6,6 These are 6
possibilities out of 36 total outcomes. The probability for rolling one of these, like 6,6 for example is 1/36 but you want to include all ways of rolling doubles.
If this was in a exam, that way of working it out takes too long so is there any quick ways so you won't waste time?
well you can think of it like this. It really doesn't matter what you get on the first dice as long as the
second dice equals the first. so the probability of the second equaling the first would be 1/6 because there are six combinations and only one of them equals the first.
At
4:14
is there a mathematical reason why the favourable outcomes line up on the diagonal?
That
is a result of how he decided to visualize this. Imagine we flip the table around a little and put it into a coordinate system. Along the x-axis you put marks on the numbers 1, 2, 3, 4, 5, 6, and you do the same on the y-axis. We are interested in rolling doubles, i.e. getting the same on both dice. If we let x denote the number of eyes on the first die, and y do the same for the second die, we are interested in the case y = x. But this is the equation of the diagonal line you refer
to.
At 2.30 Sal started filling in the outcomes of both die. This video wasn't what i was looking for but some of you might be able to help. I had a question: "Two dice are rolled, copy and complete the table below"...... Then there was a table which looked exactly like the one Sal drew. But it had been filled out differently: In the first box (Dice 1 and Dice 2) it had been filled in as 2. Sal wrote 1,1. In another box on my sheet (1 across and three down on Sal's
diagram) it had been filled out as 1. I am very confused on how they got this answer so if you understand what I'm talking about please answer. Sorry for the bad explaining!
It's because you aren't supposed to add them together. The reason for it being 1,1 is because it's supposed to symbolize two dice both equaling 1 so it would be 1,1, with each one meaning the outcome of each die.
Is there a way to find the probability of an outcome without making a chart?
Probably the easiest way to think about this would be: P(Rolling a 1 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296 P(Rolling a 2 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296 P(Rolling a 3 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296 P(Rolling a 4 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296 P(Rolling a 5 four times in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296 P(Rolling a 6 four times
in a row) = 1/6 * 1/6 * 1/6 * 1/6 = 1/1296
Adding these probabilities together, we get: 6/1296 = 1/216.
So the probability you will roll the same number four times in a row with a fair dice is 1/216. Does this help.
Find the probablility of the occurance of1on a die if it has one more of its faces marked as 1instead of 6
Here's how you'd do the problem. We know that
two of the sides have 1. The rest have other numbers. Since in total, there are 6 sides, get the amount of sides that have 1 divided by 6 (six possible outcomes) and that's your answer. So in your case, it's 2/6, which is 1/3.
I was wondering if there is another way of solving the dice-rolling probability and coin flipping problems without constructing a diagram? Is there
a way to find the solution algorithmically or algebraically?
Definitely, and you should eventually get to videos descriving it. But to show you, I will try and descrive how to do it.
So, since there is an equal chance to roll any number on a six sided die, that means the chance of rolling any one number is one out of 6 or 1/6. You can see that with a diagram. Now, rolling two different
numbers in a specific order you can tell with a diagram is 1/36. To find this out through math though you multiply probabilities of events happening if you are looking for both of them happening. so you want to roll x first AND y second. so that's 1/6*1/6, which is 1/36. And you can keep goign with this pattern.
The calculation is a bit different if you are looking for one thing to happen OR another. In simple caes it's just adding, like what are the odds of rolling a 1 OR 2 on a dice? you
add the two, which you can see on a diagram.
You still have to be careful, like if a problem asks what the odds of rolling a 1 AND a 2 on ONE die is, you can't roll both so the answer is 0. Or rolling a 1 on one die OR rolling a 2 on another. It's still 1/6 since you are rolling them separately and they don't effect each other. It can get a bit confucing, but most of the time you will be using the more simple cases. If ever in doubt use diagrams to see a few cases of what you're doing to
get an intuition
Can learners open up a black board like Sals some where and work on that instead of the space in between problems?
If you're working on a Windows pc, you would need either a touchscreen pc, complete with a stylus pen or a drawing tablet. Then you could download for free the Sketchbook Pro software for Windows and invert the colors. Voila, you have a Khan Academy style blackboard. If you don't have a pc with a touchscreen or drawing tablet, you
can use phones with drawing styluses like the Samsung Note phones to do your work on SketchBook mobile. Or just, you know, do it on plain lined, graphing, or print paper (^ u^ hope this helped.
Can someone help me "If you roll two fair six-sided dice, what is the probability that at least one die shows a
3?"
P(at least one 3)=1-P(no 3s) There are 5 ways to get no 3s
on each die, and so there are 25 ways to get no 3s, and so P(no 3s)=25/36.
So the probability you want is 1-(25/36)=11/36.
From a well shuffled 52 card's and black are removed from cards find the probability of drawing a king or queen or a red card
If the black cards are all removed, the probability of drawing a red card is 1; there are only red cards left. Therefore, the probability of (literally anything) or a red card is also 1.
