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A bowl contains ten chips numbered $1,2, ldots, 10$, respectively. Five chips are drawn at random, one at a time, and without replacement. What is the probability t...

Question

A bowl contains ten chips numbered $1,2, ldots, 10$, respectively. Five chips are drawn at random, one at a time, and without replacement. What is the probability that two even-numbered chips are drawn and they occur on even-numbered draws?

A bowl contains ten chips numbered $1,2, ldots, 10$, respectively. Five chips are drawn at random, one at a time, and without replacement. What is the probability that two even-numbered chips are drawn and they occur on even-numbered draws?



Answers

Rolling Dice Six dice are rolled. Find the probability that
two of them show a four.

You're asked. Probability. Getting some of eight when? Two days from we know we're gonna have equally likely outcomes cause we're rolling fair guys. So let's think about how the different ways in which we could get some of eight two and six, three and five girl four and four, three and six and two. These are all the ways in which we get a sum of eight. The probability of every single one of these is gonna be one out of 36. Therefore, this probability is gonna be five out of 36 because we add up all these probabilities to get the probability.

We're wondering about the probability of choosing exactly two fours when we roll six days to solve this. Let's use our formula. It's under a combination of six choose to multiplied by the probability of rolling what we want, which is a two. Now the probability rolling a two for one dices one out of six because there are six sides and only one satisfies it. And we're going to raise that to second power because we want to of them. This is that it could be multiplied by the probability of our failure. That is, if we roll the dice. There's five out of six chances of not getting what we want to six minus two power, so four. So this is just following the formula. Now we can write out what this is so combination of six choose to is gonna be six factorial over four factorial times two factorial. And then, of course, we have times 1/6 squared well deployed by 5/6 to the fourth power. Now, when you multiply all of this out, you're going to get zero point to 0094 So the probability a ruling six dice and getting exactly two fours is about 20%

So we want to know if you roll a set of two dies. What is the probability that the sum of the few dies isn't even number? Well, first we want to know how many possible outcomes do we have when we rolled to dice? Well, for the first dies, we have six options, which is the numbers one through six. And for the second option, we also have six, which is the numbers one through six again and says they are two separate dice. We have that buy the product rule. We have 36 possible pairs, and each of these Paris is equally likely to come out as our outcome. Each of these 36 parishes, Justus likely to be our outcome equally likely to be our outcome. Well, we just need therefore, witches need to show how many of these possible pairs give us an even some. Well, to get even some to get us some where the outcome is even we need to either some to even numbers, or we need to some two out numbers. Well, if you will, the first die and we haven't even number. We have three possible even outcomes, which is 24 and six and then for the second guy is the same thing. And since they're separate dies, we used a product rule and we get that nine out of thes 36 possible pairs are both even and give us, and even some for the odd is the same thing. We have three possible outcomes when we rolled the first dies and these possible outcomes are 13 implied and so are they for the second ice. And similarly, we get nine by the product rule, and we also therefore, we have nine out of these 36 possible pairs give us an even some where both of them are odd numbers. Therefore, the probability that we have the sum off the two dies be uneven. Number is nine plus nine, which is 18 over 36 because the probability that both of them are even he's nine over 36 the probability that both of them are odd is nine over 36. Therefore, the probability that we get uneven pair a pair where the summit's human is 1/2

Okay, So what is the probability that some off the numbers on two dice is even when they are rolled? Okay, So to do this, we have to kind of think, What if the first I comes up like 123 Well, so five and six. Well, if it's 123456 So if you rolled a one on your first die, then you needed to have 123 So you need one and one. You need one and three, and you need one and five for it to be even. So, if you're old too, then you need to to train to we need to and four. And you need two and six and 43 units, three and one and three and three and three and five. So it's pretty easy to see that. For each of these numbers, you'll have 333333 So it would be six chemistry number old favorable outcomes divided by the number of possible outcomes. So each of these so 123456 So this has six possible outcomes. This has sixties, has sixes, have sixes, has six. This is six So if the six some street divided by six times where six. So these 26 cats is out, so that's three divided by six, so that's one over two.


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