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(cojccmcril Find the prcbability that they will have the followirg outcomes; Express }our answer as & frattionBotn the matbes 4 ErcenBoth of the marbles are the...

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(cojccmcril Find the prcbability that they will have the followirg outcomes; Express }our answer as & frattionBotn the matbes 4 ErcenBoth of the marbles are the same colorBoth marbles are 0f different co or'(1zp1MacBook Procommtloncommand

(cojccmcril Find the prcbability that they will have the followirg outcomes; Express }our answer as & frattion Botn the matbes 4 Ercen Both of the marbles are the same color Both marbles are 0f different co or '(1zp1 MacBook Pro comm tlon command



Answers

A jar contains 2 red marbles, 6 yellow marbles, and 4 green marbles. One marble is drawn and replaced, and then a second marble is drawn. Find the probability of each outcome. (Previous course) Both marbles are green.

In this problem. I know I have a bag of marbles, and I'm looking for the probability of picking a yellow and then the other, another yellow without replacing the marbles. You can see that I have drawn out the marbles. There are three red four green to yellow, which I don't have a yellow markers, so I made them black and five blue. The first thing I'm going to do is total up the number of marbles that I have in my back, So three plus four it's seven plus two is nine plus five makes 14. That means my first fraction will have a denominator of 14 since there's 14 total marbles next. Since I want to pick a yellow, I see how maney yellow marbles there are in my back, in which case there are two. Remember, my black marbles represent yellow. Now we're going to assume that I've picked one of those yellow marbles out of the bag, so I've erased one for my second fraction. I now have to take into account that one of those yellow marbles they're gone, so there is no longer 14 marbles in my bag. They're 13 and there's no longer. Two yellow marbles left. There's only one. So my second fraction is 1/13. Since there's 13 marbles in the bag and on Lee, one of them are yellow. I'm now going to multiply straight across the top and straight across the bottom or straight across the numerator and straight across the denominator. Two times one makes two and 14 times 13 makes 182. You only change the denominator number of a fraction when you're not replacing something. So since we were not replacing the yellow marbles, that's why I went from 14 down to 13. Now, if I wanted to, I could reduce this fraction. I could divide both the numerator and the denominator by two, which gives me 1/91 probability, and there you go.

In this problem. I know I have a bag of marbles and I want to find the probability of picking a blue marble, not replacing it than picking another blue marble. You can see I have drawn out the scenario. Here. There are three red marbles for green to yellow, which I had to use black. Since I don't have a yellow marker and five blue, the first thing I'm going to do is total up the number of marbles I have, so three plus 47 plus two is nine plus five makes 14. That means my first fraction will have a denominator of 14 since that's the total number of marbles. Next. Since I want to pick a blue marble, I count how many blue marbles there are, which is five. So that becomes my numerator. Now let's assume that I choose one of those five blue marbles you can see. I've erased it from the board to represent that I've taken it out of the back. Now I ask myself how Maney marbles are in the bag. It's no longer 14 since I took one blue one out. Now I only have 13 marbles total, so my new denominator of my second fraction is 13. In the same sense, I no longer have five blue marbles to choose from. There's only four blue marbles, so my new numerator is a four. Now I'm going to multiply straight across the numerator, straight across the denominator. Five times four is 20 and 14 times 13 is 182 and that is the probability of picking a blue marble, not replacing it than picking another blue marble.

Okay and in this case was implanted to find the probability that both marbles area boot marvel are yellow. So again, there are a total number of some elements of the sample spaces. One green marble, two yellow marble and spirit marbles with simply comes out with their acquittal of six mumbles. So the final answer, Boluda quite probability is equals to your bring to mumbles. That should be a which means that that will be from here itself. So you see, two divided by 62 using the concept off combination toe to see toe is one and 62 6 times five and a little bit too. So this is one by three times five. That is equivalent when by 15 all right.

In this problem. I'm choosing marbles out of the bag, and I know I want to find the probability of picking a blue marble, not putting it back. Then picking a green mark you can see have drawn the marble situation out there. Three red four green to yellow, which I don't have yellow marker. So it's black and five blue. The first thing I'm going to do is total. The number of marbles three plus four is seven plus two is nine plus five is 14. So the first denominator will be 14 because that's the total number of marbles. Next since my first pick, I want to be blue. I see how Maney blue marbles there are, which is five. Now let's assume I have picked one of those marbles, so I'm going to erase one from the bag because I've taken it out and I didn't put it back. That now changes the total number of marbles in my bag. So for my second fraction, the denominator is no longer 14. It drops down 13 since I took one marble out. Now I want my second marble to be green, so I see how many green there are, which is for so I put a four in the numerator. Finally to finish, I multiply story across the numerator straight across the nominator. Five times four is 20 and 14 times 13 is 182. So I have now found the probability of picking a blue marble, not replacing it.


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