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Marbles. Find the 'probability of picking 3 red marbles if each A bag contains red] white _ and blue mBrBle0g Teturned to the bag before the next marble is pi...

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Marbles. Find the 'probability of picking 3 red marbles if each A bag contains red] white _ and blue mBrBle0g Teturned to the bag before the next marble is pickedNext Question

marbles. Find the 'probability of picking 3 red marbles if each A bag contains red] white _ and blue mBrBle0g Teturned to the bag before the next marble is picked Next Question



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A bag contains 3 red marbles, 4 green marbles, 2 yellow marbles, and 5 blue marbles. Once a marble is drawn, it is not replaced. Find the probability of each outcome. a blue marble, a yellow marble, and then a red marble

In this problem. I know. I'm trying to find the probability of picking a blue marble, not replacing it been a yellow marble and not replacing it. And then a red marble. I've drawn out the marble situation. You can see there are three red marbles or green to yellow, which I had to draw in black because I don't of yellow marker and five blue. The first thing I'm going to do is figure out the probability of picking a blue marble. I first start by totaling up the number of marbles. There are 14 marbles in total. Next I ask myself, how many of those 14 marbles are blue, which the answer is five. Now I'm going to assume that I've taken one of those five blue marbles out of the back. This means that there's no longer 14 marbles in the back. I've taken one out, so there's only 13 marbles. So my new denominator is 13. Now I see of those 13 how maney or yellow? Remember my black represents yellow. There are two. Let's now assume I take one of those out. This now means there are no longer 13 marbles in my bag. We have now dropped 12 marbles. And now I ask myself of those 12 how many are red? And the answer is three. To finish this, I'm going to multiply my fractions together. And to do that I multiply straight across the numerator and straight across the denominator. Five times, two times three is 30 and 14 times 13 times 12 is 2000 184. So that is our probability of picking a blue, a yellow and then a red without replacing any of the marbles 30 over.

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.

Probability of getting blue is four out of six. Probability of getting red is two out of six. We're gonna draw two marbles and we're not going to use replacement. So the probability of one being read and the other being blue is gonna be four times four over six times two over five or or we can do to over six times for over five. Regardless of which one you choose. That probability is going to be 0.267 or 26.7%.


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