Probability math question.

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Probability math question.

Mathematics
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Some questions for you: How many marbles are there in total? How many of them are red? blue?
16 marbles in total 4 red 7 blue

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Alright, so the probability of drawing a red is \(\dfrac{\text{number of reds}}{\text{total number of marbles}}\). Similarly, the probability of drawing a blue is \(\dfrac{\text{number of blues}}{\text{total number of marbles}}\)
Red probability is 4/16 simplified to 1/4 Blue probability is 7/16
right?
Yes. So in the first problem, you're drawing a red marble first, which you know has probability 1/4. When you put the marble back in, you still have 16 total marbles, so the probability of drawing a blue is 7/16. The fact that you replace the first marble means that your draws are independent. This means \[P(\text{red then blue})=P(\text{red})\times P(\text{blue})\]
so you multiply red, 4, by blue, 7. or you multiply the probabilities of 1/4 and 7/16
You multiply the probabilities. \(P(\text{red})=\dfrac{1}{4}\) and \(P(\text{blue})=\dfrac{7}{16}\), so \(P(\text{red then blue})=\dfrac{1}{4}\times\dfrac{7}{16}\).
sorry was afk my mom called. I'll get right to that :)
7/64?
Right. For the second problem, you're not putting back the first marble. So on the first draw, the probability of drawing a red stays the same, \(\dfrac{1}{4}\). On the second draw, there are now 15 marbles in total, so what's the probability of drawing a blue from these 15?
7/15
Right. As before, the probabilities are independent, so \[P(\text{red, then blue (after replacement)}=P(\text{red})\times P(\text{blue})_{\text{(after replacement)}}\]
so 1/4 x 7/15 = 7/60
Yes
The answer to part (c) should be obvious. In the first scenario, you're keeping the number of total marbles the same, but in the second it's being changed.
yeah ok. I get it. I was pretty much speeding through the lesson cause its the end of school but I do enjoy math when I slow down and understand it. Thanks so much :)
You're welcome!

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