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bealiberty47
 one year ago
Probability math question.
bealiberty47
 one year ago
Probability math question.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Some questions for you: How many marbles are there in total? How many of them are red? blue?

bealiberty47
 one year ago
Best ResponseYou've already chosen the best response.016 marbles in total 4 red 7 blue

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, 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 \(\dfrac{\text{number of blues}}{\text{total number of marbles}}\)

bealiberty47
 one year ago
Best ResponseYou've already chosen the best response.0Red probability is 4/16 simplified to 1/4 Blue probability is 7/16

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes. 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 \[P(\text{red then blue})=P(\text{red})\times P(\text{blue})\]

bealiberty47
 one year ago
Best ResponseYou've already chosen the best response.0so you multiply red, 4, by blue, 7. or you multiply the probabilities of 1/4 and 7/16

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You multiply the probabilities. \(P(\text{red})=\dfrac{1}{4}\) and \(P(\text{blue})=\dfrac{7}{16}\), so \(P(\text{red then blue})=\dfrac{1}{4}\times\dfrac{7}{16}\).

bealiberty47
 one year ago
Best ResponseYou've already chosen the best response.0sorry was afk my mom called. I'll get right to that :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right. 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right. As before, the probabilities are independent, so \[P(\text{red, then blue (after replacement)}=P(\text{red})\times P(\text{blue})_{\text{(after replacement)}}\]

bealiberty47
 one year ago
Best ResponseYou've already chosen the best response.0so 1/4 x 7/15 = 7/60

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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.

bealiberty47
 one year ago
Best ResponseYou've already chosen the best response.0yeah 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 :)
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