Michelle My Bell Posted November 25, 2015 Share Posted November 25, 2015 I need help understanding how this problem is done... Suppose there are 9 blue marbles, 4 yellow marbles and 10 green marbles in a box. What is the probability of randomly selecting three green marbles with replacement? Show fractions (don't have to simplify final fraction) and then round to 4 decimal places. I don't even know where to begin. Quote Link to comment Share on other sites More sharing options...

SparklyUnicorn Posted November 25, 2015 Share Posted November 25, 2015 10/23 is the probability of picking a green marble because there are 10 green out of 23 marbles. When you have multiple events with replacement you multiply. Replacement means each time you draw a marble you have the same probability because the conditions remain the same. This would change if you did not replace the marble you pick out of the bag. So (10/23)(10/23)(10/23) 3 Quote Link to comment Share on other sites More sharing options...

regentrude Posted November 25, 2015 Share Posted November 25, 2015 (edited) Suppose there are 9 blue marbles, 4 yellow marbles and 10 green marbles in a box. What is the probability of randomly selecting three green marbles with replacement? Show fractions (don't have to simplify final fraction) and then round to 4 decimal places. I do not know the meaning of the phrase "with replacement". If you have to pick 3 marbles consecutively and want each of them to be green: You have 23 marbles. 10 of them are green. The probability that the first marble selected is a green marble is the number of "desirable outcomes" (=10, because you have 10 green marbles in the mix) divided by the total number of marbles. So, the probability that the first marble picked is green is 10/23. You now have 22 marbles left, out of which there are 9 green ones. The probability to select from these 22 a green one is 9/22. You now have 21 marbles left, out of which there are 8 green ones. The probability to select from these 21 a green one is 8/21. The probability to have selected 3 green marbles is the product of the probability for having selected a green marble as first, second, and third marble: (10/23)*(9/22)*(8/21) If you replace the marble after the first pick and then pick again out of the total 23: The answer would be (10/23)^3, because the probability to pick a green marble is 10/23 for each of the tries. Edited November 25, 2015 by regentrude Quote Link to comment Share on other sites More sharing options...

daijobu Posted November 26, 2015 Share Posted November 26, 2015 "with replacement" is a sort of math shorthand for removing a green marble, putting it back into the box, then choosing another green marble, and then returning it to the box, then choosing a green marble again. "without replacement" is short for removing a green marble, setting it aside, and then selecting a new marble. There are fewer marbles in the box now, and the proportion of green marbles changes. Since the events (choosing a green marble) are independent, that is, the outcome of choosing the first marble does not influence the probability of choosing the next marble, you can multiply the probabilities. (I like to tell my kids that the box and marbles aren't conspiring among themselves, thinking that since we already chose 2 green marbles, it's going to make us choose some other color.) So: (10/23)^3 Quote Link to comment Share on other sites More sharing options...

MarkT Posted November 27, 2015 Share Posted November 27, 2015 "with replacement" is a sort of math shorthand for removing a green marble, putting it back into the box, then choosing another green marble, and then returning it to the box, then choosing a green marble again. "without replacement" is short for removing a green marble, setting it aside, and then selecting a new marble. There are fewer marbles in the box now, and the proportion of green marbles changes. "with replacement" is probably a "text book math" term versus math shorthand Your descriptions are spot-on! 1 Quote Link to comment Share on other sites More sharing options...

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