Probabilites or Monty Hall Problem:

Test Preparation
1

<p>I need some help with this sat2 problem:</p>

<p>Seven blue marbles and six red marbles are held in a single container. Marbles are randomly selected one at a time and not returned to the container. If the first two marbles selected are both blue, what is the probability that at least two red marbles will be chosen in the next three selections?</p>

<p>I just did [P of 2 red] x [P of 3 red] = (6/11)(5/10)(6/9) + (6/11)(5/10)(4/9). But thats wrong. I have a feeling that this is one that has a similar principle as those monty hall problems.</p>

<p>Does anyone know what I did wrong?</p>

<p>thanks</p>

2

<p>I'm not terribly sure about this, but I think the answer is (5C1*6C2) / 11C3 -or is that what you have?-. Again, I'm not sure about this. Someone more knowledgable may be able to confirm it.</p>

3

<p>hmm im getting 13/33.</p>

4

<p>maybe its 3(6/11)(5/10)(5/9) + (6/11)(5/10)(4/9).</p>

<p>u need to multiply the first part by 3 and also it should be 5/9 not 6/9 since there are 5 blue balls left</p>

<p>maybe though...</p>

5

<p>i got 19/33</p>

6

<p>(6/11)(5/10)(5/9)+(6/11)(5/10)(5/9)+(5/11)(6/10)(5/9)+(6/11)(5/10)(4/9)=570/990=19/33</p>

<p>The first part is irrelevant. All you are solving is the probablity to draw 2 reds in the next three trials starting with 5 blue and 6 red. The first term is the probability to draw the first and second as red only, the second term, to draw the first and third, the third term, to draw the second and third, and the fourth term, to draw all four red.</p>

7

<p>why do we have to break it down to drawing 1st and second vs 1st and third instead of just using drawing 2 red total?</p>

8

<p>monty hall problems have to do with the game show host and how if you have 3 doors and you eliminate one door your chance of winning the prize behind the door is not 1/2 but actually 1/6? Is that how it goes?</p>

9

<p>Hi mans Im in ninth grade and im taking the math A regents. how can i solve this problem. if u could, please give a formula :S. An urn contains four yellow marbles and three blue marbles. How many blue marbles must be added to the urn so that the probability of picking a yellow marble will be 1/3?</p>

<p>:S</p>

10

<p>x= # blues added
New total marbles = 4 (yellow) + 3 (blue) + x (blue)</p>

<p>Prob(of picking yellow0 = 4/(4+3+x)= 4/7+x = 1/3
Solve for x by cross-multiply: 12 = 7+x, x = 5</p>