2 probability questions done wrong :(

[Monkeyseat]

My teacher told me I had done these 2, well not wrong completely but needed more.

1)

http://img80.imageshack.us/img80/5889/hpim08443fz.jpg (question b)

Well I did:

Red: 29/60=0.483r 0.483r x 20=9.6r=9 or 10

White: 11/60=0.183r 0.183r x 20=3.6r=3 or 4

Yellow: 20/60=0.3r 0.3r x 20=6.6r=6 or 7

He told me I had to choose one answer and I'm not sure how/why...

2)

Finally, before I hand them in soon:

I had this table to show all the probabilities of outcomes on a dice:

http://img206.imageshack.us/img206/1958/hpim08537uj.jpg

a)

The question said what is the probability of rolling 2 6's:

The probability for getting 2 6's is shown as 1/36.

Then we had a question that said:

b)

The probability of getting a 6 on 1 dice is 1/6. The probability of getting a 6 on the other dice is 1/6.

Multiply these probabilities, compare your answer to part a.

What do you notice?

I just put;

The answers are the same, the probability of throwing 2 6's is 1/36, both methods get the same answer.

He said I needed to explain why they were the same...

Thanks.

 

[dagr8est]

1. I don't see any problem with how you did this question. It's a matter of rounding up or down since you can't have a partial bead. Since all of the results should be rounded up, it's up to you to decide which to round down.

2. Your explaination is correct but your teacher probably wants you to present it more formally. The rule is that if there are "x" ways one thing can happen, and "y" ways another thing can happen, then there are exactly "xy" ways they can happen together. There is a 1/6 chance of rolling a six on the first die. Similarily, there is a 1/6 chance of rolling a six on the second die. Therefore, there is a (1/6)(1/6)=1/36 chance of rolling a six on both dice.

 

[Monkeyseat]

1. I don't see any problem with how you did this question. It's a matter of rounding up or down since you can't have a partial bead. Since all of the results should be rounded up, it's up to you to decide which to round down.

2. Your explaination is correct but your teacher probably wants you to present it more formally. The rule is that if there are "x" ways one thing can happen, and "y" ways another thing can happen, then there are exactly "xy" ways they can happen together. There is a 1/6 chance of rolling a six on the first die. Similarily, there is a 1/6 chance of rolling a six on the second die. Therefore, there is a (1/6)(1/6)=1/36 chance of rolling a six on both dice.

Today I did this:

1)

Well if we round up it will be bigger than 20.

Red: 29/60 =about 9/20 (as it is just less than half)

White: 11/60=about 4/20 (as it is just over 1/6)

Yellow: 20/60=about 7/20 (as it is just over 1/3)

Okay thanks, I'll put:

2)

6,6 appears once in the table as there is only one six on neach dice, as there is 36 possible outcomes, the probability of 2 6's is 1/36.

The rule is that if there are "x" ways one thing can happen, and "y" ways another thing can happen, then there are exactly "xy" ways they can happen together. There is a 1/6 chance of rolling a six on the first die. Similarily, there is a 1/6 chance of rolling a six on the second die. Therefore, there is a (1/6)(1/6)=1/36 chance of rolling a six on both dice.

It is just a different technique of reaching the answer.

Would you say that is okay? I don't know how to compare it to the table. I have questions that ask the exact same thing but with colours....:

R,R appears twice in the table as Fatima has one red and Alan has 2 so Fatimas red could be drawn with any one of the other 2 reds (see table).

The rule is that if there are "x" ways one thing can happen, and "y" ways another thing can happen, then there are exactly "xy" ways they can happen together. There is a 1/3 chance of Fatima picking a red and a 1/3 chance of Alan picking a red. Therefore, there is a (1/3)(1/3)=1/9 chance of them picking red.

It is just a different technique of reaching the answer.

B,B appears eight times in the table as Fatima has 2 blues and Alan has 4 so one of Fatima's blues could be drawn with any of Alan's blues and her other blue could also be drawn with any one of Alan's blues (see table).

The rule is that if there are "x" ways one thing can happen, and "y" ways another thing can happen, then there are exactly "xy" ways they can happen together. There is a 2/3 chance of Fatima picking a blue and a 2/3 chance of Alan picking a blue. Therefore, there is a (2/3)(2/3)=4/9 chance of them picking blue.

It is just a different technique of reaching the answer.

 

[Monkeyseat]

EDIT:

Nevermind doesn't matter, handed the work in now.