Basic Probability Questions, Part 2

May 2010
5
0
As I said in the previous thread, I just need help understanding these and how to solve them. Thank you.

3.) A "special" die consisting of one spot on two sides, two spots on two sides, and three spots on two sides is tossed with a "regular" die of six sides. Find the probability that .... (hint: create an addition chart)

a. the sum of the dice is four
b. neither die is a three
c. the same number appears on each die

4.) Place an "E" beside the event if it best represents "E"mpirical probability or an "M" beside the event if it best represents "M"athematical probability.

a. predicting the weather
b. in a 7-card hand, predicting the probability of getting no aces.
c. predicting the chance of being in a car accident.
d. predicting the chance of winning the lottery
e. predicting the probability of tossing a seven with two "loaded" dice
f. predicting the chance of at least 2 people in a group having their birthdays on the same day of the year.
 

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MHF Hall of Honor
Mar 2010
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As I said in the previous thread, I just need help understanding these and how to solve them. Thank you.

3.) A "special" die consisting of one spot on two sides, two spots on two sides, and three spots on two sides is tossed with a "regular" die of six sides. Find the probability that .... (hint: create an addition chart)

a. the sum of the dice is four
b. neither die is a three
c. the same number appears on each die

4.) Place an "E" beside the event if it best represents "E"mpirical probability or an "M" beside the event if it best represents "M"athematical probability.

a. predicting the weather
b. in a 7-card hand, predicting the probability of getting no aces.
c. predicting the chance of being in a car accident.
d. predicting the chance of winning the lottery
e. predicting the probability of tossing a seven with two "loaded" dice
f. predicting the chance of at least 2 people in a group having their birthdays on the same day of the year.
I hope you don't mind my expressing some annoyance, but I gave you some advice for #3 already. Is it so hard to make a chart? Here's a big chart

(1,1) -> 2
(1,1) -> 2
(1,2) -> 3
(1,2) -> 3
(1,3) -> 4
(1,3) -> 4
(2,1) -> 3
(2,1) -> 3
(2,2) -> 4
(2,2) -> 4
(2,3) -> 5
(2,3) -> 5
(3,1) -> 4
(3,1) -> 4
(3,2) -> 5
(3,2) -> 5
(3,3) -> 6
(3,3) -> 6
(4,1) -> 5
(4,1) -> 5
(4,2) -> 6
(4,2) -> 6
(4,3) -> 7
(4,3) -> 7
(5,1) -> 6
(5,1) -> 6
(5,2) -> 7
(5,2) -> 7
(5,3) -> 8
(5,3) -> 8
(6,1) -> 7
(6,1) -> 7
(6,2) -> 8
(6,2) -> 8
(6,3) -> 9
(6,3) -> 9

And an equivalent table here

(1,1) -> 2
(1,2) -> 3
(1,3) -> 4
(2,1) -> 3
(2,2) -> 4
(2,3) -> 5
(3,1) -> 4
(3,2) -> 5
(3,3) -> 6
(4,1) -> 5
(4,2) -> 6
(4,3) -> 7
(5,1) -> 6
(5,2) -> 7
(5,3) -> 8
(6,1) -> 7
(6,2) -> 8
(6,3) -> 9

To get the probability of rolling a 9, for example, notice that in the first chart there are two ways to get 9 out of 36 possible rolls, making the probability 2/36 = 1/18. In the second chart, there is one way to get 9 out of 18 possible rolls, so the probability is again 1/18.

There may be faster ways to solve this problem, but this way is straightforward and requires just a very basic understanding of what's going on.
 
May 2010
5
0
Thank you so much, I apologize. You made that much easier for me to understand. Now I can go ahead and finish a whole section of my mathbook thanks to that reply :) You are amazing.
 

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MHF Hall of Honor
Mar 2010
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821
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Thank you so much, I apologize. You made that much easier for me to understand. Now I can go ahead and finish a whole section of my mathbook thanks to that reply :) You are amazing.
Haha, you're welcome. Glad my explanation worked for you.
 
Oct 2009
303
33
Are we sure that this is correct? The way I see it the possible outcomes of rolling die 1 (the special die) are:{1,2,3} with each having a \(\displaystyle \frac{1}{3}\) probability. And for die 2 (the regular, six-side die) our possible outcomes are:{1,2,3,4,5,6} with each number having a \(\displaystyle \frac{1}{6}\)

So for part A:

(We'll assume that in each of these pairs the first number listed is from the special die and the second number from the regular die)

The possible outcomes for achieving a sum of 4 are:{(1,3),(2,2),(3,1)}.

Okay, so this is equivalent to \(\displaystyle pr(1\cap 3)+pr(2\cap 2)+pr(3\cap 1)\)

Lastly, for C we're looking for the probability that we will roll the same number on both dies

\(\displaystyle pr(1\cap 3)= pr(1)*pr(3)=\frac{1}{3}*\frac{1}{6}=\frac{1}{18}\)

\(\displaystyle pr(2\cap 2)= pr(2)*pr(2)=\frac{1}{3}*\frac{1}{6}=\frac{1}{18}\)

\(\displaystyle pr(3\cap 1)= pr(3)*pr(1)=\frac{1}{3}*\frac{1}{6}=\frac{1}{18}\)

So then:

\(\displaystyle pr(1\cap 3)+pr(2\cap 2)+pr(3\cap 1)= \frac{1}{18}+\frac{1}{18}+\frac{1}{18}= \frac{3}{18}=\frac{1}{6}\)

For B we're looking for the probability that neither die is a 3.

So for the special die, the probability that you won't roll a 3 is \(\displaystyle pr(3')=\frac{2}{3}\) and for the regular die, the probability that you won't roll a 3 is \(\displaystyle pr(3')=\frac{5}{6}\). So now that probability that you won't roll a 3 on either die is

\(\displaystyle pr(3'\cap 3')=\frac{2}{3}*\frac{5}{6}=\frac{10}{18}=\frac{5}{9}\)

Lastly, for part C we're looking for the probability that we'll roll the same number on both dice meaning we'll get {(1,1),(2,2),(3,3)}.

\(\displaystyle pr(1\cap 1)= pr(1)*pr(1)=\frac{1}{3}*\frac{1}{6}=\frac{1}{18}\)

\(\displaystyle pr(2\cap 2)= pr(2)*pr(2)=\frac{1}{3}*\frac{1}{6}=\frac{1}{18}\)

\(\displaystyle pr(3\cap 3)= pr(3)*pr(3)=\frac{1}{3}*\frac{1}{6}=\frac{1}{18}\)

So then:

\(\displaystyle pr(1\cap 1)+pr(2\cap 2)+pr(3\cap 3)= \frac{1}{18}+\frac{1}{18}+\frac{1}{18}= \frac{3}{18}=\frac{1}{6}\)

This seems correct to me but of course I could be wrong. If I am, I'd love for someone to steer me in the right direction
 
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MHF Hall of Honor
Mar 2010
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Are we sure that this is correct?
I like your way better than mine. But I don't know why you called into question the correctness of my method, because both methods produce the same answers.
 
Oct 2009
303
33
My mistake. I thought you had slightly different numbers. I guess I was misreading.