Probabilty

[151AND8TH]

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How do you work out the following situations?

rolling of two 6-sided dice..
what is the probability of obtaining the following:
a) At least one 6
b) An even number on at least one die


a)11/36
b)3/4

Thanks

 

[will9631]

How do you work out the following situations?

rolling of two 6-sided dice..
what is the probability of obtaining the following:
a) At least one 6
b) An even number on at least one die


a)11/36
b)3/4

Thanks

I hate these, and I saw a useful formula months ago, but I've forgotten it. I think I worked out the first one though, but I had to do it the long way

first work out the possible rolls (hence I call this the long way).

1,1
1,2
1,3
1,4
1,5
1,6
2,1
2,2
2,3
2,4
2,5
2,6
.
.
.

up through 6,6; which adds up to 6*6 or 36 total combinations. This problem is assuming 1,2 and 2,1 as different rolls, which I think is usually the case with these. anyway, now you add up the number of rolls with at least one 6: 16, 61, 26, 62, 36, 63, 46, 64, 56, 65, 66; which adds to 11.

therefore, in infinite rolls, you will average getting 11 instances of at least one six out of every 36, so the answer is 11/36

For the second one all you actually have to do is realize that there are 3 possible even numbers and 3 possible odd numbers. therefore the odds of getting one odd number is .5, and the odds of getting both odds (no evens) is .5*.5 or .25. if the odds of getting no evens is 1/4, then the odds of getting at least one is 3/4, and the odds of getting both evens is 1/4 (same as getting both odds). This is therefore a much simpler problem than the first, in my mind anyway.

hope that helps,
Will

 

[Ron Jeremy]

This is a probability problem that is solved using the concept of unions and intersections. The explanations of the "hows" and "whys" of the equation I'm using can be explaned in a stats refernce guide that cost about $10. I bought the "Super Review of Statistics" at Borders for $9. The explanation is on page 111.

The equation is P(E)+P(F)-P(E)*P(F). In the first question:
P(E) is the prob of getting 6 on the 1st die
P(F) is the prob of getting 6 on the 2nd die
Therefore => (1/6)+(1/6)-(1/6)*(1/6) = 2/6 + 1/36 = 12/36 - 1/36

Final answer is 11/36.

This is a formula you will just have to memorize if you don't fully understand statistical theory. If you remeber it, the answer can be obtained in 30 seconds.

Question 2.
Using the same formula:
(1/2)+(1/2)-(1/2)*(1/2) = 1 - 1/4

Final answer is 3/4

This is a great example of why you need to review probability concepts, counting rules, and MOST IMPORTANTLY, using fractions fast. Probability and counting problems will be on the test.

will9631's method also works if you forget the formula. With only two dices, manually counting the possibilities works fine. With more dices, or more complex questions, manual counting may get too cumbersome and too time-consuming.

 

[Streetwolf]

How do you work out the following situations?

rolling of two 6-sided dice..
what is the probability of obtaining the following:
a) At least one 6
b) An even number on at least one die


a)11/36
b)3/4

Thanks

Here's another way. I hate the problems where it says 'at least'. So if it says 'at least one' a better way is to figure out the probability of it NOT happening, since that would be 'no 6' and that's only one scenario.

P(no 6 rolled) = 5/6*5/6 = 25/36
Thus P(at least one 6 rolled) = 1 - 25/36 (WHY?) = 11/36.

For the second one, you just ask what is P(no even number)?

P(no even number rolled) = 1/2*1/2 = 1/4.
Thus P(at least one even number rolled) = 1 - 1/4 (WHY?) = 3/4.

Hint: The 'WHY?' part has the same answer for both.

 

[tastybeef]

a. I did not like these "at least" questions and I'm not great at math, so the way I dealt with the question is by drawing a grid. After much practice, I can see the grid in my mind. For the visual learners like me, this helps immensely with 2-dice-throw questions.
Essentially, this turns the problem into seeing a pattern in a grid.

I've attached a doc file that explains this.

b. Easiest way is described by streetwolf.
You can also do binomial distribution, which I find a little easier to understand.
x^2 + 2xy + y^2 = 1
Where x = chance of getting even on dice 1
y = chance of getting odd on dice 2
x = y = 0.5
The basic interpretation of this equation:
When you throw 2 dice, you get 3 possibilities: even, even (x^2) ; even, odd (xy); and odd, odd (y^2).

Here you are concerned with getting at least one even, so add up the terms that have an x.
x^2 + 2xy = (0.5)(0.5) + 2(0.5)(0.5) = 0.75

 

[dentstd]

Here's another way. I hate the problems where it says 'at least'. So if it says 'at least one' a better way is to figure out the probability of it NOT happening, since that would be 'no 6' and that's only one scenario.

P(no 6 rolled) = 5/6*5/6 = 25/36
Thus P(at least one 6 rolled) = 1 - 25/36 (WHY?) = 11/36.

For the second one, you just ask what is P(no even number)?

P(no even number rolled) = 1/2*1/2 = 1/4.
Thus P(at least one even number rolled) = 1 - 1/4 (WHY?) = 3/4.

Hint: The 'WHY?' part has the same answer for both.

Streetwolf, your math posts are way too smart. I think you should stop posting. It's making me depressed. Seriously.

The smartest soln to this question would be to guess and move on. Yessa, that's what you do.

 

[151AND8TH]

Tastybeef , I have used this grid before but never actually realized the power in it!!.. this grid can single handedly solve ALL 2 dice problems!! (at least I think so)...

Have any one stumbled upon 3+ dice problems?

 

[Streetwolf]

Post them if you have any. I'm always up for a distraction from anatomy or biochem.