# Probability

• Jun 2nd 2008, 09:28 AM
ttG
Probability
Hi! Can somebody explain me please how to solve these:
a) Calculate the probability of getTing at least one die showing a 5 using two dices?
b) Calculate the prob. of getting the sum 10 and none of the dices are showing a 5?
C) Calculate the prob. of getting two 5 if you may roll twice? That is if one of the dices doesn't show a 5 this one a second time and if none of the dices are showing a 5 you role both.
(Sleepy)
• Jun 2nd 2008, 10:20 AM
janvdl
Quote:

Originally Posted by ttG
Hi! Can somebody explain me please how to solve these:
a) Calculate the probability of getTing at least one die showing a 5 using two dices?
b) Calculate the prob. of getting the sum 10 and none of the dices are showing a 5?
C) Calculate the prob. of getting two 5 if you may roll twice? That is if one of the dices doesn't show a 5 this one a second time and if none of the dices are showing a 5 you role both.
(Sleepy)

a) The question asks AT LEAST. So it does not rule out the possibility that you could throw TWO fives.

The probability of throwing one 5 and any other number from 1 to 6 on the second die is: $\displaystyle \frac{6}{36} = \frac{1}{6}$

b) Well in total we have 36 possibilities.

So now let's see in how many ways can we make 10 without using a 5.

We could get 6 and 4. And on another event we could get 4 and 6.

So that's $\displaystyle \frac{2}{36} = \frac{1}{18}$

c) I do not quite understand your question. But for this we need to use the binomial distribution. I do not know if you're familiar with it.

The probability of getting 2 fives is $\displaystyle \frac{1}{36}$

So according to our binomial distribution the probability is:

$\displaystyle {2 \choose 2} \left( \frac{1}{36} \right) ^{2} \left( \frac{5}{36} \right) ^{2-2} = \frac{5}{46656}$
• Jun 2nd 2008, 10:54 AM
ttG
My..concentration gone.. I've ment:
Calculate the probability of getting two fives if you may roll twice? That is if one of the dices doesn't show a five you roll this one a second time and if none of this dices are showing a five you roll both!
Thanks for the explanation!
• Jun 2nd 2008, 11:16 AM
janvdl
Quote:

Originally Posted by ttG
That is if one of the dices doesn't show a five you roll this one a second time and if none of this dices are showing a five you roll both!

Here's a guess on it...

It should follow the law of total probability i think.

So we have three outcomes.

1. Both are fives. (1/36)

2. One is a five, now we roll again. (5/36 + 1/6)

3. Neither is a five. Roll both again. (1/36)

Not sure if this is correct though.
• Jun 2nd 2008, 11:17 AM
ttG
hmmm.. I think so! Well, if so..I have to evaluate one after another to get the sum of possible outcomes, right?
• Jun 2nd 2008, 11:20 AM
janvdl
Quote:

Originally Posted by ttG
hmmm.. I think so! Well, if so..I have to evaluate one after another to get the sum of possible outcomes, right?

I replied. ;)
• Jun 2nd 2008, 11:23 AM
ttG
Thank u! It has to be right, at least i hope so (Worried)
• Jun 2nd 2008, 04:45 PM
arbolis
I repost my post since the other thread has been closed.
Quote:

The probability of an event is equal to the number of favorable cases divided by the number of possible cases.
For example I do your question a).

Quote:

a) Calculate the probability of getTing at least one die showing a 5 using two dices?
You can make a table to show all the possible and favorable cases. A event will be a par like http://www.mathhelpforum.com/math-he...63f6af1c-1.gif, or http://www.mathhelpforum.com/math-he...b01fe4dc-1.gif for example. As it is a 6x6 table, there are http://www.mathhelpforum.com/math-he...585e4c22-1.gif possible cases. And if you count the favorable cases, you realize that there are http://www.mathhelpforum.com/math-he...82652dca-1.gif of them. So applying the simple formula I gave you, there is a probability of http://www.mathhelpforum.com/math-he...81930e2d-1.gif to get at least a 5 pitching once two dices.
I've counted the case (5,5). I believe I'm right on the a), so the result wouldn't be $\displaystyle \frac{13}{36}$ but $\displaystyle \frac{11}{36}$. I can easily enumerates the favorable cases : (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4) and (5,6). Therefore I'm almost sure it's $\displaystyle \frac{11}{36}$. I hope someone else can help us about this uncertain point.