+0  
 
0
97
3
avatar

A standard die has the numbers 1 to 6, each appearing with a probability of 1/6. If the die is modified so that the number 6 appears with half the frequency of the other numbers, what is the new probability of rolling a 6?
Thank you.

Guest May 29, 2017
Sort: 

3+0 Answers

 #1
avatar+1221 
+1

 

 

First, let me show you a wrong approach. I almost fell for it myself:
\(\frac{1}{\frac{6}{2}}=\frac{1}{6}*\frac{1}{2}=\frac{1}{12}\)

 

This math above is simple, but this would imply that the probability of rolling a 6 is 1/12. Well, you'll see...

 

If you combine the probabilities of all events occurring, then the probability should be 1. Let's try that with our current answer. Let's test it:

 

\(\frac{1}{12}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=1\) Create a common denominator, first
\(\frac{1}{12}+\frac{2}{12}+\frac{2}{12}+\frac{2}{12}+\frac{2}{12}+\frac{2}{12}=1\) Add the numerators
\(\frac{11}{12}=1\) This is a false statement
   

 

11/12 is not equal to one 1. Therefore, the probability cannot of rolling a 6 cannot be 1/12. However, there's a trick that we can use. How can we make 11/12=1? That's right, multiply by its reciprocal, 12/11. Therefore, multiply all of your fractions by it:
 

\(\frac{12}{11}(\frac{1}{12}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6})\)

 

But wait! WE only need the probability of rolling a 6, so we only need to multiply 12/11 by 1/12!

 

\(\frac{12}{11}*\frac{1}{12}=\frac{12}{132}=\frac{1}{11}\)

 

As you'll see below, there are other methods to do this problem. Of course, the method above is what I chose. Isn't it beautiful how multiple approaches still leads to the same final answer?

TheXSquaredFactor  May 29, 2017
edited by TheXSquaredFactor  May 29, 2017
 #2
avatar
+1

Another way:

Let the probability of rolling a 6 =p, then,

The probability of rolling every other number =2p, 

Now will add up all probabilities =p + 2p + 2p + 2p + 2p + 2p =1

                                                               11p = 1

                                                                   p = 1/11 new probability of rolling a 6.

Guest May 29, 2017
 #3
avatar+76925 
+1

 

Thanks, X^2 and guest.....here's another approach

 

Let the probability that each of the numbers 1- 5  are rolled   =  1/n

 

Then..... the probability of rolling a "6"   equal 1/2 of this =  1 / (2n)

 

And  the total probability  must equal 1

 

So

 

5 (1/n)   +  1 / (2n)   =  1

 

5/n  +  1 /(2n)   = 1

 

[ 10  + 1 ] / (2n)  =  1

 

11  =  2n

 

11/2   = n

 

So......the probability of rolling a "6"  =  1 / [ 2(11/2) ]  =  1/11

 

Just as X^2  and guest   found .....!!!!!

 

 

 

cool cool cool

CPhill  May 29, 2017
edited by CPhill  May 29, 2017

28 Online Users

avatar
avatar
avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details