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# More Dice......

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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
#1
+2248
+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
+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
+90001
+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 .....!!!!!

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