+0  
 
+1
224
4
avatar+376 

Six 6-sided dice are rolled. What is the probability that exactly two of the dice show a 1 and exactly two of the dice show a 2? Express your answer as a common fraction.

RektTheNoob  Feb 2, 2018
 #1
avatar+92624 
+3

Six 6-sided dice are rolled. What is the probability that exactly two of the dice show a 1 and exactly two of the dice show a 2? Express your answer as a common fraction.

 

P(2 twos and 2 ones and the other two different ) +  P(2 sixes and 2 ones and 2 of some other number)

 

\( =\frac{6!}{2!2!}*(\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{4}{6}*\frac{3}{6}) + \frac{6!}{2!2!2!}*(\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{4}{6}*\frac{1}{6})\\ =180*\frac{12}{6^6} + 90*\frac{4}{6^6}\\ =\frac{2160}{6^6} + \frac{360}{6^6}\\ =\frac{2520}{46656}\\ =\frac{35}{648} \\\approx 0.054\)

 

I do not think this is correct, I think 2160 should be half the size at 1080.

BUT I do not know what I did wrong. Probablility is like that ... it's sneaky!!

 

----------------------------------------------------------------------------

 

Here is another way to look at it 

You have six squares an each square represents a number rolled.

There will be 6^6 = 46656  possible outcomes but I have to work out how many of those are favourable.

First there are 6C2 = 15 places where the 1s can go.

Now there are four spots left so

There are  4C2= 6 places where the 2s can go.

So far that is   15*6=90

Now there are 2 spots left.

 

If the the last two numbers are different then you can chose from 3,4,5, or 6

So there will be 4 choices for the next place and 3 fro the last place so that is    90*4*3 = 1080

 

If the the last two numbers are the same then you can chose from 3,4,5, or 6

So there will be 4 choices for the next place and 1 for the last place so that is    90*4*1 =360

 

So that is 1080+360 = 1440

 

So the probablility is \(\frac{1440}{6^6}=\frac{1440}{46656}=\frac{5}{162}\)

 

 

This is why I think that the second answer is correct. I counted the possibilities.

 

Melody  Feb 2, 2018
edited by Guest  Feb 2, 2018
edited by Melody  Feb 2, 2018
edited by Melody  Feb 2, 2018
edited by Melody  Feb 2, 2018
 #2
avatar+92624 
+1

Please note that I have edited my answer and it is open to examination  indecision

Melody  Feb 2, 2018
 #3
avatar+1048 
+3

Solution: 

 

\(\text {With six dice }\\ \small \text {There are } \dbinom{6}{2} \text { ways to chooses two dice for the “ones” }\\ \small \text{There are then} \dbinom{4}{2} \text { ways to choose two dice for the “twos” }\\ \small \text{There are } 6^2 = 36 \text{ ways to roll two dice. There is one way to roll two dice for two 1s (snake eyes) (1/36). }\\ \small \text{There is one way to roll two dice for two 2s (1/36).} \\ \small \text{Of the remaining two dice, there are 4 of 6 ways to roll each die without a 1 or 2 (2/3). }\\ \dbinom{6}{2} * (1/36) * \dbinom{4}{2} * (1/36) * (2/3) * (2/3) = 5/162 \\ \)

 

We really do get it right if we monkey around with it enough. laugh

 

 

GA

GingerAle  Feb 2, 2018

5 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.