+0  
 
0
42
1
avatar+97 

Each of five, standard, six-sided dice is rolled once. Two of the dice come up the same, but the other three are all different from those two and different from each other. The pair is set aside, and the other three dice are re-rolled. The dice are said to show a "full house" if three of the dice show the same value and the other two show the same value (and potentially, but not necessarily, all five dice show the same value). What is the probability that after the second set of rolls, the dice show a full house?

 

Plz show step-by-step solutions, greaty appreciated!

MATHEXPERTISE  Oct 21, 2018
 #1
avatar+2714 
+1

\(P[\text{pair and 3 odd dice rolled}]=\dfrac{\dbinom{5}{2}\dbinom{6}{1}\dbinom{5}{3}3!}{6^5}=\dfrac{25}{54}\)

 

The factors in the numerator correspond to

a) pick 2 dice out of the 5 to be the pair

b) pick 1 value out of 6 to be the pair's value

c) pick 3 values out of 5 for the other dice

d) permute the 3 odd values since we do care about order here since we are scaling by the entire number of possible rolls

 

 

Now we roll the three odd dice to obtain a three of a kind of any value

 

There are only 6 ways of doing this, 1 per value.

 

\(P[\text{roll 3 of a kind}]=\dfrac{6}{6^3}=\dfrac{1}{36}\)

 

we multiply these together to get the final probability

 

\(P[\text{full house or 5 of a kind}]=\dfrac{25}{55}\dfrac{1}{36}=\dfrac{25}{1944}\)

Rom  Oct 21, 2018
edited by Rom  Oct 21, 2018

17 Online Users

avatar
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.