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Three dice are rolled independently. What is the probability that the highest and the lowest value differ by 4?  

  1. $\frac{1}{3}$  
  2. $\frac{1}{6}$  
  3. $\frac{1}{9}$  
  4. $\frac{5}{18}$  
  5. $\frac{2}{9}$
asked in Probability by Veteran (319k points) 577 1442 2960 | 304 views
2/9 is correct.

6 Answers

+2 votes
Best answer
Case 1:largest is 5 , smallest 1 and middle is 2 or 3 or 4 : 3*3!

Case 2.largest is 5 , smallest 1 and middle is 1 or 5 : 3!*2/2!

Case 3:largest is 6 , smallest 2 and middle is 3 or 4 or 5 : 3*3!

Case 4:largest is 6 , smallest 2 and middle is 6 or 2: 3!*2/2!

So probability the highest and the lowest value differ by 4 =( 3*3!+3!*2/2!+3*3!+3!*2/2!)/6^3 =2/9
answered by Loyal (4.3k points) 4 24 57
selected by
+4 votes

suppose three dices are there A,B,C, three are rolled independently, see we get 4 difference between max and min when we get either (1,5) or (2,6) in any of the two dices. okay?

1st case: we get (1,5), 1 and 5 can be the result of any of the two dice in 3C2=3 ways, either (A-B) or (B-C) and (C-A) and also they can be permutated between them,so 3C2*2! ways possible, now for each way 3rd result can come either 1 or 2 or 3 or 4 or 5(6 cant be possible as min:1 and max:5) But we cant do 3C2*2!*5 because then (1,1,5), (1,5,1),(5,1,1),(1,5,5),(5,1,5),(5,5,1) all these 6 ways are counted two times.So we do like below:
When 3rd result is 1, no of ways=3!/2!=3  because its no of permutation of 1,1 and 5

When 3rd result is 5,no of ways=3!/2!=3 because its no of permutation of 1,5 and 5

when 3rd result is 2 or 3 or 4, 3C2*2!*3=18 ways possible.so totally 18+3+3=24

2nd case: we get (2,6), 2 and 6 can be the result of any of the two dice in 3C2=3 ways, either (A-B) or (B-C) and (C-A) and also they can permutate between them,so 3C2*2! ways possible, now for each way 3rd dice result can come either 2 or 3 or 4 or 5 or 6(min:2 and max:6) But we cant do 3C2*2!*5, So we do like below:

Similarly as shown above, when 3rd result is 2 or 6,no of ways=2*3!/2=6 ways possible,when 3rd result is 3 or 4 or 5,3C2*2!*3 =18 ways possible.So totally 18+6=24 ways possible.


Finally total=24+24=48 ways possible
sample space= 6^3=216
so probability= 48/216=2/9

answered by Junior (517 points) 1 3 10
edited by
+1 vote
difference highest and lowest is 4.
1 (1,2,3,4,5) 5
+
2(2,3,4,5,6)6
=(3!* 1.5.1 + 3!*1.5.1)/6.6.6
=2.6.5/6.6.6
=10/36=5/18
answered by Loyal (3.1k points) 5 36 78
+1 vote
Sno Lowest Value Highest Value  Values at dice No of ways
1 1 5 {1,1,5} 3
2 1 5 {1,2,5} 6
3 1 5 {1,3,5} 6
4 1 5 {1,4,5} 6
5 1 5 {1,5,5} 3
6 2 6 {2,2,6} 3
7 2 6 {2,3,6} 6
8 2 6 {2,4,6} 6
9 2 6 {2,5,6} 6
10 2 6 {2,6,6} 3

 

By naive defn of Prob = # outcomes which satisfy our constraint / total # of outcomes

Prob = 48 / 63 = 2 / 9 

The correct answer is (E) 2/9

 

answered by Boss (8.8k points) 3 8 12
0 votes
1/9 C is answer .  there will be 24 such permutations having diff. of 4 in max. and min. in dies

24/216=1/9
answered by (25 points) 1 2
0 votes

There will be 2 possibilities for highest and lowest differ by 4, i.e (1,5),(2,6) and they can arrange in 2! ways

Now among 3 dies 2 dice can be selected in 3C2 ways

Case 1:

for (1,5) the 3rd dice value can be any value from 1 to 5, but it cannot be 6, Because in that case difference cannot be 4

so probability is 3C2*2!*5/ (6^3)

Case 2:

for (2,6) the 3rd dice value can be any value from 2 to 6, but it cannot be 1, Because in that case difference cannot be 4

so probability is 3C2*2!*5/ (6^3)

So,  probability that the highest and the lowest value differ by 4

=3C2*2!*5/ (6^3)+3C2*2!*5/ (6^3)= 5/18

Ans is (D)

 

 

answered by Veteran (64.8k points) 35 220 622


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