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

1. $\left(\dfrac{1}{3}\right)$
2. $\left(\dfrac{1}{6}\right)$
3. $\left(\dfrac{1}{9}\right)$
4. $\left(\dfrac{5}{18}\right)$
5. $\left(\dfrac{2}{9}\right)$
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+1
2/9 is correct.

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

selected
Case 1:largest is $5$ , smallest $1$ and middle is $2$ or $3$ or $4 : 3\times 3!$

Case 2.largest is $5$ , smallest $1$ and middle is $1$ or $5 : \dfrac{3!\times 2}{2!}$

Case 3:largest is $6$ , smallest $2$ and middle is $3$ or $4$ or $5 : 3\times 3!$

Case 4:largest is $6$ , smallest $2$ and middle is $6$ or $2: \dfrac{3!\times 2}{2!}$

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

$=\dfrac{\left( 3\times 3!+\dfrac{3!\times 2}{2!}+3\times 3!+\dfrac{3!\times 2}{2!}\right)}{6^{3}} =\dfrac{2}{9}.$
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How you got 3 * 3! ? and (3! * 2 ) / 2! ? Can you please explain ...
 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

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)

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
–1 vote
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