TIFR CSE 2011 | Part A | Question: 19

Question

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

• A. $\left(\dfrac{1}{3}\right)$    
• B. $\left(\dfrac{1}{6}\right)$  
• C. $\left(\dfrac{1}{9}\right)$  
• D. $\left(\dfrac{5}{18}\right)$  
• E. $\left(\dfrac{2}{9}\right)$

Comments

2/9 is correct.

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Its a beautiful question and indeed a beautiful solution @aritra nayak

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When should we count the permutations, and when not count them?

Answer 1

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 $4,$

$ \quad=\dfrac{\left( 3\times 3!+\dfrac{3!\times 2}{2!}+3\times 3!+\dfrac{3!\times 2}{2!}\right)}{6^{3}} =\dfrac{2}{9}.$

Correct Option: E

Comments

How you got 3 * 3! ? and (3! * 2 ) / 2! ? Can you please explain ...

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@shreya would you please explain how you got the 2nd case and 4th....

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For case 1:  (1, 2/3/4 ,5) for middle place we have 3 choice and all places can be arranged in 3! ways

So total is 3*3! ways,

For case 2: (1, 1/5 ,5) for middle place we have 2 choice and all places can be arranged in 3!/2! ways

So total is 2*(3!/2!) ways,

Answer 2

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

Comments

Plain and simple approach but I guess thinking and executing this will consume atleast 5 mins in Gate

Answer 3

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

Correct Answer: $E$

Comments

It becomes simple the moment we enumerate all favourable ways

Case 1: Minimum value on dice is 1 and the maximum value on dice is 5.

Numbers on 3 dices	Total number of ways these 3 numbers can appear on the dice
1,1,5	$\frac{3!}{2!}=3$
1,2,5	$3!=6$
1,3,5	$3!=6$
1,4,5	$3!=6$
1,5,	$\frac{3!}{2!}=3$

 Total favourable ways in this case=3+6+6+6+3=24.

Case 2: Minimum value on the dice is 2 and the maximum value is 6

Now when 2 and 6 have appeared the third number can be anyone from $\{2,3,4,5,6\}$ so that the difference between the minimum and the maximum number remains 4.

Now, this case is symmetrical to case 1 where there are two possibilities when the number on the third dice has already appeared in one of the two dices and remaining 3 possibilities when the number on three dices are all distinct and hence this case also has favourable ways=24.

 

Total favourable ways=48.

Total possible outcomes=$6^3$

Probability=$\frac{48}{216}=\frac{2}{9}$

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nice ayush

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nice aritra nayak

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Thanks Ayush sir and  aritra sir

Answer 4

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)