Joey throws a dice 3 times and decides to invite

Question

Joey throws a dice 3 times and decides to invite the same number of guests for his birthday party as the sum of the outcomes of the throws. What’s the probability that he invites 12 guests?

A. 1/6
B. 1/16
C. 1/18
D. 1/36
E. 25/216

shriramvelamuri · Answer

Hi,

I tried this one. I am getting a total of 16 combinations to attain 12.

Hence, the answer would be 16/216. But I am missing something somewhere. Please could you explain.

Bunuel · Answer

Total outcomes: 
6^3.

Favorable outcomes: 
{6 - 5 - 1} - 6 cases ({6 - 5 - 1}, {6 - 1 - 5}, {5 - 6 - 1}, {5 - 1 - 6}, {1 - 6 - 5}, {1 - 5 - 6});
{6 - 4 - 2} - 6 cases;
{6 - 3 - 3} - 3 cases;
{5 - 5 - 2} - 3 cases;
{5 - 4 - 3} - 6 cases;
{4 - 4 - 4} - 1 case.

6 + 6 + 3 + 3 + 6 + 1 = 25.

P = (favorable)/(total) = 25/6^3.

Answer: E.

kelvind13 · Answer

I too arrived at the same solution, although it took me close to about 4 minutes to do it. isn't there a faster shortcut for the same?

PrepTap · Answer

As such there is no shortcut possible but the counting can be done in an orderly fashion which might reduce the time you took.

We know that the sum required is 12.
So if in the first place we have:
1 -> Sum of other two numbers = 12 -1 = 11 -> possible combinations (5+6, 6+5)
2 -> Sum of other two numbers = 12 -2 = 10 -> possible combinations (4+6, 6+4, 5+5)
3 -> Sum of other two numbers = 12 -3 = 9 -> possible combinations (3+6, 6+3, 4+5, 5+4)
4 -> Sum of other two numbers = 12 - 4 = 8 -> possible combinations (2+6, 6+2, 3+5, 5+3, 4+4)
5 -> Sum of other two numbers = 12 -5 = 7 -> possible combinations (1+6, 6+1, 2+5, 5+2, 3+4, 4+3)
6 -> Sum of other two numbers = 12 -6 = 6 -> possible combinations (1+5, 5+1, 2+4, 4+2, 3+3)

Favourable outcomes: 25
Total outcomes: 6 * 6 * 6 = 216

Probability = 25/216

Lucky2783 · Answer

Min sum=1+1+1=3
Max sum=6+6+6=18
clearly there is only 1 way to achieve these sums.

Sum=4; #ways 3
sum=17; #ways 3

So we know that the series of #ways for sum {3..10} will be just reverse of series of #ways for sum {11..18}
and each series sums to 108.

sum=9 or 12 #ways will be same.

Sum=9 ; lets look at it.

A+b+c=9
Now this is just a question of distributing 9 identical things into 3 people. Since each one will have atleast 1 thing so we have to distribute 6 things among 3 people.
8C3=28
now there will be 3 cases in which each a, b and c will have 7 things so subtract 3 from 28 To get 25.

25/216 answer

mtk10 · Answer

I solved it in 2 mins 10 secs using this approach. hope it helps.

You will have to involve bit of permutations for a shortcut, assume each setting of 3 throws as letters. e.g 444 is a letter repeating thrice 255 is a letter with two repeating elements.
Having that in mind.
List out basic settings which give you and permute for repeating elements

(1,5,6) This can be written in 3 ! ways   6 ways
(2,5,5) this can be written in 3 !/2! ways   3 ways
(2,4,6) This can be written 3 ! ways. 6 ways
(3,4,5) Same 3 ! ways 6 ways
(444) This can be written in 3!/3! ways 1 way
(6,3,3) This can be written in 3!/2! ways 3 ways

Total outcomes = 6^3 = 216
Possible outcomes = 6+6+6+3+3+1= 25
Answer is E 25/216

CrackverbalGMAT · Answer

If you are Joey and you invite 12 guests, it means that the sum of the outcomes was 12. So, basically, this problem is asking you to find the probability of the sum of the outcomes being 12, when a dice is thrown thrice.

Probability (Sum of the throws being 12) =  \frac{Number of outcomes favorable to the event }{ Total Possible Outcomes}

The first thing we can determine in this case is the number of total possible outcomes. Since the dice has been thrown thrice, the total possible outcomes will be 6*6*6 = 216. Therefore, the denominator has to be 216 or its factors.  Note that 16 is not a factor of 216, so we can eliminate answer option B .

The number of ways in which we can obtain a sum of 12 can be calculated as follows:

A sum of 12 can be obtained as {4,4,4} in one way.

A sum of 12 can be obtained as {3,4,5} in a total of 6 ways. Remember getting 3 on the first, 4 on the second and 5 on the third throw is entirely different from getting it in the reverse order. The order in which the numbers appear, matters. Hence, we try to find out the number of ways in which the objects 3,4 and 5 can be permuted. Since there are 3 distinct objects, we can permute them in 3! Ways i.e. 6 ways.

A sum of 12 can also be obtained as {2,4,6} in 6 ways and as {1,5,6} in 6 ways respectively.

A sum of 12 can be obtained as {3,3,6} in \frac{ 3!}{2!}  Ways i.e. 3 ways. Likewise with the combination of {2,5,5}.

Therefore, the total number of favourable outcomes = 1+6+6+6+3+3 = 25.
The required probability =  \frac{25 }{ 216} .  The correct answer option is E .

When you are dealing with questions like these on probability, bear in mind that  you need to consider the different permutations since the order in which the numbers turn up matters . 
While trying to arrive at the combinations,  try to fix one number and work out the combination for the other two numbers . I second  PrepTap  on this. This is the fastest approach, it also ensures that you do not miss out on any combination.

Hope that helps!

Archit3110 · Answer

total possible cases to make up sum as 12
552 ; 444 ; 543 ; 651; 642 ; 633
which can be arranged in ; 3+1+6+6+6+3 ; 25 ways
6^3 ; 216
25/216 
OPTION E

Kinshook · Answer

Given: Joey throws a dice 3 times and decides to invite the same number of guests for his birthday party as the sum of the outcomes of the throws.

Asked: What’s the probability that he invites 12 guests?

Total ways = 6^3 = 216
Favorable ways = {(6,5,1):6,(6,4,2):6,(6,3,3):3,(5,5,2):3,(5,4,3):6,(4,4,4):1} = 6+6+3+3+6+1 = 25

Probability = 25/216

IMO E

Joey throws a dice 3 times and decides to invite

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Joey throws a dice 3 times and decides to invite

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