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Four standard fair -sided dice are rolled. What is the probability that the sum of the numbers rolled is 12?

 Oct 22, 2022
 #1
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The probability is 7/16.

 Oct 22, 2022
 #2
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There are \(6^4 = 1296\) ways to roll the dice. 

 

Use stars and bars to count the number of ways to roll a 12. Each roll must be at least 1, so we subtract \(4 \times 1 = 4\) to get 8 stars and \(4 - 1 = 3\) bars.

 

This makes for \({8 + 3 \choose 3} = {11 \choose 3} = 165\) ways to roll a 12. But, we can't have cases like 6, 1, 1, 0; 6, 2, 0, 0; 7, 1, 0, 0; and 8, 0, 0, 0. 

 

There are \({4! \over 2!} = 12 \) ways for each of the first 3, and there are \({4! \over 3!} = 4 \) ways for the last one, which sums to 40. 

 

So, there are 165 - 40 = 125 ways to roll a 12, which makes for a probability of  \(\color{brown}\boxed{125 \over 1296}\) 

 Oct 22, 2022
edited by BuilderBoi  Oct 22, 2022
 #3
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That answer is incorrect.

Guest Oct 22, 2022
 #4
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YES, It is correct ! As follows:

 

1146 , 1155 , 1164 , 1236 , 1245 , 1254 , 1263 , 1326 , 1335 , 1344 , 1353 , 1362 , 1416 , 1425 , 1434 , 1443 , 1452 , 1461 , 1515 , 1524 , 1533 , 1542 , 1551 , 1614 , 1623 , 1632 , 1641 , 2136 , 2145 , 2154 , 2163 , 2226 , 2235 , 2244 , 2253 , 2262 , 2316 , 2325 , 2334 , 2343 , 2352 , 2361 , 2415 , 2424 , 2433 , 2442 , 2451 , 2514 , 2523 , 2532 , 2541 , 2613 , 2622 , 2631 , 3126 , 3135 , 3144 , 3153 , 3162 , 3216 , 3225 , 3234 , 3243 , 3252 , 3261 , 3315 , 3324 , 3333 , 3342 , 3351 , 3414 , 3423 , 3432 , 3441 , 3513 , 3522 , 3531 , 3612 , 3621 , 4116 , 4125 , 4134 , 4143 , 4152 , 4161 , 4215 , 4224 , 4233 , 4242 , 4251 , 4314 , 4323 , 4332 , 4341 , 4413 , 4422 , 4431 , 4512 , 4521 , 4611 , 5115 , 5124 , 5133 , 5142 , 5151 , 5214 , 5223 , 5232 , 5241 , 5313 , 5322 , 5331 , 5412 , 5421 , 5511 , 6114 , 6123 , 6132 , 6141 , 6213 , 6222 , 6231 , 6312 , 6321 , 6411 , Total =  125

 

Therefore, the probability is = 125 / 6^4 =125 / 1296

Guest Oct 22, 2022

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