Conditional probability - generating numbers

Question

Problem: Player generates 3 different numbers - each from 1 to 7. Numbers are generated randomly and equally likely.

Task is to find probability that at least one number is 6, if sum of generated three numbers doesn't exceed 17.

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My question:

I don't really understand whether it's assumed that the problem should be solved by Conditional probability.

Below there are two solutions I have in mind. Please advice:

1. Is this a Conditional probability assumed here ?
2. Which solution from two is right?

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I) If it's Conditional Probability:

$\Omega$ = "outcomes in total"

A = "at least one of numbers showed 6"

B = "sum of three numbers <= 17"

P(A|B) = P(AB) / P(B) = |AB| / |B|

|$\Omega$ | = 7^3

|B| = |$\Omega$| - |{477, 567, 576, 577, 657, 666, 667, 675, 676, 677, 747, 756, 757, 765, 766, 767, 774, 775, 776, 777}| = 343 - 20 = 323

|A| = |{ 116, 126, 136, 146, 156, 161, 162, 163, 164, 165, 166, 167, 176, 216, 226, 236, 246, 256, 261, 262, 263, 264, 265, 266, 267, 276, 316, 326, 336, 346, 356, 361, 362, 363, 364, 365, 366, 367, 376, 416, 426, 436, 446, 456, 461, 462, 463, 464, 465, 466, 467, 476, 516, 526, 536, 546, 556, 561, 562, 563, 564, 565, 566, 611, 612, 613, 614, 615, 616, 617, 621, 622, 623, 624, 625, 626, 627, 631, 632, 633, 634, 635, 636, 637, 641, 642, 643, 644, 645, 646, 647, 651, 652, 653, 654, 655, 656, 661, 662, 663, 664, 665, 671, 672, 673, 674, 716, 726, 736, 746, 761, 762, 763, 764}| = 114

P(A|B) = P(AB) / P(B) = |AB| / |B| = 114 / 323

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II) It's it's classic Probability:

|A| = |"at least one of numbers showed 6, sum <= 17"| = 114

|$\Omega$| = 7^3 = 343

P(A) = |A| / |Omega| = 114 / 343

Answer 1

It's a Conditional Probability problem because you were asked for a probability of an event occuring under a given condition: "if the sum doesn't..."

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There are $7^3$ outcomes in total of which $6^3$ do not include 6s.

There are $20$ outcomes whose sum exceeds 17, of which $7$ do not include 6s.

So therefore there are $7^3-20$ outcomes whose sum does not exceed 17, of which $7^3-6^3-(20-7)$ do include 6s.

Answer 2

Hint: Try to rephrase the problem in terms of simple English.

You are given that the sum is less than $17$. In general, if you have to limit your sample space then the question is likely one of conditional probability. Then your solution becomes:

$$\mathbb{P}(\text{one number is 6 | sum is } \leq17) = \frac{\mathbb{P}(\text{one number is 6 } \wedge\text{ sum is }\leq17)}{\mathbb{P}(\text{sum is }\leq17)}$$

Hence, your first solution is the correct one (I haven't checked the exact numbers, but your methodology is right).

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Note: $\Omega$ is not all possible permuations of $3-$digit numbers, since it includes repeated digits.