5 ms·

When in doubt, use brute force: 1BT 2BM 1BM 2BT 1BT 2BT 1BT 2BT <-- Counted twice 1BT 2BW 1BW 2BT 1BT 2BR 1BR 2BT 1BT 2BF 1B

by jfager 15y ago

When in doubt, use brute force: 1BT 2BM 1BM 2BT 1BT 2BT 1BT 2BT <-- Counted twice 1BT 2BW 1BW 2BT 1BT 2BR 1BR 2BT 1BT 2BF 1BF 2BT 1BT 2BS 1BS 2BT 1BT 2BU 1BU 2BT 1BT 2GM 1GM 2BT 1BT 2GT 1GT 2BT 1BT 2GW 1GW 2BT 1BT 2GR 1GR 2BT 1BT 2GF 1GF 2BT 1BT 2GS 1GS 2BT 1BT 2GU 1GU 2BT I think the issue people are having is that they want to distinguish between the two instances of '1BT 2BT', but it doesn't work like that. Let's say I numbered each of the possibilities written above 1 through 28. If I handed you a piece of paper with '1BT 2BT' written on it, could you tell me which number that corresponded to? No - it would be one of two numbers - so the numbering is additional information not given in the problem statement.

zerostar07 15y ago
Alternatively, suppose we label the combinations of child-day BM, GM, BT, GT ... BU, GU (14 items). We have a bucket with 2 of each of these combinations, as there are 2 kids (total 28 items). The problem states that we have taken one "BT" item out of the bucket (so 27 items left), and are asking how many "B*" items are left in the bucket (13). The confusion arises because one can consider the same problem with replacement, putting back the BT item in the bucket before picking a second time. It's something that's not clear if u are not thinking of enumeration.
- jfager 15y ago
  Having 2 each of BM, GM, etc. is implicitly labeling 1BM, 2BM, 1GM, 2GM, etc. If you sampled with replacement, you have to admit the possibility that you draw the same BT twice. It's a stretch to suggest that the statement is ambiguous in this way, as it would imply that the father could have the same child twice.
  - zerostar07 15y ago
    Trying to figure out why it seems wrong at first: it seems my first reaction would be emotional, that the odds of having a boy should be 0.5. However it is exactly because the total probability of a boy has to be .5 that the probability of a second boy has to be less than 1/2. The fact that you can pair "boy" with any enumerable attribute that can bring the probability down to 1/3 is ... funny.
- DougBTX 15y ago
  What would the question have to be to make the following the answer? Suppose we label the combinations of child-day BM, GM, BT, GT ... BU, GU (14 items). We have two buckets which both contain a copy of each combination (total 28, 14 in each bucket). The problem states that we have taken one "BT" item out of one bucket, and are asking how many "B" items are left in the other bucket (7 out of 14).
  - zerostar07 15y ago
    That would be two parents having one child each, one of whom says his only son was born tuesday, and asking what 's the probability the other guy has a son.

by jfager 15y ago

zerostar07 15y ago

Alternatively, suppose we label the combinations of child-day BM, GM, BT, GT ... BU, GU (14 items). We have a bucket with 2 of each of these combinations, as there are 2 kids (total 28 items). The problem states that we have taken one "BT" item out of the bucket (so 27 items left), and are asking how many "B*" items are left in the bucket (13). The confusion arises because one can consider the same problem with replacement, putting back the BT item in the bucket before picking a second time. It's something that's not clear if u are not thinking of enumeration.

jfager 15y ago
Having 2 each of BM, GM, etc. is implicitly labeling 1BM, 2BM, 1GM, 2GM, etc. If you sampled with replacement, you have to admit the possibility that you draw the same BT twice. It's a stretch to suggest that the statement is ambiguous in this way, as it would imply that the father could have the same child twice.
- zerostar07 15y ago
  Trying to figure out why it seems wrong at first: it seems my first reaction would be emotional, that the odds of having a boy should be 0.5. However it is exactly because the total probability of a boy has to be .5 that the probability of a second boy has to be less than 1/2. The fact that you can pair "boy" with any enumerable attribute that can bring the probability down to 1/3 is ... funny.
DougBTX 15y ago
What would the question have to be to make the following the answer? Suppose we label the combinations of child-day BM, GM, BT, GT ... BU, GU (14 items). We have two buckets which both contain a copy of each combination (total 28, 14 in each bucket). The problem states that we have taken one "BT" item out of one bucket, and are asking how many "B" items are left in the other bucket (7 out of 14).
- zerostar07 15y ago
  That would be two parents having one child each, one of whom says his only son was born tuesday, and asking what 's the probability the other guy has a son.