Multi part Question involving probability and combinatrics with words

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Discussion Overview

The discussion revolves around calculating probabilities related to the arrangement of the letters in the word "TARANTULA." Participants explore various events involving the letters, such as the inclusion of specific letters in certain positions and the occurrence of specific sequences. The scope includes combinatorial reasoning and probability calculations.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant proposes a method for calculating the probability that the first three letters include no A's as 6/9 * 5/8 * 4/7.
  • Another participant expresses uncertainty about calculating the probability for part (b), suggesting it might involve doubling the result from part (a) but struggles with the reasoning.
  • For part (c), a participant suggests that the probability of the fourth letter being the first A can be calculated as the product of the probabilities from part (a) and the probability of the fourth letter being A.
  • In part (d), a participant calculates the probability that the first and last letters are the same, considering repeatable letters and their probabilities.
  • For part (e), a participant initially states the probability as 1/9! but is corrected regarding the number of permutations due to repeated letters in "TARANTULA."
  • In part (f), a participant notes that there are 6 permutations of "RAT" and multiplies this by 1/9! for the probability.
  • Another participant questions how to calculate the probability for the last three letters containing no A's, suggesting it may be similar to the first three letters.
  • Clarifications are made regarding the calculation of probabilities for overlapping events in part (b) using a Venn diagram approach.

Areas of Agreement / Disagreement

Participants generally agree on some calculations, particularly for parts (a) and (c), but there is disagreement and uncertainty regarding parts (b), (d), and (e). The discussion remains unresolved with multiple competing views on the correct approaches and calculations.

Contextual Notes

Participants express uncertainty about the assumptions underlying their calculations, particularly regarding the treatment of repeated letters and the overlap in probabilities for parts (b) and (d). There are also unresolved questions about the correct application of combinatorial principles.

hb2325
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4. An experiment consists of randomly rearranging the 9 letters of the word
TARANTULA into a sequence of 9 letters, where all possible orders of these 9 letters are equally
likely. Find the probability of each of the following events:

( a ) the first three letters include no A's;
( b ) the first three letters or the last three letters (or both) include no A's;
( c ) the fourth letter is the first A;
( d ) the first letter and the last letter are the same;
( e ) the word `TARANTULA' is obtained;
( f ) the sequence contains the word `RAT'.Attempt at solutions :

a) 6/9 * 5/8 * 4/7 (Probability of first non A * another non A letter * another non A letter)

b) 6/9 * 5/8 * 4/7 + ( I am stuck I don't get it - I think it might just be 2 * anser of part a but I'm unable to think it through, I know though that if first 3 and last 3 have no A's, then middle will have all A's so it becomes more weird )

c) 6/9 * 5/8 * 4/7 * 3/6 ( Prob. in a * probability of 4th letter being A in the scenario of question a)

d) 5/9 * 1/8 + 5/9 * 2/8 (Probability of choosing a repeatable letter * probability of second letter coming up for both T and A since R U L N are not repeatable)

e) 1/9!

f) After working it out on paper it seems there are 6 permutations of RAT so 6 * 1/9!
I am still not sure if these are correct and I know there must be a better way of doing these using combination formula so I would be greatful if someone could help me out.

Thanks!
 
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hb2325 said:
4. An experiment consists of randomly rearranging the 9 letters of the word
TARANTULA into a sequence of 9 letters, where all possible orders of these 9 letters are equally
likely. Find the probability of each of the following events:

( a ) the first three letters include no A's;
( b ) the first three letters or the last three letters (or both) include no A's;
( c ) the fourth letter is the first A;
( d ) the first letter and the last letter are the same;
( e ) the word `TARANTULA' is obtained;
( f ) the sequence contains the word `RAT'.Attempt at solutions :

a) 6/9 * 5/8 * 4/7 (Probability of first non A * another non A letter * another non A letter)

b) 6/9 * 5/8 * 4/7 + ( I am stuck I don't get it - I think it might just be 2 * anser of part a but I'm unable to think it through, I know though that if first 3 and last 3 have no A's, then middle will have all A's so it becomes more weird )

c) 6/9 * 5/8 * 4/7 * 3/6 ( Prob. in a * probability of 4th letter being A in the scenario of question a)

d) 5/9 * 1/8 + 5/9 * 2/8 (Probability of choosing a repeatable letter * probability of second letter coming up for both T and A since R U L N are not repeatable)

e) 1/9!

f) After working it out on paper it seems there are 6 permutations of RAT so 6 * 1/9!
I am still not sure if these are correct and I know there must be a better way of doing these using combination formula so I would be greatful if someone could help me out.

Thanks!


For ( b ) the required probability is:

P(first 3 contain no A's) + P(last 3 contain no A's) - P(both the first and last 3 contain no A's)

A Venn diagram will show that the sum of the first two terms counts "both the first and last 3 contain no A's" twice, so that probability needs to be removed once.

Your answers to ( a) and ( c) look OK.

( e) is wrong since there are 6x2 (since you have 3 "A"s and 2 "T"s in "TARANTULA") of the 9! permutations that give "TARANTULA".( d) probability that the first letter and last letter are A is (3/9)x(2/8), and that they are T is (2/9)x(1/8)

CB
 
Last edited:
Firstly thanks for taking the time to help me.

About b) The problem is I don't understand what P(Last 3 contain no A's) would be in terms of say how first 3 is 6/9 * whatever and so on? Would it also be 6/9*5/8*4/7? I'm kinda stumped on how to take it.

Also Part e) Yeah I overlooked the 3 A's and 2 T's part so still wouldn't it be 3x2 and not 6x2 :S Where does the 6 come from?

Thank
 
hb2325 said:
Firstly thanks for taking the time to help me.

About b) The problem is I don't understand what P(Last 3 contain no A's) would be in terms of say how first 3 is 6/9 * whatever and so on? Would it also be 6/9*5/8*4/7? I'm kinda stumped on how to take it.

The probability that the last 3 contain no A's is the same as the probability that the first three contain no A's

CB
 

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