Multi part Question involving probability and combinatrics with words

[hb2325]

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!

 

[CaptainBlack]

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

 

[hb2325]

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

 

[CaptainBlack]

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

 

[blue_raver22]

I appreciate your attempt at solving these probability questions involving combinatorics. Your solutions for parts a, c, d, e, and f are correct. For part b, you are on the right track by adding the probabilities of the two scenarios (first three letters with no A's and last three letters with no A's). However, instead of multiplying by 2, you should add the probabilities. This is because the two scenarios are not mutually exclusive - they can both occur at the same time. So the correct solution for part b would be: (6/9 * 5/8 * 4/7) + (4/9 * 3/8 * 2/7) = 1/6.

For a more efficient way to solve these types of problems, you can use the combination formula. For example, for part a, you can calculate the number of ways to choose 3 non-A letters out of 6 (since there are 6 non-A letters in TARANTULA) and divide it by the total number of ways to arrange 9 letters. So the probability for part a would be: (6C3/9P3) = 20/84 = 5/21.

Similarly, for part b, you can calculate the number of ways to choose 3 non-A letters out of 6 and multiply it by 2 (since there are two scenarios - first three letters and last three letters). So the probability for part b would be: 2 * (6C3/9P3) = 10/21.

Using the combination formula can also help you solve part f more easily. You can calculate the number of ways to arrange RAT within a sequence of 9 letters (which is 7, since there are 7 remaining letters after placing RAT in the sequence) and divide it by the total number of ways to arrange 9 letters. So the probability for part f would be: (7/9P7) = 1/9. This is the same as your solution, but using the combination formula can save you time and make it easier to calculate for larger words.

I hope this helps and good luck with your future probability and combinatorics problems!