MHB 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