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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!

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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