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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!
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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