Solve Probability Problem: 4 Letters from MISSISSIPPI

[neomage]

OK I've been working on a math problem for a while now, and the answer just isn't coming any help would be appreciated.
This is for an independent study class. I am basically teaching myself various math subjects that we don't get in school.


here it is:

Four letters are chosen at random from the word MISSISSIPPI. Determine the probability that at least three I's are chosen.

now I know the answer should be (29/330) according to my book, but I can't come up with this answer.

It seems to me that the logical thing would be to do this:

(4/11)*(3/10)*(2/9)*(1/8)= (1/330) the probability of getting all I's
then
(4/11)*(3/10)*(2/9)*(1-(1/8))= (7/330)

(7/330)+(1/330)= (8/330)= (4/165)

I've tried many other things to come up with (29/330) but I can't get that answer

any help would be greatly appreciated.

 

[mathman]

Your second term (7/330) has to be multiplied by 4, since anyone of the 4 I's can be omitted.

 

[neomage]

I was trying so many permutations and combinations that I was going crazy. If I would have just stopped and thought about it for a second and tried not to make it so complicated i might have gotten it. THanks for the help

 

[metacristi]

Another way to solve the problem (simpler) is using combinatorics.

There are 11 letters and 4 of them are drawed,therefore the total number of possible combinations to draw 4 of them is:

N=C₁₁⁴=11!/[(11-4)!*4!]=330

How many possibilities exist that have at least 3 'I'?

We have:

I I I ? ;where for ? we have 1 'I',1 'M',4 'S' and 2 'P' as possibilities.Therefore we have 1+1+4+2=8 valid possibilities.

Now the position of ? (1 'I',1'M',4 'S' and 2 'P') in the above string can be on the first place,the second or the third place also.

For every such case we have 7 distinct possibilities left (1+4+2) (this is because 'I' do not give distinct possibilties when ? is on the first,the second or the third place).

Finally the required probability is:

p=[8+7+7+7]/N=29/330 q.e.d.