# Thread: Annoying permutations question

1. ## Annoying permutations question

hey mathematicians, I need some help with permutations and probability

1. How many ways can you arrange the letters in the world DEPRESSING

a) without restriction

This one is easy. 10!/2!x2! = 907200

b) What is the probability that the S is the last letter?

Can't do this one. The answer is 1/10. (NOTE: There are two S's and two E's.)

2. How many ways can you arrange the numbers 1 2 3 4 5 6 7 8 9 so that the odd numbers appear in ascending order?

3. A spinner is has parts A B C D E on it so that the probability of spinning on any letter is 1/5. (equally likely). Hence the probability of spinning A, B, C, D and E is 20%. How many times does the spinner have to be spun so that Michael can say that the chances of having at least one E is 99%?

4. A dice has numbers 1 2 3 4 5 6 on it. It is biased so that the probability of rolling a 1 is 50% and equally likely to roll on the other numbers. What is the probability of rolling a six?

I don't get this. The probability of a 1 is 50%. Does that mean the probability for each of the numbers will be (100 - 50)/5 = 10% for the other numbers. If so, then the chances of rolling a six will be 10%.
What do you think?

THANKS GENIUSES

2. Originally Posted by differentiate
1. How many ways can you arrange the letters in the world DEPRESSING
b) What is the probability that the S is the last letter?
Can't do this one. The answer is 1/10. (NOTE: There are two S's and two E's.)
I disagree with the answer $\frac{1}{10}$.
The correct answer is $\frac{1}{5}$
There are $\frac{9!}{2!}$ ways for the arrangement to end in S.
So ${\displaystyle{\frac{9!}{2!}}/{\frac{10!}{2!^2}}=\frac{1}{5}}$.