Annoying permutations question

• Sep 14th 2009, 03:36 AM
differentiate
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
• Sep 14th 2009, 04:19 AM
Plato
Quote:

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 ${\displaystyle{\frac{9!}{2!}}/{\frac{10!}{2!^2}}=\frac{1}{5}}$.