Question regarding Permutations (Prize distribution problem)

Question

There are three different prizes that will be distributed randomly. There are thirty people, which includes Martin and Jose, will be participated in this ceremony.

a) How many different ways are there to distribute the prize?

b) What is the probability if Martin obtains more than one prize?

c) What is the probability if Jose doesn't obtain any prize?

I understand that a) would be a simple permutation of $\displaystyle30P3 = \frac{30!}{27!} = 24360$ ways to distribute the prize.

I am not sure what I think for b) and c) is correct, but I will explain my way to think:

• For a), The possibility to have two prizes would be $\displaystyle2P3 = \frac{3!}{2!} = 6$ and the possibility to have three prizes would be $\displaystyle3P3 = \frac{3!}{0!} = 6$ So the entire probability would be $\displaystyle\frac{12}{24360}$
• Similarly, for c), I try to calculate the opposite event where Jose obtains prizes. So the whole possibility would be $\displaystyle\frac{3P1+3P2+3P3}{24360}=\frac{3+6+6}{24360}=\frac{15}{24360}$, so the event where Jose won't obtain prize will be $\displaystyle1-\frac{15}{24360}$

Is there any mistakes in my way to think?

Answer 1

The second question hints that the prizes are not 1st, 2nd and 3rd but are rather separate individual prizes.

a) Each prize can go to anyone among $30$ people. Hence $N_1=30\times 30\times 30=30^3$

b) Martin may get any $2$ of the $3$ prizes and anyone among the other $29$ can get the last prize. This gives us $N_2={3 \choose 2}\times 29= 87$. But there is the extra case where Martin gets all the prizes so $N_2=87+1=88$. Hence probability is $N_2\over N_1$

c) If Joseph doesn't get the prize, each prize has $29$ people it can possibly go to. Hence $N_3=29\times 29\times 29=29^3$. This gives us probability as $N_3\over N_1$