Probability - need the answers to know if im on the right track while practicing

1. in how many ways can a hockey team consisting of 11 players be chosen from a squad of 14 players

2. 80 disc marked with the numbers 1 to 80 are placed in a box and one disc is drawn from it. determine the probability that the number on the disc will be a multiple of 5

3. A bag contain 4 red marbles, 3 blue marbles and 7 green marbles. A marble is drawn from the bag. What is the probability that it will be a red marble.

4. In how many ways can one select a committee of four republicans, three democrats and two independents from a group of 10 distinct republicans, 12 distinct Democrats and 4 distinct independents?

Re: Probability - need the answers to know if im on the right track while practicing

A better idea, show us what you have done (to prove you have shown some effort) and we will check over it..,

Re: Probability - need the answers to know if im on the right track while practicing

Q1 sol

n/r!(n-r)!

14/11(14-11)

14*13*12/3 = 728

Q2 Not sure where to even sart

Q3 sol

4+3+7 = 14

1/14

Q4 sol

10/4(10-4) * 12/3(12-3) * 4/2(4-2)

10/4(6) * 12/3(9) * 4/2(2)

5040/24 * 1320/6 * 12/2

answer seem too big not sure if correct

Re: Probability - need the answers to know if im on the right track while practicing

Q1: $\displaystyle \dfrac{14!}{11!(14-11)!} = \dfrac{14\cdot 13\cdot 12}{3\cdot 2\cdot 1} = 364$

Q2: Every five numbers is a multiple of 5. Since 80 is divisible by 5, you have a $\displaystyle \dfrac{1}{5}$ chance of drawing a multiple of 5. But, in general, you need to find out how many multiples of 5 are among the numbers 1 to 80 (there are $\displaystyle \dfrac{80}{5} = 16$). Then, the chances of drawing a multiple of 5 are $\displaystyle \dfrac{16}{80} = \dfrac{1}{5}$.

Q3: Four of the fourteen marbles are red. So, you have a $\displaystyle \dfrac{4}{14} = \dfrac{2}{7}$ chance of drawing a red marble.

Q4: Use the exclamation mark for factorial.

$\displaystyle \dfrac{10!}{4!6!}\dfrac{12!}{3!9!}\dfrac{4!}{2!2!} = \dfrac{10\cdot 9\cdot 8\cdot 7}{4\cdot 3\cdot 2\cdot 1}\cdot \dfrac{12\cdot 11\cdot 10}{3\cdot 2\cdot 1}\cdot \dfrac{4\cdot 3}{2\cdot 1} = 277,200$