Question

MEDAL!!
There are 20 players on a hockey team. 5 of them are rookies. Determine the probability that none of the rookies will be randomly selected to be included in the team's group of 6 starters.

A. 15/114
B. 1/3,876
C. 1/38,760
D. 1001/7752

Answer 1

would my answer be A?

Answer 2

@AllTehMaffs

Answer 3

\[\frac{ \left(\begin{matrix}5 \\ 0\end{matrix}\right)\left(\begin{matrix}15 \\ 5\end{matrix}\right) }{ \left(\begin{matrix}20 \\ 5\end{matrix}\right) }\]

Answer 4

this is because you really have two groups... rookies and non-rookies.

there are 5 rookies and 15 non-rookies.

you are sampling witthout replacement and order doesn't matter so you use combinations.

for the denominator, there are 20 players, and you need to choose 5.

for the numerator, there are 5 rookies, of which you choose 0 and there are 15 non-rookies, of which you choose 5.

I hope that helps!

Answer 5

yea:) so my answer would be A correct?

Answer 6

?

Answer 7

.19369....

which is not A

Answer 8

hmmmm, should I solve that fraction you gave me, and will that give me my answer?

Answer 9

oops, hold on I may have made a calc error...

Answer 10

ok

Answer 11

it =21699 which equals 0.44017 but cant figure out howw to put it into a fraction

Answer 12

i found my mistake... there are 6 player starting, not 5. just a sec...

Answer 13

kk

Answer 14

=.129127...

= > still not A

I suggest you compute the combinations. do you have a calculator to do that?

Answer 15

oe sec

Answer 16

15/20 *14/19 *13/18 *12/17 *11/16 *10/15 = 1001/7752

Answer 17

Yeah, that's another way to do it! Great for you!

Answer 18

thanks!

Answer 19

you're so welcome!