An easier question to go along for the keeners.

A baseball card decreased in value 20% in its first year and 10% in its second year. What was the total percent decrease of the card's value over the two years?

canuckclint wrote:John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?

OA to come soon!

its 5/18 for me ...

Total number of ways one can choose 5 people from 9 = 9c5

If John and Peter are already there, then we need to select 3 from remaining 7 = 7C3.

hence answer = (7C3)/(9C5)= 5/18..

hope it makes sense.. do let me know if u have any doubts..

If John and Peter are already there, then we need to select 3 from remaining 7 = 7C3.

I don't get the above statement.

The statement of sudhir3127 means:

As John and Peter have already been selected, so you need to chose only 3 other players out of remaining total of 7 (9 - John - Peter), which is 7C3.

Yes but the thing is they are not chosen by looking at them.
They are chosen by random.

theres only one way in which Peter can be chosen (only one person called Peter) and only one way in which John can be chosen (only one person called John)

Probability its 5/18

and for another one
total % decrease will be 28%

initailly 100
after 1st yr 80
after iind yr 80 - 8 = 72
total decrease = 100-72 = 28

sudhir3127 wrote:

hence answer = (7C3)/(9C5)= 5/18..

Why is it 7C3 / 9C5 ?

canuckclint wrote:John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?

OA to come soon!

Please post answer choices and the source - thanks!

A.1/9
B.1/6
C.2/9
D.5/18
E.1/3

OA D

For me, the Source is [spoiler]MGMAT CAT # 2[/spoiler]

Stuart Kovinsky wrote:

canuckclint wrote:John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?

OA to come soon!

Please post answer choices and the source - thanks!

bhumika.k.shah wrote:

sudhir3127 wrote:

hence answer = (7C3)/(9C5)= 5/18..

Why is it 7C3 / 9C5 ?

7c3 becoz peter and jhon can be selected in only one way and rest 3 has to be selected from 7 available players(favourable outcomes)

9C5 because out of 9 , 5 has to be selceted (total possible selection or outcomes)

P=favourable outcome/tot outcome

There is 1 john prob of selecting him = 1c1
There is 1 peter; prob of selecting him = 1c1
There are 7 other players = 7c3

Total options = 9c5


Hence 1c1*1c1*7c3/9c5

equals 5/18

Second question
Value after 1st decrease = 0.8x
Value after 2nd decrease 0.9 *0.8x = 0.72x

Therefore total decrease = x-0.72x

That is 28%

This might be a stupid question, but what signaled you to think that this is a combination question? was it because the question said "five-player teams among 9"? The reason I got it wrong was because I saw the "probability" and was fixed on getting the answer using the concept of probabilities. Is there a way to answer this such questions using the rule of probability?

Thanks