In a class, 60% of students play soccer or basketball. If 10% play bot

Question

In a class, 60% of students play soccer or basketball. If 10% play both sports and 60% do not play soccer, what is the probability that a student chosen at random from the class plays neither soccer nor basketball?

A. 0.3
B. 0.4
C. 0.5
D. 0.6
E. 0.7

A. 0.3
B. 0.4
C. 0.5
D. 0.6
E. 0.7

BrentGMATPrepNow · Accepted Answer

If 60% of the students play soccer or basketball, then the other 40% play neither soccer nor basketball.
So, P(selected students plays neither soccer nor basketball) = 40% = 0.4
Answer: B

Not convinced? Think of it this way: 
There are 100 people in a room.
When I say, "raise your hand if you play soccer or basketball," 60 people raise their hand. 
What does this say about the 40 students who aren't raising their hands? 
It tells us that those 40 people play neither soccer nor basketball.

ASIDE: Some students might believe that " people who play soccer or basketball" does not include those who play both sports. 
This belief is incorrect. On the GMAT, the word OR is inclusive, which means people who play both soccer and basketball would also raise their hand.

Fighter1095 · Answer

S. Not s total

B 0.1. 0.2. 0.3

NotB 0.3. 0.4. 0.7

Tot 0.4 0.6. 1

P = (f/T)

P(neither) = (0.4)/1 = 0.4

Posted from my mobile device

Akp880 · Answer

In a class, 60% of students play soccer or basketball. If 10% play both sports and 60% do not play soccer, what is the probability that a student chosen at random from the class plays neither soccer nor basketball?

A. 0.3
B. 0.4
C. 0.5
D. 0.6
E. 0.7

A. 0.3
B. 0.4
C. 0.5
D. 0.6
E. 0.7

Explanation:

60% do not play soccer i.e P(Soccer)=0.4

10% play both sports i.e P(Soccer & Basketball)=0.1

P(only soccer)=0.4-0.1=0.3

60% of students play soccer or basketball i.e 
P(Soccer or Basketball)=0.6
0.6= P(only Soccer)+P(only basketball)+P(Soccer & Basketball)
0.6=0.3+P(basketball)+0.1
P(basketball)=0.2

P(neither soccer nor basketball) =1- P(either soccer or basketball)
=1-0.6
=0.4
 
Hence B is the correct answer.

CrackverbalGMAT · Answer

Using set theory notations, P(A or B) is P(A U B).

The Universal set = 100% = n(AUB) + n(AUB)', where n(A U B)' denotes 'neither' or those who don't take part in any of the activities.

Therefore 60 + n(AUB)' = 100

n(AUB)' = 40

Therefore the required probability = 40/100 = 0.4

Option B

Arun Kumar

Nik11 · Answer

The answer is B; It would have been C if the problem had stated "..60% plays soccer, bask or both.." In this case you should have written 40 + x = 60 ---> 20/100 instead of 30 + x -10 = 60

Hope this help who has chosen C

Fdambro294 · Answer

I think you might have intended something different than what you wrote.

“60% plays soccer, basketball, or both”

is equal to

60% plays soccer OR Basketball = (# overall who Play Soccer) + (# overall who Play Basketball) - (# who play BOTH)

and another way to look at it

60% plays soccer OR basketball = (# play soccer ONLY) + (# play basketball ONLY) + (# play BOTH soccer and basketball)

Posted from my mobile device

sayan640 · Answer

KarishmaB  , I dint include people who play both soccer and basket ball in the 60% of students. Dont you think the langugae of the questions is a bit ambiguous ?  MartyMurray

mayankkamal999 · Answer

Is it okay if I say that there are 60% students who play soccer or basket ball then taking complement i.e. 100-60=40% students who don't play either. So 40/100 = 0.4.
It this way correct?

In a class, 60% of students play soccer or basketball. If 10% play bot

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In a class, 60% of students play soccer or basketball. If 10% play bot

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