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 soccer only?

A. 0.1
B. 0.2
C. 0.3
D. 0.4
E. 0.5

yashikaaggarwal · Answer

Let Total no. of students = 100

the probability that a student chosen at random from the class plays basketball only = Students playing only Soccer/ total no. of students = 30/100 = 0.3

Answer is C

Pranithare · Answer

Hi Yashika,
can you please explain how you got the split between non soccer as 20 and 40.

yashikaaggarwal · Answer

See total is 100 (I let that total is 100) so, if In a class, 60% of students play soccer or basketball. 60% of 100 = 60 play game, rest 40 are lazy (like me

) -------------------------(1) If 10% play both sports => 10 people play both game ----------------------------(2) and 60% do not play soccer= 60 do not play soccer ------------------(3) => so 40 does play soccer, with or without basketball ---------------------(4) Put all 4 in Matrix, you will get value.

CrackverbalGMAT · Answer

Let there be a total of 100 players.

From Set Theory, Total = P(A U B) + P(A U B)' = Only A + Only B + P(A ∩ B) + P(A U B)' ... Equation (1)

Given P(A ∩ B) = 10% of 100 = 10

Also given that those who don't play soccer = Thse play only Hockey + P(S U H)' = 60% of 100 = 60

Putting these values in Equation (1), we get 100 = Only Soccer + 10 + 60

Those who play only soccer = 100 - 70 = 30

Probability = \frac{Favorable \space Outcomes}{Total \space Outcomes}= \frac{30}{100} = 0.3

Option C
 
Arun Kumar

EgmatQuantExpert · Answer

Given

• In a class, 60% of students play soccer or basketball.
• 10% play both sports.
• 60% do not play soccer.

To Find

• The probability that a student chosen at random from the class plays soccer only.

Approach and Working Out

• Let us assume, people who play only soccer is A%, only basketball = B%, both the sports = C% and none of the sports = D%
 o A + B +C + D = 100%

• 60% of students play soccer or basketball.
 o A + B + C = 60%
o D = 100 – 60 = 40%

• 10% play both sports,
 o C = 10%
o A + B = 100 – C – D
o A + B = 50%

• 60% do not play soccer.
 o B + D = 60%
o B = 60 – D
o B = 20%
o A = 30%

• Player who plays soccer only is A which is 30% and that will be the probability as well.

Correct Answer: Option C

MathRevolution · Answer

From Set Theory, Total = P(A U B) + P(A U B)' = Only A + Only B + P(A ∩ B) + P(A U B)'

Let the total number of players be 100.

=> 10% play both sports = P(A ∩ B) =  \frac{10}{100}  * 100 = 10

=> 60% do not play soccer = Only basketball + P(A U B)' =  \frac{60}{100}  * 100 = 60

=> 100 = Only soccer + 10 + 60

=> Only soccer = 30

=> Probability:  \frac{30 }{ 100}  = 0.3

Answer C

twobagels · Answer

If 0.6 from the class do not play soccer, 0.4 do play soccer.

0.1 play both soccer and basketball.

0.4 - 0.1 = 0.3

0.3 is the probability that a student selected at random from the class plays soccer only.

Gambrosio · Answer

I have the right solution, but I am not sure wether this is the right approach.

60% do not play soccer - 40% do play soccer, we just don't know if soccer and basketball or just soccer.

We know that 10% play both sports.

Since soccer players are soccer = {soccer and basketball} + {soccer without basketball},
leaves us with: 0.4 = 0.1 + {soccer}

So: just soccer = 0.3

My problem here is that I did not use the first information; 60% play soccer or basketball, at all. Does anyone know if this is still a proper way to handle such problems?

Bunuel · Answer

That's correct. Sometimes a question includes information that's not needed to solve it.

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

Prep Toolkit

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