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Suppose T be total students in class

(Given) 0.4 T do not play both games.

So, T - 0.4 T = O.6 T play either Basketball or Soccer or both.

Also. (Given) 0.1 T play both games.

Therefore , play either games (only one game ) : 0.6 T - 0.1 T = 0.5 T

hence, Desired Probability = 0.5 T / T = 0.5

Hope it helps.
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Bunuel
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 only one of the sports?

A. 0.2
B. 0.3
C. 0.4
D. 0.5
E. 0.6

Say the class has 100 students. 60 play soccer or basketball. 10 play both.
Attachment:
Screenshot 2022-04-29 at 11.54.38.png
Screenshot 2022-04-29 at 11.54.38.png [ 30.12 KiB | Viewed 1927 times ]
Then number of people lying in yellow, green or blue regions are 60. We know that 10 lie in the green region.
Then number of people lying in the yellow or blue regions are 50. They are the ones who play either only soccer or only basketball.

50% students play only sport hence probability that the chosen student plays only one of these sports is 0.5
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