at a particular school with 200 male students, 58 play football 40 plays basketball and 8 plays both. what is the probablity that a randomly selected male student plays neither
0.55
0.47
0.47 because 58+40+8=106, 200-106=94/200= 0.47
0.45
To find the probability that a randomly selected male student plays neither football nor basketball, we need to subtract the probability of playing either football or basketball from 1.
Let's calculate it step by step:
Step 1: Calculate the probability of playing football.
Given that 58 male students play football, we can write it as P(F) = 58/200.
Step 2: Calculate the probability of playing basketball.
Given that 40 male students play basketball, we can write it as P(B) = 40/200.
Step 3: Calculate the probability of playing both football and basketball.
Given that 8 male students play both sports, we can write it as P(F ∩ B) = 8/200.
Step 4: Calculate the probability of playing either football or basketball.
Using the principle of inclusion-exclusion, we calculate the probability as:
P(F ∪ B) = P(F) + P(B) - P(F ∩ B)
P(F ∪ B) = (58/200) + (40/200) - (8/200)
Step 5: Calculate the probability of playing neither football nor basketball.
Finally, we subtract the probability of playing either football or basketball from 1:
P(neither) = 1 - P(F ∪ B)
Now, let's calculate it:
P(neither) = 1 - [(58/200) + (40/200) - (8/200)]
P(neither) = 1 - (98/200)
P(neither) = 1 - 0.49
P(neither) = 0.51
Therefore, the probability that a randomly selected male student plays neither football nor basketball is 0.51, or 51%.