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Mathematics 8 Online
OpenStudy (anonymous):

Of a group of 500 people, 60 are fit enough to climb Mount Shasta in California. The group has 300 women and 200 men. 12% of the men are fit to climb the mountain. What is the probability that a person picked at random from the group is a male or is fit to climb Mount Shasta?

OpenStudy (anonymous):

so 12% of 200 =24 so 24 men and 36 women so out of 300 women 12% were fit im confused HELP!!!

OpenStudy (anonymous):

Is that supposed to say and or or in the final question?

OpenStudy (anonymous):

What is the probability that a person picked at random from the group is a male or is fit to climb Mount Shasta?

OpenStudy (anonymous):

male or fit?

OpenStudy (anonymous):

so 24 are male and 60 are fit

OpenStudy (anonymous):

If it was an "and " then the answer would be %4.8 But if it were an "and" then the probability of getting a male would be %40.0 And the probability of getting a fit person would be 12%

OpenStudy (anonymous):

my choices are 59/125 13/25 56/125

OpenStudy (anonymous):

59/125 = 47.2% 13/25 = 52% 56/125=44.8%

OpenStudy (anonymous):

56/125

OpenStudy (anonymous):

@eliassaab can u help please?

OpenStudy (anonymous):

It is 59/125. Let me tell you why \[ P(Man \cup Fit) = \\P(Man) + P(Fit) - P( Man \cap Fit)=\\ \frac {200}{500} + \frac {60}{500} - \frac {24}{500}=\frac{59}{125} \]

OpenStudy (anonymous):

how do you get the 24/500 ?

OpenStudy (anonymous):

its not 56/125

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