BeccaB003
  • BeccaB003
Question about probability. I appreciate the help!
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
BeccaB003
  • BeccaB003
This table is an example of the principle of independence. This table is not an example of the principle of independence. There is not enough information to answer this question.
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BeccaB003
  • BeccaB003
I need help understanding principle independence. @jim_thompson5910 Thanks!
jim_thompson5910
  • jim_thompson5910
How many people have a membership?

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BeccaB003
  • BeccaB003
40
jim_thompson5910
  • jim_thompson5910
this is out of 70 people total so P(has membership) = 40/70 = 4/7
jim_thompson5910
  • jim_thompson5910
what is the probability a person attends one or more of the classes offered?
BeccaB003
  • BeccaB003
31--that is if we are talking about people with and without memberships. Only with memberships is 17 and without membership is 14.
jim_thompson5910
  • jim_thompson5910
so the probability someone attends 1 or more classes is 31/70
jim_thompson5910
  • jim_thompson5910
now IF the two events (shown below) * has membership * attends 1 or more classes are independent, then P( has membership AND attends 1 or more classes) = P(has membership) * P(attends 1 or more classes)
jim_thompson5910
  • jim_thompson5910
does that look familiar?
BeccaB003
  • BeccaB003
Yes, it does. So, P(40) * P(31) =1240 Right?
BeccaB003
  • BeccaB003
And the answer would be: This table is an example of the principle of independence. (choices listed above)
jim_thompson5910
  • jim_thompson5910
P( has membership AND attends 1 or more classes) = P(has membership) * P(attends 1 or more classes) P( has membership AND attends 1 or more classes) = (4/7) * (31/70) P( has membership AND attends 1 or more classes) = 62/245 do you see how I got that?
BeccaB003
  • BeccaB003
oh! *slaps forehead* Yes, sorry I wasn't thinking and ignored what you'd said before about the problem. I understand it now.
jim_thompson5910
  • jim_thompson5910
how many people fit these requirements has membership AND attends 1 or more classes
BeccaB003
  • BeccaB003
Isn't that just what we solved for? 62/245? And how does this correlate to the principle of independence?
jim_thompson5910
  • jim_thompson5910
look for the "has membership" row and the "1 or more classes" column what number is there?
BeccaB003
  • BeccaB003
17
LynFran
  • LynFran
@jim_thompson5910 when you finish here can u please visit the link in ur notification this question needs a 2nd opinion thanks
jim_thompson5910
  • jim_thompson5910
since it's 17 out of 70 total the actual probability P( has membership AND attends 1 or more classes) should be 17/70
jim_thompson5910
  • jim_thompson5910
and not 62/245
jim_thompson5910
  • jim_thompson5910
you only multiply IF the two events are independent thinking in reverse, if you can multiply and get the same result as looking in the table, then the events are independent
jim_thompson5910
  • jim_thompson5910
but 17/70 is not equal to 62/245 so they are not independent events
BeccaB003
  • BeccaB003
Thank you so much! You explained everything very well. So, this table is not an example of the principle of independence because the two events don't equal each other?
jim_thompson5910
  • jim_thompson5910
correct, there is some connection between the two events (one event is dependent on the other somehow)
BeccaB003
  • BeccaB003
Okay, thank you so much. You are amazing!
jim_thompson5910
  • jim_thompson5910
you're welcome

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