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Mathematics 13 Online
OpenStudy (helpmeplease14):

How do you know when a problem have independent, dependent, or mutually exclusive events?

jimthompson5910 (jim_thompson5910):

Two events A and B are independent if P(A and B) = P(A) * P(B) they dependent if P(A and B) = P(A) * P(B|A)

jimthompson5910 (jim_thompson5910):

Two events A and B are mutually exclusive if P(A and B) = 0 in other words, this says that it is impossible for the 2 events to occur at the same time

OpenStudy (helpmeplease14):

I understand that but look for example

jimthompson5910 (jim_thompson5910):

Ask yourself this: if one name is picked out, is it placed back into the hat?

OpenStudy (helpmeplease14):

?

jimthompson5910 (jim_thompson5910):

Where's your question?

OpenStudy (helpmeplease14):

This

jimthompson5910 (jim_thompson5910):

I meant the question you had about what I posted

OpenStudy (anonymous):

5 people in all and 2 girls

OpenStudy (anonymous):

i guess 2/5

OpenStudy (helpmeplease14):

I got that

OpenStudy (anonymous):

good

OpenStudy (anonymous):

is that the answer

jimthompson5910 (jim_thompson5910):

no it's not the answer

jimthompson5910 (jim_thompson5910):

what is the probability of just picking lisa?

OpenStudy (helpmeplease14):

1/5

jimthompson5910 (jim_thompson5910):

yes

OpenStudy (anonymous):

it is unlikly because more boys then girls so it is 1/5

jimthompson5910 (jim_thompson5910):

now after you pick lisa, you can't pick her again so the 5 names to pick from reduces to 4 after Lisa is picked (and her name taken out of the hat), what is the probability of picking Chrissie?

OpenStudy (helpmeplease14):

But it says what is the probability of Lisa and another girl

OpenStudy (anonymous):

1/4 i guess

jimthompson5910 (jim_thompson5910):

yes we're focusing on each one at a time at first

OpenStudy (helpmeplease14):

OK

OpenStudy (helpmeplease14):

1/5

jimthompson5910 (jim_thompson5910):

nope, remember that lisa's name was taken out to make sure we didn't pick her again

OpenStudy (helpmeplease14):

1/4

jimthompson5910 (jim_thompson5910):

yep

jimthompson5910 (jim_thompson5910):

you will then multiply 1/5 and 1/4 to get 1/20

OpenStudy (helpmeplease14):

Oh

jimthompson5910 (jim_thompson5910):

since I'm using the rule P(A and B) = P(A) * P(B|A) which means the events are dependent

OpenStudy (helpmeplease14):

OK thx

jimthompson5910 (jim_thompson5910):

the probability of picking Chrissie depends on the outcome of picking Lisa

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