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Mathematics 20 Online
OpenStudy (anonymous):

Please help with this, I won't give up!! *Use the data in the following table, which summarizes results from 171 pedestrian deaths that were caused by accidents. If two different deaths are randomly select the replacement, find the probability they both involved intoxicated drivers. Pedestrian Intoxicated? Yes No Drivers yes 25 34 Intoxicated? No 97 15 Please help me, Thanks!!! ;))

jimthompson5910 (jim_thompson5910):

how many drivers were intoxicated?

OpenStudy (anonymous):

122?

jimthompson5910 (jim_thompson5910):

look in the first row

jimthompson5910 (jim_thompson5910):

that's where the drivers were intoxicated

jimthompson5910 (jim_thompson5910):

add up those numbers to get 25+34 = 59

jimthompson5910 (jim_thompson5910):

so there were 59 intoxicated drivers total

OpenStudy (anonymous):

Oh!

jimthompson5910 (jim_thompson5910):

So what is the probability of selecting one random intoxicated driver?

OpenStudy (anonymous):

That's my problem. How do I select the Random?

OpenStudy (anonymous):

Add them all?

OpenStudy (anonymous):

it will be then 59/171?

jimthompson5910 (jim_thompson5910):

you are correct, the probability of selecting an intoxicated driver is 59/171

OpenStudy (anonymous):

Ooh good! :)

jimthompson5910 (jim_thompson5910):

so if you're selecting with replacement, then P(2 intoxicated drivers) = (59/171)*(59/171) if you're selecting without replacement, then P(2 intoxicated drivers) = (59/171)*(58/170)

OpenStudy (anonymous):

Ohh waoo!! It will be without replacement..Then the answer is 0.1177% Which is correct. OMG!!

jimthompson5910 (jim_thompson5910):

you got it, very nice

OpenStudy (anonymous):

Thank you so muchh!!!! Now I am ready for my quiz! jeje!

jimthompson5910 (jim_thompson5910):

you're welcome

jimthompson5910 (jim_thompson5910):

good luck

OpenStudy (anonymous):

Thanx!! =D

jimthompson5910 (jim_thompson5910):

np

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