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Mathematics 33 Online
OpenStudy (hockeychick23):

The estimated probability of a bowler getting strike during a particular frame is 41%. If several simulations of the bowler bowling 2 frames were performed, what percentage of the simulations would the bowler be most likely to get a strike in each of the two frames? A. 61% B. 38% C. 18% D. 81%

OpenStudy (perl):

if we assume that the events are independent

OpenStudy (perl):

we can multiply probabilities

OpenStudy (hockeychick23):

So its 41% * 41% ?

OpenStudy (perl):

let me think about it, one moment

OpenStudy (perl):

yes i think thats right

OpenStudy (hockeychick23):

ok, when i multiplied them i got 16.81%

OpenStudy (hockeychick23):

I'm just unsure if i should be taking the percentage thats closest to 41 which would be 38 or not

OpenStudy (perl):

18% could be the answer, it might be a typo

OpenStudy (perl):

i could try this with a real simulation

OpenStudy (hockeychick23):

oh ok cool, how do i do that?

OpenStudy (perl):

yes i checked i got .41 * .41 = .1681 the closest answer looks like 18%

OpenStudy (hockeychick23):

ok thanks!

OpenStudy (perl):

also the questions says , what percentage would the bowler be most likely to bowl, so it is not asking for exact percentage, only whats closest

OpenStudy (hockeychick23):

oh yea thanks(:

OpenStudy (perl):

by the way, i was kidding. a real simulation is going bowling :)

OpenStudy (hockeychick23):

hahaha awesome(:

OpenStudy (perl):

you could simulate this with excel *

OpenStudy (hockeychick23):

oh cool

OpenStudy (perl):

but the nice thing about probability, we don't have to simulate . we can use our knowledge of how probability works . there are situations where we would have to use a simulation, since we simply dont know anything about it.

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