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Mathematics 21 Online
OpenStudy (rizags):

please help! question below.

OpenStudy (rational):

Jim can win if Helen flips a tail and Jim flips a head : 1/2*1/2 or if Helen flips a tail again and Jim flips a head : (1/2*1/2)*1/2*1/2 or if Helen flips a tail again and Jim flips a head : (1/2*1/2*1/2*1/2)*1/2*1/2 ....

OpenStudy (rational):

so the probability for Jim winning is given by \[\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\cdots\]

OpenStudy (rizags):

how do i find that sum?

OpenStudy (rational):

its a converging geometric series

OpenStudy (rizags):

I know what both those terms mean, but I am unfamiliar with how to calculate it, sorry...

OpenStudy (rational):

a = 1/4 r = 1/4 plug and chug

OpenStudy (rizags):

It's \[\huge \frac{\frac{1}{4}}{1-\frac{1}{4}}\] or \[\huge \frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}\]

OpenStudy (rational):

looks good

OpenStudy (rizags):

Great! Thanks, so this is not a fair game, right?

OpenStudy (rational):

Clearly the one who flips the coin first has an advantage

OpenStudy (rizags):

ok, got it. The final answer would thus be \[\huge (1,3)\] right?

OpenStudy (rational):

Yep!

OpenStudy (rizags):

ok, great! thanks

OpenStudy (rational):

np :)

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