document.write( "Question 1164546: a) Plot the pmf and cdf of the geometric random variable X with probability of success p = 1/3.
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Algebra.Com's Answer #854202 by mccravyedwin(420)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "This concerns getting the probablity that we will get our first success on the\r\n" );
document.write( "kth trial, with a probability of success on each trial being p = 1/3.\r\n" );
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document.write( "The other tutor was correct for part b), which was: \r\n" );
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document.write( "b)\r\n" );
document.write( "\"E%28X%29=1%2Fp=%281%5E%22%22%29%2F%281%2F3%29=3\"\r\n" );
document.write( "\"Var%28X%29=%28%281-p%29%2Fp%5E2%29=%282%2F3%29%2F%281%2F9%29=%282%2F3%29%2A9=2%2A3=6\"\r\n" );
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document.write( "a) Instead of plotting, I'll just list the values:\r\n" );
document.write( "PMF(probability mass function) \"P%28X=k%29=%281-p%29%5E%28k-1%29%2Ap\"\r\n" );
document.write( "CDF(cumulative distribution function) \"P%28X%3C=k%29=1-%281-p%29%5Ek\"\r\n" );
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document.write( " k|PMF(k) |CDF(k)\r\n" );
document.write( " 1|0.3333 |0.3333    \r\n" );
document.write( " 2|0.2222 |0.5556\r\n" );
document.write( " 3|0.1481 |0.7037\r\n" );
document.write( " 4|0.0988 |0.8025\r\n" );
document.write( " 5|0.0658 |0.8683\r\n" );
document.write( " 6|0.0439 |0.9122\r\n" );
document.write( " 7|0.0293 |0.9415\r\n" );
document.write( " 8|0.0195 |0.9610\r\n" );
document.write( " 9|0.0130 |0.9740\r\n" );
document.write( "10|0.0087 |0.9827\r\n" );
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document.write( "c)\r\n" );
document.write( "\"P%28X%3Ek%29=%281-p%29%5Ek\"  \r\n" );
document.write( "\"P%28X%3E4%29=%281-1%2F3%29%5E4=%282%2F3%29%5E4=%282%2F3%29%5E4=16%2F81=0.1975\" (approximately)\r\n" );
document.write( " \r\n" );
document.write( "Edwin
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