Question

According to Harper's Index, 40% of all federal inmates are serving time for drug dealing. A random sample of 20 federal inmates is selected. (a) What is the probability that 9 or more are serving time for drug dealing? (Round your answer to three decimal places.) (b) What is the probability that 3 or fewer are serving time for drug dealing? (Round your answer to three decimal places.) (c) What is the expected number of inmates serving time for drug dealing? (Round your answer to one decimal place.)

Answer 1

Solution :

The Binomial Distribution probability mass function is

P(X = x) = nC_x * p^x * q^{(n - x)}

where

n: number of trials / number of items considered
p: probability of success
q: probability of failure and q = ( 1 - p )
x: number of success

( b )

$\mathrm{n = 20; \:\: p = 0.4;\:\: q = 0.6 }$

$\mathrm{P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)}$

$=20\mathrm{C}_{0}*(0.4)^0*(0.6)^{(20-0)}+20\mathrm{C}_{1}*(0.4)^1*(0.6)^{(20-1)}+20\mathrm{C}_{2}*(0.4)^2*(0.6)^{(20-2)}+20\mathrm{C}_{3}*(0.4)^3*(0.6)^{(20-3)}$

^{$=1*(0.4)^0*(0.6)^{20}+20*(0.4)^1*(0.6)^{19}+190*(0.4)^2*(0.6)^{18}+1140*(0.4)^3*(0.6)^{17}$}

$= 0.00004 + 0.00049 + 0.00309 + 0.01235$

$= 0.01597\approx 0.016$

$= 0.016$

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( c )

$\mathrm{n = 20 \:\:and\:\: p = 0.4 }$

$\mathrm{Expected\:number\:of\:inmates\:serving\:time\:for\:drug\:dealing}$

$\mathrm{=n*p}$

$\mathrm{=20*(0.4)}$

$\mathrm{=8}$