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3-integers-are-chosen-at-random-from-the-first-20-integers-The-probability-that-their-ptoduct-is-even-is-

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Question Number 81926 by zainal tanjung last updated on 16/Feb/20
3 integers are chosen at random from the first 20 integers. The probability that their ptoduct is even, is
$$\mathrm{3}\:\mathrm{integers}\:\mathrm{are}\:\mathrm{chosen}\:\mathrm{at}\:\mathrm{random}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{20}\:\mathrm{integers}.\:\mathrm{The}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{their}\:\mathrm{ptoduct}\:\mathrm{is}\:\mathrm{even},\:\mathrm{is} \\ $$
Commented by mr W last updated on 16/Feb/20
p=1−(C_3 ^(10) /C_3 ^(20) )=1−((120)/(1140))=89.5% or p=((C_3 ^(10) +C_2 ^(10) C_1 ^(10) +C_1 ^(10) C_2 ^(10) )/C_3 ^(20) )=((1020)/(1140))=89.5%
$${p}=\mathrm{1}−\frac{{C}_{\mathrm{3}} ^{\mathrm{10}} }{{C}_{\mathrm{3}} ^{\mathrm{20}} }=\mathrm{1}−\frac{\mathrm{120}}{\mathrm{1140}}=\mathrm{89}.\mathrm{5\%} \\ $$$${or} \\ $$$${p}=\frac{{C}_{\mathrm{3}} ^{\mathrm{10}} +{C}_{\mathrm{2}} ^{\mathrm{10}} {C}_{\mathrm{1}} ^{\mathrm{10}} +{C}_{\mathrm{1}} ^{\mathrm{10}} {C}_{\mathrm{2}} ^{\mathrm{10}} }{{C}_{\mathrm{3}} ^{\mathrm{20}} }=\frac{\mathrm{1020}}{\mathrm{1140}}=\mathrm{89}.\mathrm{5\%} \\ $$
Answered by MJS last updated on 16/Feb/20
the product is uneven when no even number is among the chosen numbers the probability to get an uneven number is for the 1^(st) number ((10)/(20))=(1/2) for the 2^(nd) number (9/(19)) for the 3^(rd) number (8/(18))=(4/9) 3 uneven numbers (1/2)×(9/(19))×(4/9)=(2/(19))≈10.53% ⇒ the probability that the product is even is ((17)/(19))≈89.47%
$$\mathrm{the}\:\mathrm{product}\:\mathrm{is}\:\mathrm{uneven}\:\mathrm{when}\:\mathrm{no}\:\mathrm{even}\:\mathrm{number} \\ $$$$\mathrm{is}\:\mathrm{among}\:\mathrm{the}\:\mathrm{chosen}\:\mathrm{numbers} \\ $$$$\mathrm{the}\:\mathrm{probability}\:\mathrm{to}\:\mathrm{get}\:\mathrm{an}\:\mathrm{uneven}\:\mathrm{number}\:\mathrm{is} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{number}\:\frac{\mathrm{10}}{\mathrm{20}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{number}\:\frac{\mathrm{9}}{\mathrm{19}} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{number}\:\frac{\mathrm{8}}{\mathrm{18}}=\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\mathrm{3}\:\mathrm{uneven}\:\mathrm{numbers}\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{9}}{\mathrm{19}}×\frac{\mathrm{4}}{\mathrm{9}}=\frac{\mathrm{2}}{\mathrm{19}}\approx\mathrm{10}.\mathrm{53\%} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\mathrm{product}\:\mathrm{is}\:\mathrm{even} \\ $$$$\mathrm{is}\:\frac{\mathrm{17}}{\mathrm{19}}\approx\mathrm{89}.\mathrm{47\%} \\ $$

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