Question and Answers Forum

All Questions      Topic List

UNKNOWN Questions

Previous in All Question      Next in All Question      

Previous in UNKNOWN      Next in UNKNOWN      

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%

$$\mathrm{the}\:\mathrm{product}\:\mathrm{is}\:\mathrm{uneven}\:\mathrm{when}\:\mathrm{no}\:\mathrm{even}\:\mathrm{number} \\ $$$$\mathrm{is}\:\mathrm{among}\:\mathrm{the}\:\mathrm{chosen}\:\mathrm{numbers} \\ $$$$\mathrm{the}\:\mathrm{probability}\:\mathrm{to}\:\mathrm{get}\:\mathrm{an}\:\mathrm{uneven}\:\mathrm{number}\:\mathrm{is} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{number}\:\frac{\mathrm{10}}{\mathrm{20}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{number}\:\frac{\mathrm{9}}{\mathrm{19}} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{number}\:\frac{\mathrm{8}}{\mathrm{18}}=\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\mathrm{3}\:\mathrm{uneven}\:\mathrm{numbers}\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{9}}{\mathrm{19}}×\frac{\mathrm{4}}{\mathrm{9}}=\frac{\mathrm{2}}{\mathrm{19}}\approx\mathrm{10}.\mathrm{53\%} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\mathrm{product}\:\mathrm{is}\:\mathrm{even} \\ $$$$\mathrm{is}\:\frac{\mathrm{17}}{\mathrm{19}}\approx\mathrm{89}.\mathrm{47\%} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com