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Question Number 182199 by depressiveshrek last updated on 05/Dec/22

Let S={1, 2, 3, 4, 5, 6, 7} If we multiply atleast 2 numbers from this set with each other, what are the chances of the product to turn out to be divisible by 3?

$${Let}\:{S}=\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5},\:\mathrm{6},\:\mathrm{7}\right\} \\ $$$${If}\:{we}\:{multiply}\:{atleast}\:\mathrm{2}\:{numbers} \\ $$$${from}\:{this}\:{set}\:{with}\:{each}\:{other},\:{what} \\ $$$${are}\:{the}\:{chances}\:{of}\:{the}\:{product}\:{to} \\ $$$${turn}\:{out}\:{to}\:{be}\:{divisible}\:{by}\:\mathrm{3}? \\ $$

Answered by Acem last updated on 06/Dec/22

This solution is for product only 2 numbers I will solve the question later P(product_(div. by 3) )= ((11)/(21)) a×b, b×a ≡ one way, a×a is refused according to the question The method: S_1 = {1, 2, 4, 5, 7} , S_2 = {3, 6} P(product_(div. by 3) )= ((C_1 ^( 5) C_( 1) ^( 2) + C_( 2) ^( 2) _(S_2 ) )/((n(n−1))/2)) ; n= 7

$$\:{This}\:{solution}\:{is}\:{for}\:{product}\:{only}\:\mathrm{2}\:{numbers} \\ $$$$\:{I}\:{will}\:{solve}\:{the}\:{question}\:{later} \\ $$$$\:{P}\left({product}_{{div}.\:{by}\:\mathrm{3}} \right)=\:\frac{\mathrm{11}}{\mathrm{21}}\: \\ $$$$\:{a}×{b},\:{b}×{a}\:\equiv\:{one}\:{way}, \\ $$$$\:{a}×{a}\:{is}\:{refused}\:{according}\:\:{to}\:{the}\:{question} \\ $$$$ \\ $$$$\:{The}\:{method}: \\ $$$$\:{S}_{\mathrm{1}} =\:\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{4},\:\mathrm{5},\:\mathrm{7}\right\}\:\:\:\:,\:\:{S}_{\mathrm{2}} =\:\left\{\mathrm{3},\:\mathrm{6}\right\} \\ $$$$ \\ $$$$\:\:{P}\left({product}_{{div}.\:{by}\:\mathrm{3}} \right)=\:\frac{{C}_{\mathrm{1}} ^{\:\mathrm{5}} \:{C}_{\:\mathrm{1}} ^{\:\mathrm{2}} \:+\:\underset{{S}_{\mathrm{2}} } {\underbrace{{C}_{\:\mathrm{2}} ^{\:\mathrm{2}} }}}{\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}}\:\:\:\:\:\:\:\:\:\:;\:{n}=\:\mathrm{7} \\ $$

Commented by mr W last updated on 06/Dec/22

the question is “take at least two numbers”, not “take only two numbers” as you assumed.

$${the}\:{question}\:{is}\:``{take}\:{at}\:{least}\:{two}\: \\ $$$${numbers}'',\:{not}\:``{take}\:{only}\:{two}\: \\ $$$${numbers}''\:{as}\:{you}\:{assumed}. \\ $$

Commented by Acem last updated on 06/Dec/22

Yes, i didn′t notice that well, thank you Sir for the correction

$${Yes},\:{i}\:{didn}'{t}\:{notice}\:{that}\:{well},\:{thank}\:{you}\:{Sir} \\ $$$$\:{for}\:{the}\:{correction} \\ $$

Answered by mr W last updated on 06/Dec/22

1) take 7 numbers total: 1 way divisible by 3: 1 way 2) take 6 numbers total: 7 ways divisible by 3: 7 ways 3) take 5 numbers total: C_5 ^7 =21 ways divisible by 3: 1×C_3 ^5 +2×C_4 ^5 =20 4) take 4 numbers total: C_4 ^7 =35 ways divisible by 3: 1×C_2 ^5 +2×C_3 ^5 =30 5) take 3 numbers total: C_3 ^7 =35 ways divisible by 3: 1×C_1 ^5 +2×C_2 ^5 =25 6) take 2 numbers total: C_2 ^7 =21 ways divisible by 3: 1+2×C_1 ^5 =11 take at least 2 numbers: total: 1+7+21+35+35+21=120 ways divisible by 3: 1+7+20+30+25+11=94 ways p=((94)/(120))≈78.3%

$$\left.\mathrm{1}\right)\:{take}\:\mathrm{7}\:{numbers} \\ $$$${total}:\:\mathrm{1}\:{way} \\ $$$${divisible}\:{by}\:\mathrm{3}:\:\mathrm{1}\:{way} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:{take}\:\mathrm{6}\:{numbers} \\ $$$${total}:\:\mathrm{7}\:{ways} \\ $$$${divisible}\:{by}\:\mathrm{3}:\:\mathrm{7}\:{ways} \\ $$$$ \\ $$$$\left.\mathrm{3}\right)\:{take}\:\mathrm{5}\:{numbers} \\ $$$${total}:\:{C}_{\mathrm{5}} ^{\mathrm{7}} =\mathrm{21}\:{ways} \\ $$$${divisible}\:{by}\:\mathrm{3}:\:\mathrm{1}×{C}_{\mathrm{3}} ^{\mathrm{5}} +\mathrm{2}×{C}_{\mathrm{4}} ^{\mathrm{5}} =\mathrm{20} \\ $$$$ \\ $$$$\left.\mathrm{4}\right)\:{take}\:\mathrm{4}\:{numbers} \\ $$$${total}:\:{C}_{\mathrm{4}} ^{\mathrm{7}} =\mathrm{35}\:{ways} \\ $$$${divisible}\:{by}\:\mathrm{3}:\:\mathrm{1}×{C}_{\mathrm{2}} ^{\mathrm{5}} +\mathrm{2}×{C}_{\mathrm{3}} ^{\mathrm{5}} =\mathrm{30} \\ $$$$ \\ $$$$\left.\mathrm{5}\right)\:{take}\:\mathrm{3}\:{numbers} \\ $$$${total}:\:{C}_{\mathrm{3}} ^{\mathrm{7}} =\mathrm{35}\:{ways} \\ $$$${divisible}\:{by}\:\mathrm{3}:\:\mathrm{1}×{C}_{\mathrm{1}} ^{\mathrm{5}} +\mathrm{2}×{C}_{\mathrm{2}} ^{\mathrm{5}} =\mathrm{25} \\ $$$$ \\ $$$$\left.\mathrm{6}\right)\:{take}\:\mathrm{2}\:{numbers} \\ $$$${total}:\:{C}_{\mathrm{2}} ^{\mathrm{7}} =\mathrm{21}\:{ways} \\ $$$${divisible}\:{by}\:\mathrm{3}:\:\mathrm{1}+\mathrm{2}×{C}_{\mathrm{1}} ^{\mathrm{5}} =\mathrm{11} \\ $$$$ \\ $$$$\boldsymbol{{take}}\:\boldsymbol{{at}}\:\boldsymbol{{least}}\:\mathrm{2}\:\boldsymbol{{numbers}}: \\ $$$${total}:\:\mathrm{1}+\mathrm{7}+\mathrm{21}+\mathrm{35}+\mathrm{35}+\mathrm{21}=\mathrm{120}\:{ways} \\ $$$${divisible}\:{by}\:\mathrm{3}:\:\mathrm{1}+\mathrm{7}+\mathrm{20}+\mathrm{30}+\mathrm{25}+\mathrm{11}=\mathrm{94}\:{ways} \\ $$$${p}=\frac{\mathrm{94}}{\mathrm{120}}\approx\mathrm{78}.\mathrm{3\%} \\ $$

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