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Question Number 11600 by tawa last updated on 28/Mar/17

A number is selected at random from the integers 1 to 100 exclusive,   What is the chance of choosing at random a two digit number which is  (a) Divisible by 4  (b) The multiple of 5

$$\mathrm{A}\:\mathrm{number}\:\mathrm{is}\:\mathrm{selected}\:\mathrm{at}\:\mathrm{random}\:\mathrm{from}\:\mathrm{the}\:\mathrm{integers}\:\mathrm{1}\:\mathrm{to}\:\mathrm{100}\:\mathrm{exclusive},\: \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{chance}\:\mathrm{of}\:\mathrm{choosing}\:\mathrm{at}\:\mathrm{random}\:\mathrm{a}\:\mathrm{two}\:\mathrm{digit}\:\mathrm{number}\:\mathrm{which}\:\mathrm{is} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Divisible}\:\mathrm{by}\:\mathrm{4} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{5} \\ $$

Answered by mrW1 last updated on 29/Mar/17

numbers to select: 1,2,3,...,100 ⇒101 pieces    two digit numbers divisible by 4:  12,16,...,96 ⇒22 pieces    multiple of 5:  5,10,...,95⇒19 pieces    (a): ((22)/(101))=21.8%  (b): ((19)/(101))=18.8%

$${numbers}\:{to}\:{select}:\:\mathrm{1},\mathrm{2},\mathrm{3},...,\mathrm{100}\:\Rightarrow\mathrm{101}\:{pieces} \\ $$$$ \\ $$$${two}\:{digit}\:{numbers}\:{divisible}\:{by}\:\mathrm{4}: \\ $$$$\mathrm{12},\mathrm{16},...,\mathrm{96}\:\Rightarrow\mathrm{22}\:{pieces} \\ $$$$ \\ $$$${multiple}\:{of}\:\mathrm{5}: \\ $$$$\mathrm{5},\mathrm{10},...,\mathrm{95}\Rightarrow\mathrm{19}\:{pieces} \\ $$$$ \\ $$$$\left({a}\right):\:\frac{\mathrm{22}}{\mathrm{101}}=\mathrm{21}.\mathrm{8\%} \\ $$$$\left({b}\right):\:\frac{\mathrm{19}}{\mathrm{101}}=\mathrm{18}.\mathrm{8\%} \\ $$

Commented by tawa last updated on 29/Mar/17

Wow, God bless you sir.

$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by tawa last updated on 29/Mar/17

Sir, the exclussive, i thought it mean 1 and 100 will not be there

$$\mathrm{Sir},\:\mathrm{the}\:\mathrm{exclussive},\:\mathrm{i}\:\mathrm{thought}\:\mathrm{it}\:\mathrm{mean}\:\mathrm{1}\:\mathrm{and}\:\mathrm{100}\:\mathrm{will}\:\mathrm{not}\:\mathrm{be}\:\mathrm{there} \\ $$

Commented by mrW1 last updated on 29/Mar/17

then (a)=((22)/(99)) and (b)=((19)/(99))

$${then}\:\left({a}\right)=\frac{\mathrm{22}}{\mathrm{99}}\:{and}\:\left({b}\right)=\frac{\mathrm{19}}{\mathrm{99}} \\ $$

Commented by tawa last updated on 29/Mar/17

Thank you sir.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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