Question Number 76651 by john santu last updated on 29/Dec/19
![if a and b two numbers do not have to be chosen randomly and with returns from the set {1,2,3,4,5}. what probability that (a/b) is an integer ?](https://www.tinkutara.com/question/Q76651.png)
$${if}\:{a}\:{and}\:{b}\:{two}\:{numbers}\:{do}\:{not}\:{have}\: \\ $$$${to}\:{be}\:{chosen}\:{randomly}\:{and}\:{with}\: \\ $$$${returns}\:{from}\:{the}\:{set}\:\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\}.\:{what}\: \\ $$$${probability}\:{that}\:\frac{{a}}{{b}}\:{is}\:{an}\:{integer}\:? \\ $$
Answered by benjo 1/2 santuyy last updated on 29/Dec/19
![suppose that A is the occurence of a/b integer . now n(A)= 10 and n(S) = 5×5 = 25 hence P(A) = 10/25 = 2/5.](https://www.tinkutara.com/question/Q76693.png)
$${suppose}\:{that}\:{A}\:{is}\:{the}\:{occurence}\: \\ $$$${of}\:{a}/{b}\:{integer}\:.\: \\ $$$${now}\:{n}\left({A}\right)=\:\mathrm{10}\:{and}\:{n}\left({S}\right)\:=\:\mathrm{5}×\mathrm{5}\:=\:\mathrm{25} \\ $$$${hence}\:{P}\left({A}\right)\:=\:\mathrm{10}/\mathrm{25}\:=\:\mathrm{2}/\mathrm{5}.\: \\ $$
Commented by benjo 1/2 santuyy last updated on 30/Dec/19
![1/1, 2/1,3/1,4/1^](https://www.tinkutara.com/question/Q76760.png)