Two distinct numbers are selected simultaneously and at random from the set (1, 2, 3, 4, 5). What is the probability that the smaller one divides the larger one? Express your answer as a common fraction.

i am afriad i may mis a case when calculating this. thank you for help!

Guest Apr 6, 2021

edited by
Guest
Apr 6, 2021

#1**0 **

Start with the smaller one:

you have 1, which has 4 larger numbers that can divide it

you have 2, which has 2 larger numbers that can divide it

you have 3, which has 0 larger numbers that can divide it

you have 4, which has 0 larger numbers that can divide it

you have 5, which has 0 larger numbers that can divide it

Lastly, there are $5^2=25$ total combinations, and only 6 of which work. Our end result is $\boxed{\frac{6}{25}}$

RiemannIntegralzzz Apr 6, 2021

#2**+1 **

no... i sadly got this wrong. the solution it gave me:

There are 10 pairs of integers that we can potentially select. The easiest way to do this is to simply write them all out: (1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), and (4,5). The 4 pairs with 1 as an element all obviously work, as does (2,4), but none of the others do. That means that 5 out of 10 pairs work which gives us a probability of **1/2**. thank you for helping, though! :)

Guest Apr 6, 2021

#3**+1 **

Ohhhhhhh I see what happened, their solution, for the total cases, it only uses the cases $(a,b)$ where a is smaller than b, so there are only 10, and there is only 1 number larger than 2 that divides it 4, i thought there was a 6 for some reason, so the final probability is $\frac{5}{10}=\boxed{\frac{1}{2}}$

RiemannIntegralzzz
Apr 6, 2021