If two distinct members of the set $\{ 3, 7, 21, 27, 35, 42, 51 \}$ are randomly selected and multiplied, what is the probability that the p

You need to check your (7,35), (7,42) and (7,51) pairs Melody.  I get the following:

probability = 9/21 or 3/7

.

Hi LooLoo

I hope that you haven't reposted this without letting people know.  If it find that you have I will be very annoyed :/

But assuming that you haven't I will see about getting you an answer

If two distinct members of the set $\{ 3, 7, 21, 27, 35, 42, 51 \}$ are randomly selected and multiplied, what is the probability that the product is a multiple of 63? Express your answer as a common fraction.

the factors of 63 are  1,3,9,7 AND 63

7 times any multiple of 9 would work   that is  27,     (7,27)  that is the only one

21 = 9*3   that would work with any multiple of 3     (21,3)(21,27)(21,42)(21,51)

3 times any mulptile of 21 would work     (3,21)got that  (3,42)

27=3*9  times any mult of 7       (7,35)     (7,42)    (7,51)

35=5*7  times any mult of 9        got those

42=3*2*7  any mult of 3               (42,51)

that is it I think so there are  10 of them    You need to check this

Prob =  $${\frac{{\mathtt{10}}}{{\left({\frac{{\mathtt{7}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}}} = {\frac{{\mathtt{10}}}{{\mathtt{21}}}} = {\mathtt{0.476\: \!190\: \!476\: \!190\: \!476\: \!2}}$$

Prob =   $${\frac{{\mathtt{10}}}{{\mathtt{21}}}}$$

Yes, I made a stupid error.  Thanks Alan

I knew that there could easily be mistakes which is why I told the asker to check my answer.

I am sure your answer is correct.  I am not going to bother with correcting mine