Amy plays one game of snooker and one game of billards each friday. The probability that she wins every friday is as follows

Snooker: Win=1/3

Billards: Wind=3/4

If she plays one game of each for a number of fridays, calculate how many games Amy did not win at either if she won at both on 21 fridays

So I did

1/4=21/x (x being the amount of games played)

so x=84

then since 1/6 is the probability that she loses both I did

1/6=z/84 z being the number of fridays

z=14

correct?