1. ## Probability

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?

2. Hello, Mukilab!

I'll show you a method I often use.

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: Win = 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.

The probability that Amy wins both games on a particular Friday is:
. . $P(\text{win both}) \:=\:\tfrac{1}{3}\cdot\tfrac{3}{4} \:=\:\tfrac{1}{4}$

Since she won both games 21 times, we assume she played on 84 Fridays.

Place the information in a chart . . .

. . $\begin{array}{c||c|c||c|}
& \text{win B} & \text{lose B} & \text{Total} \\ \hline \hline
\text{win S} & 21 & & \\ \hline
\text{lose S} & & & \\ \hline\hline
\text{Total} & & & 84 \end{array}$

We know that she wins $\tfrac{1}{3}$ of her snooker games.
. . $\text{win S} \:=\:\tfrac{1}{3}(84) \:=\:28$

We know that she wins $\tfrac{3}{4}$ of her billiards games.
. . $\text{win B} \:=\:\tfrac{3}{4}(84) \:=\:63$

Enter that information in our chart:

. . $\begin{array}{c||c|c||c|}
& \text{win B} & \text{lose B} & \text{Total} \\ \hline \hline
\text{win S} & 21 & & 28\\ \hline
\text{lose S} & & & \\ \hline\hline
\text{Total} & 63 & & 84 \end{array}$

Now you can supply the missiing numbers and answer the question.

3. Thanks, that method is far clearer!