1. Probability question.

Hi could any one help me with the following question please.

A poll is carried out in the run up to a general election which give the following results:

Age Group---Conservatives---Labour---Liberal Democrats---Others

18-34--------32.7%----------30.7%-----------29.0%------7.6%
35-64--------35.5%----------28.9%-----------24.8%------10.8%
65+ ---------43.9%----------31.0%-----------16.0%------9.1%

The age groups represented are:

18-34-------35-64------65+

21.9%-------54.1%-----24.0%

(a) Find the probability that a randomly chosen voter who takes part in the poll but whose age is not know voted Conservative.
(b) Suppose the voter did vote Conservative, find the probability that he/she is at least 65 years old.

I think for (b) you would use the conditional probability formula but how would (a) be calculated?

2. Hello, axa121!

A poll is carried out prior to a general election which give the following results:

. . $\begin{array}{|c|c|c|c|c||c|}
\text{Ages} & \text{Cons.} & \text{Labor} & \text{Lib.Dem.} & \text{Others} & \text{Total} \\ \hline \hline

18-34 & 32.7\% & 30.7\% & 29.0\% & 7.6\% & 21.9\% \\ \hline
35-64 & 35.5\% & 28.9\% & 24.8\% & 10.8\% & 54.1\% \\ \hline

65+ & 43.9\% & 31.0\% & 16.0\% & 9.1\% & 24.0\% \\ \hline \end{array}$

(a) Find the probability that a randomly chosen voter who takes part in the poll
but whose age is not known voted Conservative.

A more complete table looks like this:

. . $\begin{array}{|c|c|c|c|c||c|}
\text{Ages} & \text{Cons.} & \text{Lab.} & \text{Lib.Dem.} & \text{Others} & \text{Total} \\ \hline \hline

\text{18-34} & 0.071613 & 0.007233 & 0.063510 & 0.016644 & 0.219 \\ \hline

\text{35-64} & 0.192055 & 0.156349 & 0.134168 & 0.058428 & 0.541 \\ \hline

\text{65+} & 0.105360 & 0.074400 & 0.038400 & 0.021840 & 0.240 \\ \hline\hline
\text{Total} & 0.369028 & 0.297982 & 0.236078 & 0.096912 & 1.000 \\ \hline
\end{array}$

$P(\text{Cons.}) \:=\:0.369028$

(b) Suppose the voter did vote Conservative.
Find the probability that he/she is at least 65 years old.

$P(\text{(65+)}\,|\,\text{(Cons)}) \;=\;\dfrac{P(\text{(65+)} \wedge \text{(Cons)})}{P(\text{Cons})} \;=\;\dfrac{0.105360}{0.369028} \;=\;0.285506791$

3. Ah thanks. I was trying something completely different.