91Ó°ÊÓ

What percent of voters participated in the local election?

Consider events:

I - a randomly chosen person is an independent

L - a randomly chosen person is a Liberal

C - a randomly chosen person is a conservative

E - a randomly chosen person participated in the elections.

Given probabilities,

$P\left(I\right)=0.46$

$P\left(L\right)=0.30$

$P\left(C\right)=0.24$

$P(E/I)=0.35$

$P(E/L)=0.62$

$P(E/C)=0.58$

$\text{a)}P(Iâˆ\pounds E)=\text{? b)}P(Lâˆ\pounds E)=\text{?c)}P(Câˆ\pounds E)=\text{? d}P\left(E\right)=\text{?}$

Calculate (d) first.

$I,L\text{, and}C\text{are mutually exclusive (competing hypothesis), therefore:}$

$P\left(E\right)=P(Eâˆ\pounds I)P\left(I\right)+P(Eâˆ\pounds L)P\left(L\right)+P(Eâˆ\pounds C)P\left(C\right)$

$=0.35â‹\dots 0.46+0.62â‹\dots 0.30+0.58â‹\dots 0.24$

$=0.4862$

Percent of voters who participated in the local election is$0.4862$$0.4862$

what is the probability that he or she is an Independent?

By using the definition of conditional probability, and transforming it to obtain $P\left(IE\right)=P(E/I)P\left(I\right)$

$P(Iâˆ\pounds E)=\frac{P\left(IE\right)}{P\left(E\right)}$

$=\frac{P(Eâˆ\pounds I)â‹\dots P\left(I\right)}{P\left(E\right)}$

$=\frac{0.35â‹\dots 0.46}{0.4862}$

$=0.3311$

The probability that he or she is Independent will be$0.3311$

what is the probability that he or she a Liberal?

Similarly

$P(Lâˆ\pounds E)=\frac{P\left(LE\right)}{P\left(E\right)}$

$=\frac{P(Eâˆ\pounds L)â‹\dots P\left(L\right)}{P\left(E\right)}$

$\frac{0.62.0.30}{0.4862}$

$0.3826$

The probability that he or she a Liberal is$0.3826$

what is the probability that he or she is a Conservative?

Similarly for (c ),

$P(Câˆ\pounds E)=\frac{P\left(CE\right)}{P\left(E\right)}$

$=\frac{P(Eâˆ\pounds C)â‹\dots P\left(C\right)}{P\left(E\right)}$

$=\frac{0.58â‹\dots 0.24}{0.4862}$

$0.2863$

The probability that he or she is a Conservative is$0.2863$