Question

when polled, 50.08% of voters reported their intention to vote for polemic party, while 49.92% reported that they will vote for team theory. however, these values are not accurate, because some of the voters lied in the poll. it is known that 6% of team theory supporters are liars (if asked, they say they support polemic party), and that 18% of polemic party are liars (they claim that they support team theory). a) let p be the probability that a voter selected at random is a supporter of polemic party, and let q be the probability that a randomly selected polemic party supporter (someone who intends to vote for polemic party) is a liar. express the probability that a person is both a liar and a polemic party supporter in terms of p and q . probability that a voter is both a liar and a polemic party supporter

Answer 1

The probability that a person is both a liar and a supporter of the polemic party can be calculated by multiplying the probability of being a polemic party supporter by the probability of being a liar.

Step-by-step explanation:

Let's define p as the probability that a randomly selected voter is a supporter of the polemic party. We are given that 50.08% of voters reported their intention to vote for the polemic party, so p = 0.5008.

Next, let's define q as the probability that a randomly selected polemic party supporter is a liar. We are given that 18% of polemic party supporters are liars, so q = 0.18.

To find the probability that a person is both a liar and a polemic party supporter, we can multiply the probabilities p and q: p * q = 0.5008 * 0.18 = 0.090144.

Therefore, the probability that a voter is both a liar and a polemic party supporter is 0.090144, or 9.01%.