Question

A total of 46% of voters in a certain city classify themselves as Independents; 30% as Liberals and 24% as Conservatives. In a recent election, 35% of the Independents, 62% of the Liberals, and 58% of the conservatives voted. A voter is chosen at random. Given that this person voted, what is the probability that he or she is

(a) an Independent?

(b) a Liberal?

(c) an Conservative?

(d) What fraction of the voters participated in the election?

Answer 1

Given that:
solution - given that =- Mobability of independents PCI) - 0.46 (467) Probability of liberal P(L) = 30% => 0.3 pcc) = 24% =) 0.24 probability of lcoted independents who P(VII) = 35% PCV/1=0.35 Probability 06 liberal who loved PC VIL)- 62% = 0.62 probability of conscicabiver who pcvlc) - 587, voted @ an Independent let us consider, probability of Voled people to be independene- realy) - report P(Ilv). P(I) P(VII) P(V)

Where, Exclusive $ : I,L,C core Exhaustive mutually Events Therefore, PCV) = P(VII). PCI)+ POUILD PCL) + PC VIC) PCC - 0.35 (0.96) +062 (0.3) + 0.58 (0.24) = 0.161 + 0.186+ 0.1392 | PCV) 0.4662 | * PCIlv) = P(1) P(Vlo) PCI) 0.3560.46) 0.4862 hits 0.161 0.4862 = 0.331139449 PC Ill) ♡ 0.3341

a liberal Probability 06 Koted people to be liberal PCLIV) = PCL) PCVIL) P(V) PCLLV) = 0.62 (0.3) 0.4862 0.186 0.4862 0.382558618 PCLW) 0.3826 © an conservative pobability of Koted people to be conservative Pcclv) PC(lv) = PCC) PCVIC) P(V)

= 0.58 C0.24) 0.4862 = 0.1392 0.4862 = 0.286301933 P(clv)a 0.2863 participated @ what fraction of the voters in the election ? Fraction of lotee pas ticipaled Election, PCV) in the Docal PCV) = PCVII) PCI) + P(VILO PCL) + PCvic) pce = 0.3560.46) + 0.62 (0.3) + 0.58 C0.24) = 0.161 + 0.186 + 0.1392 PCV) = 0.4862 ( Thank you.