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MAT 230 EXAM TWOThis document is proprietary to Southern New Hampshire University. It and the problems withinmay not be posted on any non-SNHU website.Keith Ellison1
Directions: Type your solutions into this document and be sure to show all steps for arriving atyour solution. Just giving a final number may not receive full credit.Problem 1This question has 2 parts.Part 1:Suppose thatFandXare events from a common sample space withP(F)6= 0 andP(X)6= 0.(a) Prove thatP(X) =P(X|F)P(F) +P(X|¯F)P(¯F).Hint:Explain whyP(X|F)P(F) =P(X∩F) is another way of writing the definition of conditional probability, and then usethat with the logic from the proof of Theorem 4.1.1.From the conditional probability,P(X—F)=P(XF)/P(F) P(X—F)P(F)=P(XF) P(X—F)=P(XF’)/P(F) P(X—F)P(F)=P(XF’)P(X)=P(X) P(X)=P(X(FF)) =P((XF)(XF))the probability is independent,P((XF)(XF))=P(XF)+P(XF) =P(X)P(F)+P(X)P(F) P(X)=P(XF)+P(XF)(b) Explain whyP(F|X) =P(X|F)P(F)/P(X) is another way of stating Theorem 4.2.1 Bayes’Theorem.From the conditional density function,P(F—X)=P(XF)/P(X)P(X—F)P(F)=P(XF) Used in the conditional equation, P(F—X)=P(XF)/P(X) P(F—X)=P(X—F)P(F)/P(X)Part 2:A website reports that 70% of its users are from outside a certain country. Out of their usersfrom outside the country, 60% of them log on every day. Out of their users from inside the country,80% of them log on every day.(a) What percent of all users log on every day? Hint: Use the equation from Part 1 (a).Use ’X’ as the total percentage of users that log on everyday F is the percentage of usersfrom outside the country who log on everyday FFbe the percentage of users from insidethe country logging on everyday. The probability of the total percentage of users logging oneveryday; P(X)=100xProbability of users log from outside the country; P(F)=70xProbability of users log from inside the country; P(F)=30xProbability of users log from outside the country everyday; P(X—F)P(F)=60/100(70x)=42xProbability of users log from inside the country everyday, P(X—F)P(F)=80/100(30x)=24xTotal probability of users logging on everyday is given by using the proof of Part 1(a),P(X)=P(X—F)P(F)+P(X—F)P(F) P(X)=42x+24x=66x P(X)=66x.