SAMPLE PROBLEM PROBLEM SET - EXAM P/CAS 1

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1 SAMPLE PROBLEM SET - EXAM P/CAS 1 1 SAMPLE PROBLEM PROBLEM SET - EXAM P/CAS 1 1. A life insurer classifies insurance applicants according to the folloing attributes: Q - the applicant is male L - the applicant is a homeoner Out of a large number of applicants the insurer has identified the folloing information: 40% of applicants are male, 40% of applicants are homeoners and 20% of applicants are female homeoners. Find the percentage of applicants ho are male and do not on a home. A.1 B.2 C.3 D.4 E.5 2. A test for a disease correctly diagnoses a diseased person as having the disease ith probability.85. The test incorrectly diagnoses someone ithout the disease as having the disease ith a probability of.10. If 1% of the people in a population have the disease, hat is the chance that a person from this population ho tests positive for the disease actually has the disease? A!!& B!( C!(& D &!! E!!! 3. A class contains 8 boys and 7 girls. The teacher selects 3 of the children at random and ithout replacement. Calculate the probability that number of boys selected exceeds the number of girls selected. & &' $' A $$(& B '& C & D $$(& E '& B 4. \ is a continuous random variable ith density function 0ÐBÑ œ -/ ß B. Find TÒ\ $l\ Ó. $ A / B / C / D / / E / / 5. To players put one dollar into a pot. They decide to thro a pair of dice alternately. The first one ho thros a total of 5 on both dice ins the pot. Ho much should the player ho starts add to the pot to make this a fair game? A B C D E

2 2 SAMPLE PROBLEM SET - EXAM P/CAS 1 6. A carnival sharpshooter game charges $25 for 25 shots at a target. If the shooter hits the bullseye feer than 5 times then he gets no prize. If he hits the bullseye 5 times he gets back $10. For each additional bullseye over 5 he gets back an additional $5. The shooter estimates that he has a.2 probability of hitting the bullseye on any given shot. What is the shooter's expected gain if he plays the game (nearest $1? A & B! C & D! E & 7. Three individuals are running a one kilometer race. The completion time for each individual is a random variable. \ 3 is the completion time, in minutes, for person 3. \ À uniform distribution on the interval Ò ß $Ó \ À uniform distribution on the interval Ò( ß $Ó \ $ À uniform distribution on the interval Ò ß $$Ó The three completion times are independent of one another. Find the probability that the earliest completion time is less than 3 minutes. A.89 B.91 C.94 D.96 E Let \ and ] be continuous random variables ith joint density function BC for!ÿbÿ and!ÿcÿ \ 0ÐBßCÑ œ š. What is TÒ Ÿ ] Ÿ \Ó!ß otherise? $ $ $ A $ B C % D E % 9. Bob and Doug are both 100-metre sprinters. Bob's sprint time is normally distributed ith a mean of seconds and Doug's sprint time is also normally distributed, but ith a mean of 9.90 seconds. Both have the same standard deviation in sprint time of 5. Assuming that Bob and Doug have independent sprint times, and given that there is.& chance that Doug beats Bob in any given race, find 5. A.040 B.041 C.042 D.043 E A loss random variable is uniformly distributed on the integers from 0 to 11. An insurance pays the loss in excess of a deductible of 5.5. Find the expected amount not covered by the insurance. A 2 B 3 C 4 D 5 E 6

3 SAMPLE PROBLEM SET - EXAM P/CAS 1 3 SAMPLE PROBLEM PROBLEM SET SOLUTIONS 1. TÒQÓ œ %ß TÒQ Ó œ 'ß TÒLÓ œ %ß TÒL Ó œ 'ß TÒQ LÓ œ ß We ish to find TÒQ L Ó. From probability rules, e have ' œ T ÒL Ó œ T ÒQ L Ó T ÒQ L Ó, and ' œ T ÒQ Ó œ T ÒQ LÓ T ÒQ L Ó œ T ÒQ L Ó. Thus, T ÒQ L Ó œ % and then T ÒQ L Ó œ. The folloing diagram identifies the component probabilities. The calculations above can also be summarized in the folloing table. The events across the top of the table categorize individuals as male ( Q or female ( Q, and the events don the left side of the table categorize individuals as homeoners ( L or non-homeoners ( L. TÐQÑœ%, given TÐQÑœ %œ' TÐLÑ œ % TÐQ LÑ É TÐQ LÑ œ, given given TÐLÑœ %œ' Anser: B œtðlñ TÐQ LÑœ% œ TÐQ LÑœTÐQÑ TÐQ LÑœ% œ 2. We define the folloing events: H - a person has the disease, XT - a person tests positive for the disease. We are given TÒXTlHÓ œ & and T ÒX T lh Ó œ! and T ÒHÓ œ!. We ish to find T ÒHlX T Ó. Using the formulation for conditional probability e have TÒHlXTÓ œ TÒH XTÓ TÒXTÓ. But T ÒH X T Ó œ T ÒX T lhó T ÒHÓ œ Ð&ÑÐ!Ñ œ!!&, and T ÒH X T Ó œ T ÒX T lh Ó T ÒH Ó œ Ð!ÑÐÑ œ!. Then,!!& T ÒX T Ó œ T ÒH X T Ó T ÒH X T Ó œ!(& p T ÒHlX T Ó œ!(& œ!(.

4 4 SAMPLE PROBLEM SET - EXAM P/CAS 1 2. continued The folloing table summarizes the calculations. TÒHÓ œ!, given Ê TÒH Ó œ TÒHÓ œ TÒH XTÓ œ T ÒX T lhó T ÒHÓ œ!!& T ÒX T Ó œ T ÒH X T Ó T ÒH X T Ó œ!(& TÒH XTÓ œ T ÒX T lh Ó T ÒH Ó œ! TÒH XTÓ!!& T ÒHlX T Ó œ TÒXTÓ œ!(& œ!(. Anser: B & &x & % $ 3. There are Œ œ œ œ %&& ays of selecting $ children from a group of $ $x x $ & ithout replacement. The number of boys selected exceeds the number of girls selected if either (i $ boys and! girls are selected, or (ii boys and girl are selected. ( There are Œ Œ ays in hich selection (i can occur, and $! œ x (x $x &x!x (x œ &' ( there are Œ Œ ays in hich selection (ii can occur. œ x (x x 'x x 'x œ ' &' ' $' The probability of either (i or (ii occurring is %&& œ '& Anser: E T Ò \ $Ó 4. TÒ\ $l\ Óœ TÒ\ Ó TÒ\ Ó œ ' B -/.B œ -/ TÒ \ $Ó œ ' $ B $, -/.B œ -Ð/ / Ñß $ -Ð/ / Ñ TÒ\ $l\ Óœ -/ œ /. Note that e can find -, from œ ' 0ÐBÑ.B œ ' -/ B.B œ -/ p - œ / ; but this is not necessary for this exercise. Anser: A

5 SAMPLE PROBLEM SET - EXAM P/CAS Player 1 thros the dice on thros 1, 3, 5,... and the probability that player ins on thro 5 is Ð for 5 œ!ßßß$ß (there is a probability of throing a total of 5 on any 5 one thro of the pair of dice. The probability that player 1 ins the pot is % âœ œ Ñ Player 2 thros the dice on thros 2, 4, 6,.. The probability that player 2 ins the pot on thro 5 is Ð Ñ 5 for 5 œßß$ßand the probability that player 2 ins is 1 $ 1 & 1 1 â œ œ œ Ñ If player 1 puts - dollars into the pot, then his expected gain is -Ñ and player 2's expected gain is Ð -Ñ In order for the to players to have the same expected gain, e must have -Ñ œ!, so that -œ Anser: C 6. No. of bullseyes:! $ % & ' ( & Prize:!!!!!! &!! &\ & À &! &! &! &!! Let \œ number of bullseyes. \ has a binomial distribution ith 8œ&, :œ, and & B & B IÒ\Ó œ &. :ÐBÑ œ Š B ÐÑ ÐÑ Note that for 5 bullseyes or more the prize is &\ &. We can find the expected prize by first finding IÒ&\ &Ó and adjusting for the factors corresponding to \ œ!ß ß ß $ß % Therefore, Expected prize œ IÒ&\ &Ó & :Ð!Ñ! :ÐÑ & :ÐÑ! :Ð$Ñ &:Ð%Ñ &! & & % & $ œ &IÒ\Ó & & Š! ÐÑ ÐÑ! Š ÐÑÐÑ & Š ÐÑ ÐÑ & $ & % Ð!ÑŠ $ ÐÑ ÐÑ & Š % ÐÑ ÐÑ œ (. The expected gain is ( & œ &. 7. 0\ Ð>Ñœ œ& for Ÿ>Ÿ$ß J\ Ð>ÑœTÒ\ Ÿ>Óœ&Ð> Ñ for Ÿ>Ÿ$ 0\ Ð>Ñ œ œ & Ÿ > Ÿ $ ß J\ Ð>Ñ œ T Ò\ Ÿ >Ó œ &Ð> (Ñ ( Ÿ > Ÿ $ for 7 for 2 0\ $ Ð>Ñ œ % œ & for Ÿ > Ÿ $$ ß J\ $ Ð>Ñ œ T Ò\ $ Ÿ >Ó œ &Ð> Ñ for Ÿ > Ÿ $$ TÒ738Ð\ß\ß\Ñ $Óœ TÒ738Ð\ß\ß\Ñ $ $ $Ó œ TÒÐ\ $Ñ Ð\ $Ñ Ð\ $ $ÑÓ œ Ò J\ Ð$ÑÓ Ò J\ Ð$ÑÓ Ò J\ $ Ð$ÑÓ œ Ò &Ð$ ÑÓ Ò &Ð$ (ÑÓ Ò &Ð$ ÑÓ œ!'&. Anser: B

6 6 SAMPLE PROBLEM SET - EXAM P/CAS 1 8. The region of probability is shon in the shaded figure belo The probability is ' ' B ' ' $ $! BÎ BC.C.B BÎ BC.C.B œ $ $ œ. Alternatively, the probability is ' ' C! C BC.B.C œ œ $. Anser: D 9. F HµRÐß 5 F H! ÑTÒF HÓœTÒF H!ÓœTÒ Óœ& 5È 5È! Ê œ '%& T Ò^ '%&Ó œ & p œ!%$ 5 È (since 5. Anser: D 10. The amount not covered is Loss! $ % & ' ( â Amt.! $ % & && && â && Not Covered Prob. â The expected amount not covered by the insurance is Ð ÑÒ! $ % & &&Ð'ÑÓœ%. Anser: C

CREDIBILITY - PROBLEM SET 1 Limited Fluctuation Credibility

CREDIBILITY - PROBLEM SET 1 Limited Fluctuation Credibility CREDIBILITY PROBLEM SET 1 Limited Fluctuation Credibility 1 The criterion for the number of exposures needed for full credibility is changed from requiring \ to be ithin IÒ\Ó ith probability Þ*, to requiring