Exam 1 Practice Questions I solutions, 18.05, Spring 2014

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1 Exam Practice Questions I solutions, 8.5, Spring 4 Note: This is a set of practice problems for exam. The actual exam will be much shorter.. Sort the letters: A BB II L O P R T Y. There are letters in all. We build arrangements by starting with slots and placing the letters in these slots, e.g A B I B I L O P R T Y Create an arrangement in stages and count the number ( ) of possibilities at each stage: Stage : Choose one of the slots to put the A: ( ) Stage : Choose two of the remaining slots to put the B s: ( ) 8 Stage : Choose two of the remaining 8 slots to put the B s: ( ) 6 Stage 4: Choose one of the remaining 6 slots to put the L: ( ) 5 Stage 5: Choose one of the remaining 5 slots to put the O: ( ) 4 Stage 6: Choose one of the remaining 4 slots to put the P: ( ) Stage 7: Choose one of the remaining slots to put the R: ( ) Stage 8: Choose one of the remaining slots to put the T: ( ) Stage 9: Use the last slot for the Y: Number of arrangements: ( )( )( )( )( )( )( )( )( ) Note: choosing out of is so ( simple ) we could have immediately written instead of belaboring the issue by writing. We wrote it this way to show one systematic way to think about problems like this.. Build the pairings in stages ( ) and count the ways to build each stage: 6 Stage : Choose the 4 men:. 4( ) 7 Stage : Choose the 4 women: 4 We need to be careful because we don t want to build the same 4 couples in multiple ways. Line up the 4 men M, M, M, M 4 Stage : Choose a partner from the 4 women for M : 4. Stage 4: Choose a partner from the remaining women for M : Stage 5: Choose a partner from the remaining women for M : Stage 6: Pair the last women with M 4 :

2 Exam Practice I, Spring 4 ( )( ) 6 7 Number of possible pairings: 4!. 4 4 Note: we could have done stages -6 in on go as: Stages -6: Arrange the 4 women opposite the 4 men: 4! ways.. We are given P (A c \ B c )/ and asked to find P (A [ B). A c \ B c (A [ B) c ) P (A [ B) P (A c \ B c ) /. 4. D is the disjoint union of D \ C and D \ C c. So, P (D \ C)+P (D \ C c )P (D) ) P (D \ C c)p (D) P (D \ C) (We never use P (C).5.) C D D C c The following tree shows the setting p p Know Guess /c /c Correct Wrong Correct Wrong Let C be the event that you answer the question correctly. Let K be the event that you actually know the answer. The left circled node shows P (K \ C) p. Both circled nodes together show P (C) p + ( p)/c). So, P ( K jc) P (K \ C) P (C) p p + ( p)/c Or we could use the algebraic form of Bayes theorem and the law of total probability: Let G stand for the event that you re guessing. Then we have, P (CjK),P (K) p, P (C) P (CjK)P (K)+P (CjG)P (G) p + ( p)/c. So, P (Kj C) P (C jk)p (K) P (C) p p + ( p)/c 6. Sample space Ωf(, { ), (, ), (, ),..., (6, 6) g } { f(i, j) j i, j,,, 4, 5, 6 g. } (Each outcome is equally likely, with probability /6.) A { f(, ), (, )g, } B f(, { 6), (, 5), (, 4), (4, ), (5, ), (6, )g } C { f(, ), (, ), (, ), (, 4), (, 5), (, 6), (, ), (, ), (4, ), (5, ), (6, )g} P (A \ C) (a) P ( AjC) /6 P (C) /6..

3 Exam Practice I, Spring 4 P (B \ C) (a) P (BjC) P (C) /6 /6.. (c) P (A) /6 6 P (AjC), so they are not independent. Similarly, P (B) 6/6 6 P (BjC), so they are not independent. 7. We show the probabilities in a tree: / /4 /4 Know Eliminate Total guess / / /4 /4 Correct Wrong Correct Wrong Correct Wrong For a given problem let C be the event the student gets the problem correct and K the event the student knows the answer. The question asks for P (KjC). We ll compute this using Bayes rule: P (KjC) P (C jk) P (K) / P (C) / + / + / % 8. We have P (A [ B).4.58 and we know because of the inclusion-exclusion principle that P (A [ B) P (A) + P (B) P (A \ B). Thus, P (A \ B) P (A) + P (B) P (A [ B) (.4)(.) P (A)P (B) So A and B are independent. 9. We will make use of the formula Var(Y ) E(Y ) E(Y ). First we compute E[X ] x xdx Thus, and E[X ] x xdx E[X 4 ] x 4 xdx. Var(X) E[X 4 ] (E[X]) 9 8 Var(X ) E[X 4 ] ( E[X ] ) 4.

4 Exam Practice I, Spring 4 4. Make a table X: prob: (-p) p X. From the table, E(X) ( p) + p p. Since X and X have the same table E(X ) E(X) p. Therefore, Var(X) p p p( p).. Let X be the number of people who get their own hat. Following the hint: let X j represent whether person j gets their own hat. That is, X j if person j gets their hat and if not. We have, X X j, so E(X) E(X j ). j j Since person j is equally likely to get any hat, we have P (X j ) /. Thus, X j Bernoulli(/) ) E(X j ) / ) E(X).. For y,, 4,..., n, P (Y y) P (X y ( ) ( ) n n ) y/.. The CDF for R is r r F R (r) P (R r) e u du e u e r. Next, we find the CDF of T. T takes values in (, ). For < t, F T (t) P (T t) P (/R < t) P (/t > R) F R (/t) e /t. d We differentiate to get f T (t) (e /t) e dt 4. The jumps in the distribution function are at,, 4. The value of p(a) at a jump is the height of the jump: a 4 p(a) /5 /5 /5 5. We compute 5 P (X 5) P (X < 5) λe λx dx ( e 5λ ) e 5λ. (b) We want P (X 5jX ). First observe that P (X 5, X ) P (X 5). From similar computations in (a), we know t /t P (X 5) e 5λ P (X ) e λ.

5 Exam Practice I, Spring 4 5 From the definition of conditional probability, P (X 5 X ) P (X 5, X ) P (X ) P (X 5) e 5λ P (X ) Note: This is an illustration of the memorylessness property of the exponential distribution. 6. Transforming Normal Distributions (a) Note, Y follows what is called a log-normal distribution. F Y (a) P (Y a) P (e Z a) P (Z ln(a)) Φ(ln(a)). Differentiating using the chain rule: d f y (a) F Y (a) d da da Φ(ln(a)) a φ(ln(a)) π a e (ln(a)) /. (b) (i) The. quantile for Z is the value q. such that P (Z q. ).. That is, we want Φ(q. ). q. Φ (.). (ii) We want to find q.9 where F Y (q.9 ).9 Φ(ln(q.9 )).9 q.9 e Φ (.9). (iii) As in (ii) q.5 e Φ (.5) e. 7. (a) Total probability must be, so 4 f(x, y) dy dx c(x y + x y ) dy dx c, (Here we skipped showing the arithmetic of the integration) Therefore, c. 4 (b) P ( X, Y ) f(x, y) dy dx (c) For a and b. we have F (a, b) 5 c 8 c(x y + x y ) dy dx a b ( a b f(x, y)dy dx c 6 + a b ) 9

6 Exam Practice I, Spring 4 6 (d) Since y is the maximum value for Y, we have ( ) 9a F X (a) F (a, ) c 6 + a 9 c a a 7 (e) For x, we have, by integrating over the entire range for y, f X (x) ( ) f(x, y) dy cx + c 7 x 9 x. This is consistent with (c) because d (x /7) x /9. dx (f) Since f(x, y) separates into a product as a function of x times a function of y we know X and Y are independent. 8. (Central Limit Theorem) Let T X + X X 8. The central limit theorem says that T N(8 5, 8 4) N(45, 8 ) Standardizing we have ( ) T P (T > 69) P > 8 8 P (Z > ).975 The value of.975 comes from the rule-of-thumb that P (jzj < ).95. A more exact value (using R) is P (Z > ).977.

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