Too-Hard Probability Questions

Too-Hard Probability Questions 1. MATH 3 10 S7 A jar contains four m arbles: t hree red, on e w hit e. Tw o marbles are drawn w ith replacement. (i.e. A marble is randomly selected, the color noted, the marble replaced in the jar, then a second marble is drawn.) a. b. c. d. e. f. 2. a. b. d. f. List a sample space containing four outcomes. List a sample space with sixteen outcomes. Write the probability of each of the four outcomes in (a). What are the prob abilit ies of th e outcomes in (b)? What is t he probability the colors of the t w o marbles matc h? Wh at is the probabilit y the same marble is draw n t w ice? We are Let " H" Let " D" Let " A" playing w ith a short deck, as shown at right . be the event t he card drawn is a heart. be the event t he card draw n is a diamond. be t he event the card is an ace. P(H) = P(D) = P(H or D) = c. P(H and D) = e. Are H and D independent events? g. AÌ 2Ì 3Ì 4Ì AË 2Ë 3Ë 4Ë AÊ 2Ê 3Ê 4Ê AÍ 2Í 3Í 4Í P(A) = P(H or A) = P(H and A) = Are H and A independent events? 3. If three cards are draw n f rom the deck in #2 , on e at a time, w hat is t he probabil it y t hat a. the 1 st card is t he ace of heart s, t he 2 nd is t he 2 of diamonds, and the 3 rd is the 3 of c lubs? b. all t hree cards are aces? 4. An airpl ane is built to be able to f ly on one engine. If the plane' s t w o engines operat e independently, and each has a 1% chance of f ailing in any given f our-hour flight , w hat is t he chance t he plane w ill fail t o complete a four-hour flight to Oklahoma due t o engine f ailure? 5. A pair of fair, standard dic e are rolled. What is th e probabilit y t he sum of the dic e is 5? 6. Fifty marbles are to be drawn f rom the jar in problem #1 w ith replacement. If the f irst four marbles drawn are red, w hat is the probability the next marble draw n w ill not be red? 7. A probabilit y ex periment has f our possible outc omes: e1 , e2 , e3 , e4 . Th e out com e e1 is four tim es as likely as each of t he three remaining outcomes. Find the probability of e1 . 8. What are the odds in favor of rolling a sum of seven in one roll of a pair of fair st andard dice? 9. If P(A) = ½ and P(B) = ½ and P(B| A) = 1/3 , f ind: a. P(A and B) b. P(A or B) c. P(A| B) * 10. The deck of sixt een cards show n in #2 is t horoughly shuff led. Three cards are draw n from t he top of the deck, one at a time. What is t he probabil it y t he t hird card is an ace? (Hint: There is a really simple, direct solution.) * 11. “ The Birthday Problem” (famous) In a roomful of 3 0 people, what is the probabil it y t hat at least tw o people hav e the same birt hday ? Assume birt hday s are uniformly dist ributed and there is no leap year complication. (Hint: what is the probability that they all have different birthdays?) * 12. A 1 -inch-diameter coin is throw n on a table covered wit h a grid of lines tw o inches apart. What is th e probabilit y t he coin lands in a square w it hout touching any of the lin es of t he grid? (Hint: in order that the coin not touch any of the grid lines, w here must the cent er of the c oin be?) To o -Har d Pr obabilit y An s w e rs : 1 a. { RR, RW , W R, W W } 1 c . 9 /1 6 3 / 1 6 3 /1 6 1 / 1 6 respect ively S7 1b. { R1 R1 , R2 R1 , R3 R1 , W 1 R1 , R1 R2 , R2 R2 , R3 R2 , W 1 R2 , R1 R3 , R2 R3 , R3 R3 , W 1 R3 , R1 W 1 R2 W 1 R3 W 1 W 1W 1 } 1d. The ou tc om es det ailed in the sample space in 1 b are equally likely; each h as P = 1 / 1 6 . 1 e . P(c olor s m at c h) = P(RR ) + P(W W ) = 9 /1 6 + 1/16 = 10/16 or 5/8 1f . P(same m arble t w ice) = P(R1 R1 ,R2 R2 ,R3 R3 ,W 1 W 1 } = 4/16 (using 1 b; SS in 1 a is no help at all) ...or, you c an reason t hus: P(same marb le tw ice) = P(second m arble is same as t he f irst) = 1/4 because th ere are 4 m arbles in t he jar on t he seco nd dr aw , and only one is t he same m arble as t he 1 st . 2 a . P(H ) = P( Ì) = P(A Ì, 2 Ì, 3 Ì, 4 Ì} = 4 /1 6 ...or... P( Ì) = P(D ) = P( Ë) = P( Ì) = 1 /4 P(A) = P({A Ì, A Ë, A Ê, A Í} ) = 4 /1 6 = 1 /4 P( Ì, 1 of the 4 equal l y li kel y sui ts) = 1/ 4 2b. P(H or D) = P(H) + P(D) because the events 1 / 4 + 1 / 4 = 1 / 2 H and D are disjoint . 2c . P(H or A ) = P(H) + P(A) – P(H and A ) = 1/4 + 1/4 – 1/16 = 7/16 2d. P(H and D) = 0 2e. P(H & A ) = P(A Ì) = 1 / 1 6 (see 2b) 2 f . H & D are not indepen dent , t hey are mu tu ally e x c lu s iv e. If one occurs, the other cannot! 3. 2g . P(A Ì) = P(A) @P( Ì) ... so: y es, t hey are independent. A l so, P( Ì) = 4 / 1 6 = 1 / 4 = P( Ì| A ). Ì has same P if A. a . P( AÌ) P( 2 Ë | AÌ gone) P( 3 Ê | 2 Ë& AÌ gone) = (1/16) (1/15) (1/14) b. P( AA A) = (4/ 16 ) (3/ 15 ) (2/ 14 ) ... by reaso ning sim ilar to part a. .01 4. .99 E1 fails E1 OK .01 E2 fails ! E2 OK E2 fails E2 OK The plan e w ill fail t o m ake t he f light due t o engin e failu re only i f B OTH engi nes fail (because the plane can fl y on one engine. ) P(flight fails) = P(BOTH engines f ail) = P(1st f ails) @P( 2 nd f ai ls ) = . 0 0 0 1 5. P(s um = 5) = P(rol l in g 1 4 o r 2 3 o r 3 2 o r 4 1) = 4/36 = 1/9 6. Every t im e a m a rb le is ta k en fro m th i s j ar ( assumi ng previ ously draw n marbles are repl aced), th e prob abilit y of o b t aining a r ed m a rb le is 3 /4 . T h e re fo re, P(not red) = 1/4. 7. 4 p + p + p + p = 1 Y 7 p = 1 Y p = 1/7. Y P(e 1 ) = 4p = 4(1/7) = 4/7 8. Th ere ar e six wa y s to ro ll a s u m o f 7 : 1 6 , 25 , 34, 43 , 52, 61 . P(sum = 7) = 6/36 or 1/6 (not t he quest ion! ) There are six fav orable ou tc om es in t his SS w ith 36 equally likely o ut com es, so 2 9 are u nf avorab le. The odds in f av o r o f a s u m o f 7 a re 6 :2 9 (Bec aus e t hey a re 2 9 :6 a g a in s t....) 9. a. P(A and B) = P(A) P(B| A) = (½ ) (a) = b. P(A or B) = P(A ) + P(B) – P(A an d B ) = c. P(A and B) P(A| B) = )))))))))) P(B) = 1 / )))6 = a 1 /6 ½ + ½ – 1 /6 = 5 /6 ½ W e not e th at A & B are NOT ind ependen t. P(A| B) P(A) (show i ng B has an effect on A !) ¼ = P(A) CP(B) Also P(A and B) = P(A) P(B| A) = 1 / 6 1 1 . It is dif fic ult to calcu late d irect ly t he ch ance o f at least t w o m atc hing b irth days, becau se you have to allow for so many possibilities: just tw o matching, th ree matchi ng, tw o pai rs matching, etc. etc . Th e COM PLEM ENT of th is ev en t is , ho w ev er , qu it e s im pl e. If t he re ar e N OT at le as t t w o m at chi ng bi rt hd ay s, t he n t he re ar e N ON E! P(all differen t) = 36 5 3 64 36 3 ))) ))) ))) 365 365 365 C C C 336 ))) 365 (H ere it i s appropri ate to use a cal culator– caref ully. ) This tu rns out to b e under 30 % . Therefore, the pro bability that at least t w o birth days mat ch is over 7 0% ! 1 2 . Wh ere does the coin have to land in order to w in? Wh at d eterm ines t he loc atio n of th e coin ? Wh ere mu st the center of the coin be? Draw a picture of w here it can be. The answer is one-fourth.