1-3 조건부확률

1.3-1. A common test for AIDS is called the ELISA (enzyme-linked immunosorbent assay) test. Among 1

million people who are given the ELISA test, we can expect results similar to those given in the following table:

 

If one of these 1 million people is selected randomly, find the following probabilities: (a) P(B1), (b) P(A1), (c)

P(A1 |B2), (d) P(B1 |A1). (e) In words, what do parts (c) and (d) say?

 

(a)

P(B1) = 5000/1,000,000

 

(b)

P(A1) = 78,515/1,000,000

 

(c)

P(A1|B1) = P(A1 ∩ B1) / P(B1) = 4,885 /5000

 

(d)

P(B1|A1) = P(A1 ∩ B1) / P(A1) = 4,885/ 78,515

 

(e)

-> 표본공간(분모) 부분이 다르다.

 

1.3-2. The following table classifies 1456 people by their gender and by whether or not they favor a gun law.

Compute the following probabilities if one of these 1456 persons is selected randomly: (a) P(A1), (b) P(A1 | S1), (c)

P(A1 | S2). (d) Interpret your answers to parts (b) and (c).

 

(a)

P(A1) = 1041/1456

 

(b)

P(A1|S1) = P(A1 ∩ S1) / P(S1) = 392/633

 

(c)

P(A1|S2) = P(A1 ∩ S2) /P(S2) = 649/823

 

(d) A1이 어떤 사건과 함께 일어나는지, 그리고 표본공간에 따라 값이 바뀐다.

 

1.3-3. Let A1 and A2 be the events that a person is lefteye dominant or right-eye dominant, respectively. When a

person folds his or her hands, let B1 and B2 be the events that the left thumb and right thumb, respectively, are on

top. A survey in one statistics class yielded the following table:

 

 

If a student is selected randomly, find the following probabilities: 

(a) P(A1 ∩ B1),

(b) P(A1 ∪ B1),

(c) P(A1 |B1),

(d) P(B2 |A2).

(e) If the students had their hands folded and you hoped to select a right-eye-dominant student,

would you select a “right thumb on top” or a “left thumb on top” student?Why?

 

(a)

P(A1 ∩ B1) = 5/35

(b) P(A1 ∪ B1) = 19 + 12 -5 /35

(c) P(A1 |B1) = P(A1 ∩ B1)/P(B1) = 5/19

(d) P(B2|A2) = P(B2 ∩ A2) /P(A2) = 9/23

(e) left, probability is bigger

 

1.3-4. Two cards are drawn successively and without replacement from an ordinary deck of playing cards.

Compute the probability of drawing 

(a) Two hearts.

(b) A heart on the first draw and a club on the second draw.

(c) A heart on the first draw and an ace on the second draw.

Hint: In part (c), note that a heart can be drawn by getting the ace of hearts or one of the other 12 hearts.

 

(a) 13/52 x 12/51

(b) 1/4 x 12/51

(c) 12/52 x 4/51 + 1/52 x 3/51 (하트 에이스 외에 하트를 뽑을 확률 + 하트에이스 뽑고 에이스 뽑을 확률)

 

1.3-5. Suppose that the alleles for eye color for a certain male fruit fly are (R,W) and the alleles for eye color

for the mating female fruit fly are (R,W), where R and W represent red and white, respectively. Their offspring

receive one allele for eye color from each parent.

(a) Define the sample space of the alleles for eye color for the offspring.

(b) Assume that each of the four possible outcomes has equal probability. If an offspring ends up with either

two white alleles or one red and one white allele for eye color, its eyes will look white. Given that an

offspring’s eyes look white, what is the conditional probability that it has two white alleles for eye color?

 

(a) {RR,RW,WR,WW}

(b) 1/3

P(모두 흰색 |자식 흰색) = P(WW) / P(WW) + P(WR) +P(RW) = 1/3

 

1.3-6. A researcher finds that, of 982 men who died in 2002, 221 died from some heart disease. Also, of the 982

men, 334 had at least one parent who had some heart disease. Of the latter 334 men, 111 died from some heart

disease. A man is selected from the group of 982. Given that neither of his parents had some heart disease, find the

conditional probability that this man died of some heart disease.

 

 

 

110/(982-334)

 

 

1.3-7. An urn contains four colored balls: two orange and two blue. Two balls are selected at random without

replacement, and you are told that at least one of them is orange. What is the probability that the other ball is also

orange?

 

4개의 공(2개 오렌지, 2개 파랑)에서 

1번쨰 오렌지 2번째 오렌지 나올 확률 = 2/4 x 1/3 =1/6

1번째 오렌지 2번째 파랑 나올 확률 = 2/4 x 2/3 = 2/6

1번쨰 파랑 2번째 오렌지 나올 확률 = 2/4 x 2/3 =2/6

1번쨰 파랑 2번째 파랑 나올 확률 = 2/4 x 1/3 =1/6

 

P(둘 중 하나는 오렌지) = (1/6) + (2/6) + (2/6) = 5/6

P(모두오렌지가 나올 확률) = 1/6

P (모두 오렌지/ 하나가 오렌지 일 경우) = 1/6 / 5/6 =1/5

 

1.3-8. An urn contains 17 balls marked LOSE and 3 balls marked WIN. You and an opponent take turns selecting a

single ball at random from the urn without replacement. The person who selects the third WIN ball wins the game.

It does not matter who selected the first two WIN balls.

(a) If you draw first, find the probability that you win the game on your second draw.

(b) If you draw first, find the probability that your opponent wins the game on his second draw.

(c) If you draw first, what is the probability that you win? Hint: You could win on your second, third, fourth, . . . , or tenth draw, but not on your first.

(d) Would you prefer to draw first or second?Why?

 

(a)

3/20 x 2/19 x 1/18

 

(b)

{나, 상대, 나, 상대} 이런식으로 뽑는다면, 상대가 2번째에 뽑아서 이기려면 4번째에 win이 나와야 한다.

그렇다면 4번째는 win으로 고정하고, 1~3번째의 경우의 수를 센다.

{win, win, lose}가 임의의 순서대로 나오면 된다.

\(\frac{ {_{3}\mathrm{C}_2 \times  _{17}\mathrm{C}_{1}}{ _{20}\mathrm{C}_{3}}\)

(비복원추출은 조합의 경우로 나타낼 수 있다. )

그리고 남은 카드 17개 중의 1개(win은 1개만 남았다)인 확률 1/17을 곱한다.

 

 

(c)

자신이 처음에 뽑는다면, 홀수 번째에 또 카드를 뽑는다. 그렇다면 홀수번째에 승리할 수 있는 기회가 온다.

3번째에 승리할 확률 {win, win, win} 

\(\frac{1}{_{20}\mathrm{C}_{2}} \times \frac{1}{18}\)

-> \(\frac{1}{_{20}\mathrm{C}_{2}} \times \frac{1}{18}\) 

5번째에 승리할 확률 {{win, win, lose, lose}, win} 마지막의 win은 고정한 체 앞의 4개만 랜덤하게 돌린다. 

7번째에 승리할 확률:{{win, win, lose, lose,lose,lose}, win} 

...

위와 같은 방법을 계속한다.

 

계산이 너무 많아서 생략한다..

답은 0.4605 이라고 한다.

 

 

(d) 두번쨰가 좋다

1-0.4605(첫번쨰 이길확률, (c)에서 구함)

 

1.3-9. An urn contains four balls numbered 1 through 4. The balls are selected one at a time without replacement.

A match occurs if the ball numbered m is the mth ball selected. Let the event A_i denote a match on the ith draw,

 

 

 

   

(a)

공이 1,2,3,4가 있다. 이를 나열하는 갯수는 4!, 분모는 4!이 된다.

{1,{2, 3, 4}} 라고 하면, 1이 첫번째에 고정되고, 나머지 3개만 순서를 바꿀 경우의 수는 3!이다.

마찬가지로 {1, {2}, 3, 4} 2는 2번째에 고정되고 나머지만 순서를 바뀔 확률도 3!이므로

각 i에 대하여 3!/4!이 성립한다.

 

(b)

{1, {2,3,4}} = 3!/4!이었다. 1이 고정된 경우였다.

1이 고정된 것이 일어난 후에 2도 고정될 확률은 {1, {{2}, 3,4}}이므로 3,4만 순서를 바꾼다. 이는 2!이 된다.

P(A_1 ∩ A_2) = 2!/4!이 된다.

 

P(A_1 ∩ A_2) = P(A_1)P(A_2|A_1) = 2!/3! x 3!/4! = 2!/4!

 

(c) 

P(A_1 ∩ A_2 ∩ A_3) = P(A_1) x P(A_2|A_1) x P(A_3|A_1 ∩ A_2)

3!/4! x 2!/3! x 1!/2! = 1/4!

 

(d)

P(A_1) + P(A_2) +P(C) + P(D) - P(A_1 ∩ A_2) - P(A_1 ∩ A_3) - P(A_1 ∩ A_4) - P(A_2 ∩ A_3) - P(A_2 ∩ A_4) - P(A_3 ∩ A_4) + P(A_1 ∩ A_2 ∩ A_3) + P(A_1 ∩ A_2 ∩ A_4) + P(A_2 ∩ A_3 ∩ A_4) -  P(A_1 ∩ A_2  ∩ A_3 ∩ A_4 )

 

각각의 확률은 등확률이므로

3!/4! x 4 - 2!/4! x 6 + 1!4! x4 -1/4!

= 1 - 1/2 - 1/3! - 1/4!

 

(e)

모르겠다

 

1.3-10. A single card is drawn at random from each of six well-shuffled decks of playing cards. Let A be the event

that all six cards drawn are different. 

(a) Find P(A).

(b) Find the probability that at least two of the drawn cards match.

 

(a)

P(A) = 1 x 51/52 x 50/52 x 49/52 x 48x/52 x 47/52

(b) 1- P(A)

 

1.3-11. Consider the birthdays of the students in a class of size r. Assume that the year consists of 365 days.

(a) How many different ordered samples of birthdays are possible (r in sample) allowing repetitions (with

replacement)?

(b) The same as part (a), except requiring that all the students have different birthdays (without replacement)?

(c) If we can assume that each ordered outcome in part (a) has the same probability, what is the probability

that at least two students have the same birthday?

(d) For what value of r is the probability in part (c) about equal to 1/2? Is this number surprisingly small? Hint: Use a calculator or computer to find r.

 

(a) \(365^r\)

-> 모두 365개 중에 하나를 가질 수 있다.

(b) \(_{365}\mathrm{P}_{r}\)

-> 비복원추출이므로 앞에 나온 것은 못한다.

(처음 365, 두번째 364, -> 365 x 364 x 363...)

(c) 1-P(A) = \(1-\frac{_{365}\mathrm{P}_{r}}{365^r}\)

P(A)는 전체가 모두 다를 확률

(d) \(1-\frac{_{365}\mathrm{P}_{r}}{365^r} = \frac{1}{2}\) 를 계산

계산기나 컴퓨터를 사용하라고 한다. 코드 짜보자.

 

순열계산코드

def soon(n,r):
    answer=1
    for i in range(365,365-r,-1):
        answer*=i
    return answer

for i in range(365):
    a=soon(365,i)/365**i
    if(a<0.5):
        print(i,a)
        break

23 0.4927027656760146

 

-> 답지에 있는 23과 비슷하게 나온다.

 

1.3-12. You are a member of a class of 18 students. A bowl contains 18 chips: 1 blue and 17 red. Each student
is to take 1 chip from the bowl without replacement. The student who draws the blue chip is guaranteed an A for
the course.

(a) If you have a choice of drawing first, fifth, or last, which position would you choose? Justify your choice
on the basis of probability.
(b) Suppose the bowl contains 2 blue and 16 red chips.What position would you now choose?

 

(a)

첫번째에 뽑을 확률 1/18

다섯번째에 뽑을 확률 = 17/18 x 16/17 x 15/16 x 14/15 x 1/14

마지막에 뽑을 확률 = 17/18 x 16/17 x 15/16 x 14/15 x .... x 2/3 x 1/2 x 1

 

모두 1/18이므로 상관없다.

 

(b)

위와 같은 원리로 1/9가 된다. 모두 똑같다

 

 

1.3-13. In the gambling game “craps,” a pair of dice is rolled and the outcome of the experiment is the sum of
the points on the up sides of the six-sided dice. The bettor wins on the first roll if the sum is 7 or 11. The bettor loses
on the first roll if the sum is 2, 3, or 12. If the sum is 4, 5, 6, 8, 9, or l0, that number is called the bettor’s “point.” Once
the point is established, the rule is as follows: If the bettor rolls a 7 before the point, the bettor loses; but if the point
is rolled before a 7, the bettor wins.

(a) List the 36 outcomes in the sample space for the roll of a pair of dice. Assume that each of them has a
probability of 1/36.
(b) Find the probability that the bettor wins on the first  roll. That is, find the probability of rolling a 7 or 11,
P(7 or 11).
(c) Given that 8 is the outcome on the first roll, find the probability that the bettor now rolls the point 8 before
rolling a 7 and thus wins. Note that at this stage in the game the only outcomes of interest are 7 and 8. Thus
find P(8 |7 or 8).
(d) The probability that a bettor rolls an 8 on the first roll and then wins is given by P(8)P(8 | 7 or 8). Show that
this probability is (5/36)(5/11).

(e) Show that the total probability that a bettor wins in the game of craps is 0.49293. Hint: Note that the bettor
can win in one of several mutually exclusive ways:by rolling a 7 or an 11 on the first roll or by establishing
one of the points 4, 5, 6, 8, 9, or 10 on the first roll and then obtaining that point on successive rolls
before a 7 comes up.

 

(a) {(1,1), (1,2),(2,1), (1,3), (3,1), (1,4), (4,1) ... (6,6)}

(b)

7이 나올 경우의 수:(1,6),(2,4),(3,3), (4,2), (6,1) 

11이 나올 경우의 수: (5,6), (6,5) 

-> 8/36

(c)

P(8) =( 2,6), (3,5), (5,3), ( 4,4), (6,2) ->5가지

P(7) = (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)  ->6가지

P(8 or 7) = 11가지

-> 5/11(5+6) = P(8)/( P(8) + P(7))

(d) 7,8이 나오고 8이 연속해서 나와야 한다. 이는 곱의 정리에 의해서 곱셈으로 나타낼 수 있다.

(e) 

 

1.3-14. Paper is often tested for “burst strength” and “tear strength.” Say we classify these strengths as low,
middle, and high. Then, after examining 100 pieces of paper, we find the following:

If we select one of the pieces at random, what are the probabilities that it has the following characteristics:
(a) A1?
(b) A3 ∩ B2?
(c) A2 ∪ B3?
(d) A1, given that it is B2?
(e) B1, given that it is A3?

 

(a) 7+11+12 / (7+11+12+11+21+9+13+9+7) =0.3

(b) 9/ (7+11+12+11+21+9+13+9+7) =0.09

(c) (11+21+9+12+7)/ (7+11+12+11+21+9+13+9+7) =0.60

(d) P(A1|B2) = P(A1 ∩  B2) /P(B2) = 11/(11+21+9) =0.2682926829268293

(e) P(B1|A3) = P(B1 ∩ A3) /P(A3) = 13/(13+9+7) = 0.4482758620689655

 

1.3-15. An urn contains eight red and seven blue balls. A second urn contains an unknown number of red ballsff
and nine blue balls. A ball is drawn from each urn at random, and the probability of getting two balls of the same
color is 151/300. How many red balls are in the second urn?

 

첫번째 상자

P(빨강_2) = 8/15

P(파랑_2) = 7/15

 

두번째 상자

P(빨강_2) = x - 9 / x

P(파랑_2) = 9/x

x= 전체 갯수

 

P(빨강_1) x P(빨강_2) + P(파랑_1) x P(파랑_2) = 151/300

\(\frac{8}{15} \times \frac{x-9}{x} + \frac{9}{x} \times \frac{7}{15} = \frac{151}{300}\)

\(8 \times \frac{x-9}{x} + 9x \times 7 = \frac{151}{20}\)

\(20 \times (x-9) + 20 \times 9x \times 7 = 151x\)

\( 9x = 180\)

\( x=20\)

전체 갯수는 20개, 파란 공은 9개, 빨간공은 11개이다.

 

1.3-16. Bowl A contains three red and two white chips, and bowl B contains four red and three white chips. A
chip is drawn at random from bowl A and transferred to bowl B. Compute the probability of then drawing a red
chip from bowl B.

 

A에서 빨강을 뽑을 확률 P(A) = 3/5

A에서 하양을 뽑을 확률 P(B) = 2/5

 

A에서 빨강을 뽑고 B에서 빨강을 뽑을 확률 = (3/5) x (5/8)

A에서 하양을 뽑고 B에서 빨강을 뽑을 확률 = (2/5) x (4/8)

23/40