Typical Test Problems (with solutions)

1. Typical Test Problems (with solutions)

2. Try solving these problems and compare with the solutions 1. One bag contains 4 white balls and 2 black balls; another contains 3 white and 5 black. If one ball is drawn from each bag, find the probability that (a) both are white; (b) both are black; (c) one is white and one is black. Answer: (a) 4/6*3/8 = 1/4; (b) 2/6*5/8=5/24; (c) (4/6)(5/8)+(2/6)(3/8)= 13/24 2. A and B play a game in which they alternately toss a pair of dice. The one who is first to get a total of 7 wins the game. Find the probability that (a) one who tosses first will win the game; (b) one who tosses second will win the game. Reminder: Infinite Geometric Progression Answer. (a) (1/6)+(5/6)²(1/6)+(5/6)4(1/6)...=6/11 (b) (5/6)(1/6) + (5/6)³(1/6)+....=5/11

3. 3. A machine produces a total of 12,000 bolts a day, which are on average 3% defective. Find the probability that out of 600 bolts chosen at random, 12 will be defective. Solution, Of 12000 bolts, 360 are defective and 11640 are not. P= C11640,588 C360,12 /C12000,600 = 0.0347277 (Eq. 1) Some comments: This problem caused some controversy. Some of you suggested that the probability of finding a defective bolt is 0.03. Then the probability could be estimated through the Binomial distribution p(12 out of 600) = C600,12 0.03120.97588~0.036 (Eq. 2). This is very close indeed to the result of Eq. 1. May be this tiny difference is just a result of an error? To demonstrate that this is not true, we consider a simpler example. Suppose that there are 10 balls, 6 are black and 4 are white. What is the probability that of 4 balls chosen at random, 2 are white.

4. The conventional solution is p= C4,2C6,2/C6,4=0.429. Trying to be silly, we can also use a Binomial formula assuming that the probabilities are 3/5 and 2/5: p = C4,2(2/5)2(3/5)2= 0.346. This time the difference is significant. In addition, we can now understand the source of the problem, The Binomial distribution can be used only when the probabilities of two outcomes do not depend on the number of previous trials. This condition is not fulfilled in our example. The probability of choosing a white ball depends on the outcomes of previous selections. The same is true for the “bolts” problem, although the differences are much less significant. Suggestion : try to investigate how the results derived with Eqs. 1 and 2 depend on the total number of bolts n (initially we had n=12000). Change n from 100 to 100000. Use Matematica. The function below utilizes Eq. 1 and can be useful. We still assume that 600 bolts are selected, and the probability of 12 defective bolts is questioned. f[n_]:= Binomial[0.03*n,12]*Binomial[0.97 n, 588]/Binomial[n,600]//N Please, let me know about the outcome. Good luck.

5. 4. The probabilities that a husband and wife will be alive 30 years from now are given by 0.7 and 0.8 respectively. Find the probability that in 20 years (a) both (b) neither (c) at least one, will be alive. Answer (a) 0.56; (b) 0.06 (c) 0.94 5. Four different mathematics books, six different physics books, and two different chemistry books are to be arranged on a shelf. How many different arrangements are possible if (a) the books in each particular subject must all stand together. (b) only the mathematics books must stand together Answer: (a) 6!*4!*2*3!= 207360; (b) 12-4+1= 9; 9!4!= 8709120

6. 6. An urn contains 6 red and 8 blue marbles. Five marbles are drawn at random from it without replacement. Find the probability that 3 are red and 2 are blue. Answer: C6,3 C8,2/C14,5=0.27972 Some commentsThis problem caused a very productive discussion in one of my classes. The counterargument to our solution sounded as follows:Let’s consider one possible selection, namely rrrbb. It’s probability is: P(r,r,r,b,b)= (6/14)(5/13)(4/12)(8/11)(7/10). For any other sequence (such as r-b-r-b-r) the probability is the same. For instance, P(r,b,r,b,r)= (6/14)(8/13)(5/12)(7/11)(4/10) = P(r,r,r,b,b) = 0.027972 . Based on this, it was suggested that this probability (which turned to be 1/10 of the previous result), gives the correct answer to our problem. This error has a historical parallel , the famous D‘ Alambert’s assertion. D’Alambert did not notice that the “Head and Tail” event for 2 coins corresponds to two outcomes: HT and TH. However, D’Alambert wrongly suggested that the probability of “H and T” is 1/3, while our students calculated the probabilities of each sequence correctly. They simply did not sum up the probabilities of various outcomes corresponding to the event in question (3 red and 2 blue). In our case, the number of outcomes is the number of possible arrangements of 3 r and 2 b marbles: C5,3=C5,2=10. Thus, the total probability is p(total) =10 P(r,r,r,b,b) = 0.27972.

7. 7 (a) Find the probability of getting the sum 7 on at least one of three tosses of a pair of fair dice. (b) How many tosses (n) is needed in order that this probability (getting 7 in at least 1of n tosses) is greater than 0.95. Answer: (a) 1-(5/6)³=91/216;(b) (5/6)ⁿ<0.05,n≥17. 8. The odds in favor of A winning a game of chess against B are 3:2. If three games are to be played, what are the odds (a) in favor of A winning at least 2 games; (b) against A of losing the first two games to B Hint: First, translate odds in the probability. You will find out that the probability of A winning against B is 3/5. Answer: (a) 3(3/5)²(2/5)+(3/5)³= 81/125 (81:44)(b) 4/25 (4:21)

8. 9. A pair of dice is tossed repeatedly. (a) Find the probability that the sum of 11 occurs for the first time on the 6-th toss. Answer : (1/18)*(17/18)5= 0.262 10. Find the probability of scoring a total of 7 points (a) once, (b) at least once, (c) twice, in two tosses of a pair of dice. Answer: (a) 2*(1/6)(5/6)= 5/18 (b) 1-25/36 = 11/36 (c) 1/36

9. 11. A box contains 9 tickets numbered from 1 to 9, inclusive. If 3 tickets are drawn from the box 1 at a time, find the probability that they are alternatively either odd-even-odd or even-odd even. Solution: Total number of choices: P_{9,3}=9!/6!=504 (Notice that the order here is important).The number of the possible choices for the first event: 5*4*4, For the second, 4*5*3. The answer: (5*4*4+4*5*3)/504= 5/(18) 12. Three cards are drawn from an ordinary deck of 52 cards. Find the probability that (a) all cards are of the same suit (b) at least two aces are drawn Answer: (a) There are Binomial[13, 3] (I use this Mathematica’s name for Cn,m) ways of choosing 3 cards from one suit (say hearts). It should be multiplied by 4, the total number of suits: P= 4.*Binomial[13, 3]/Binomial[52, 3]=0.052; (b) There are Binomial[4,2]*48 ways to chose 2 cards out of 4 aces and one card which is not an ace, and Binomial[4,2] ways to choose three aces out of four : P= (Binomial[4,2]*48 + Binomial[4,3])/Binomial[52, 3]=0.013

10. 13. 15 students are evenly distributed among three groups. If three of them are siblings, (a) what is the probability that all three are in the same group? (b) What is the probability that each group gets one? Solution: (a) The total number of outcomes: N=15!/(5!5!5!). The positive outcomes are: 3 12!/(2!5!5!) (If you put all 3 in one of the groups, than you have to distribute other 12 in the groups of 5,5,and 2. And then it should be multiplied by 3.) (b) 3! 12!/(4!4!4!) 14. In how many ways can 4 boys and 4 girls pair off? In how many ways can they stay in a row in alternating gender? Answer: (a) 4! (b) 2*4! 4!

11. 15 A marble is drawn at random from a box containing 10 red, 30 white, 20 blue and 15 orange marbles. Find a probability that it is (a) orange or red (b) nor red or blue Answer: (a) 10/75+15/75= 1/3 (b) 45/75= 3/5 16. Two marbles are drawn at random with replacement from a box containing 10 red, 30 white, 20 blue and 15 orange marbles. Find a probability that (a) the first is blue and the second is orange (b) they are either red or white or both (red and white). Hint: "both" consists of 2 outcomes, depending on the order. Answer: (a) 20*15/(75*75)= 4/75 (b) ((10/75)²+(30/75)²+2*10*30)/(75*75)= 64/225