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Single Maths B Probability & Statistics: Exercises & Solutions
1. QUESTION:
Describe the sample space and all 16 events for a trial in which two coins are thrown and each shows either a head or a tail.
SOLUTION:
The sample space is S = {hh, ht, th, tt}. As this has 4 elements there are 2^4 = 16 subsets, namely φ, hh, ht, th, tt, {hh, ht}, {hh, th}, {hh, tt}, {ht, th}, {ht, tt}, {th, tt}, {hh, ht, th}, {hh, ht, tt}, {hh, th, tt}, {ht, th, tt} and finally {hh, ht, th, tt}.
- QUESTION: A fair coin is tossed, and a fair die is thrown. Write down sample spaces for
(a) the toss of the coin; (b) the throw of the die; (c) the combination of these experiments.
Let A be the event that a head is tossed, and B be the event that an odd number is thrown. Directly from the sample space, calculate P(A ∩ B) and P(A ∪ B).
SOLUTION:
(a) {Head, T ail} (b) { 1 , 2 , 3 , 4 , 5 , 6 } (c) {(1 ∩ Head), (1 ∩ T ail),... , (6 ∩ Head), (6 ∩ T ail)}
Clearly P(A) = 12 = P(B). We can assume that the two events are independent, so
P(A ∩ B) = P(A)P(B) =
Alternatively, we can examine the sample space above and deduce that three of the twelve equally likely events comprise A ∩ B. Also, P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 34 , where this probability can also be determined by
noticing from the sample space that nine of twelve equally likely events comprise A ∪ B.
- QUESTION: A bag contains fifteen balls distinguishable only by their colours; ten are blue and five are red. I reach into the bag with both hands and pull out two balls (one with each hand) and record their colours.
(a) What is the random phenomenon? (b) What is the sample space? (c) Express the event that the ball in my left hand is red as a subset of the sample space.
SOLUTION:
(a) The random phenomenon is (or rather the phenomena are) the colours of the two balls. (b) The sample space is the set of all possible colours for the two balls, which is
{(B, B), (B, R), (R, B), (R, R)}.
(c) The event is the subset {(R, B), (R, R)}.
4. QUESTION:
M&M sweets are of varying colours and the different colours occur in different proportions. The table below gives the probability that a randomly chosen M&M has each colour, but the value for tan candies is missing.
Colour Brown Red Yellow Green Orange Tan Probability 0.3 0.2 0.2 0.1 0.1?
(a) What value must the missing probability be? (b) You draw an M&M at random from a packet. What is the probability of each of the following events? i. You get a brown one or a red one. ii. You don’t get a yellow one. iii. You don’t get either an orange one or a tan one. iv. You get one that is brown or red or yellow or green or orange or tan.
SOLUTION:
(a) The probabilities must sum to 1.0 Therefore, the answer is 1− 0. 3 − 0. 2 − 0. 2 − 0. 1 − 0 .1 = 1− 0 .9 = .1. (b) Simply add and subtract the appropriate probabilities. i. 0.3 + 0.2 = 0.5 since it can’t be brown and red simultaneously (the events are incompatible). ii. 1 − P(yellow) = 1 − 0 .2 = 0.8. iii. 1 − P(orange or tan) = 1 − P(orange) − P(tan) = 1 − 0. 1 − 0 .1 = 0.8 (since orange and tan are incompatible events). iv. This must happen; the probability is 1.
- QUESTION: You consult Joe the bookie as to the form in the 2.30 at Ayr. He tells you that, of 16 runners, the favourite has probability 0.3 of winning, two other horses each have probability 0.20 of winning, and the remainder each have probability 0.05 of winning, excepting Desert Pansy, which has a worse than no chance of winning. What do you think of Joe’s advice?
SOLUTION:
Assume that the sample space consists of a win for each of the 16 different horses. Joe’s probabilities for these sum to 1.3 (rather than unity), so Joe is incoherent, albeit profitable! Additionally, even “Dobbin” has a non-negative probability of winning.
- QUESTION: Not all dice are fair. In order to describe an unfair die properly, we must specify the probability for each of the six possible outcomes. The following table gives answers for each of 4 different dice.
Probabilities Outcome Die 1 Die 2 Die 3 Die 4 1 1/3 1/6 1/7 1/ 2 0 1/6 1/7 1/ 3 1/6 1/6 1/7 -1/ 4 0 1/6 1/7 -1/ 5 1/6 1/6 1/7 1/ 6 1/3 1/7 2/7 1/
(a) There are 216 equally likely triples and of these only 10 have a sum ≤ 5 so P (X 1 + X 2 + X 3 ≤
- = 10/216. (b) The smallest of three numbers is bigger than i only when all three are so P (min(X 1 , X 2 , X 3 ) ≥ i) = P (X 1 ≥ i, X 2 ≥ i, X 3 ≥ i) = (7 − i)^3 / 216 (picture this group as a cube within the bigger cube of all 216 states). (c) Of 36 triples with X 3 = 2 only 3 have X 1 + X 2 < 4 and of 36 triples with X 3 = 3, 26 have X 1 + X 2 < 9 so that P ((X 3 )^2 > X 1 + X 2 ) =
j P^ (X^1 +^ X^2 < j (^2) , X 3 = j) = 137/216.
- QUESTION: You play draughts against an opponent who is your equal. Which of the following is more likely: (a) winning three games out of four or winning five out of eight; (b) winning at least three out of four or at least five out of eight?
SOLUTION:
Let X and Y be the numbers of wins in 4 and 8 games respectively. For 4 games there are 2^4 = 16 equally likely outcomes e.g. W LW W which has 3 wins so X = 3. Using our basic counting principles there will be
j
outcomes containing j wins and so P (X = 3) = 4 × 0. 54 = 0.25. Similarly with 8 games there are 2^8 = 256 equally likely outcomes and this time P (Y = 5) = 56× 0. 58 = 0 .2188 so the former is larger. For part (b) remember that X ≥ 3 means all the outcomes with at least 3 wins out of 4 etc and that we sum probabilities over mutually exclusive outcomes. Doing the calculations, P (X ≥ 3) = 0 .25 + 0.0625 = 0.3125 is less than P (Y ≥ 5) = 0.2188 + 0.1094 + 0.0313 + 0.0039 = 0.3633 – we deduce from this that the chance of a drawn series falls as the series gets longer.
- QUESTION: Count the number of distinct ways of putting 3 balls into 4 boxes when: MB all boxes and balls are distinguishable; BE the boxes are different but the balls are indentical; FD the balls are identical, the boxes are different but hold at most a single ball. See if you can do the counting when there are m balls and n boxes.
SOLUTION:
#(MB)= 4^3 = 64, #(BE)=
3
= 20, #(FD)=
3
= 4. The general cases are nm,
(m+n− 1 m
(i.e. ar- rangements of balls and fences),
( (^) n m
11. QUESTION:
A lucky dip at a school fˆete contains 100 packages of which 40 contain tickets for prizes. Let X denote the number of prizes you win when you draw out three of the packages. Find the probability density of X i.e. P (X = i) for each appropriate i.
SOLUTION:
There are
3
choices of three packages (in any ordering). There are
3
choices of three packages without prizes. Hence P (X = 0) =
3
3
≈ 0 .2116. If a single prize is won this can happen in
1
2
ways. Hence P (X = 1) =
1
2
3
≈ 0 .4378 and similarly P (X = 2) =
2
60 1
3
≈ 0 .2894 and P (X = 3) =
3
3
≈ 0 .0611 (there is some small rounding error in the given values).
- QUESTION: Two sisters maintain that they can communicate telepathically. To test this assertion, you place the sisters in separate rooms and show sister A a series of cards. Each card is equally likely to depict either a circle or a star or a square. For each card presented to sister A, sister B writes down ‘circle’, or ‘star’ or ‘square’, depending on what she believes sister A to be looking at. If ten cards are shown, what is the probability that sister B correctly matches at least one?
SOLUTION:
We will calculate a probability under the assumption that the sisters are guessing. The probability of at least one correct match must be equal to one minus the probability of no correct matches. Let Fi be the event that the sisters fail to match for the ith card shown. The probability of no correct matches is P(F 1 ∩ F 2 ∩... ∩ F 10 ), where P(Fi) = 23 for each i. If we assume that successive attempts at matching cards are independent, we can multiply together the probabilities for these independent events, and so obtain
P(F 1 ∩ F 2 ∩... ∩ F 10 ) = P(F 1 )P(F 2 )... P(F 10 ) = (
)^10 = 0. 0173.
Hence the probability of at least one match is 1 − 0 .0173 = 0.9827.
- QUESTION: An examination consists of multiple-choice questions, each having five possible answers. Suppose you are a student taking the exam. and that you reckon you have probability 0.75 of knowing the answer to any question that may be asked and that, if you do not know, you intend to guess an answer with probability 1/5 of being correct. What is the probability you will give the correct answer to a question?
SOLUTION:
Let A be the event that you give the correct answer. Let B be the event that you knew the answer. We want to find P(A). But P(A) = P(A∩B)+P(A∩Bc) where P(A∩B) = P(A|B)P(B) = 1× 0 .75 = 0. 75 and P(A ∩ Bc) = P(A|Bc)P(Bc) = 15 × 0 .25 = 0.05. Hence P(A) = 0.75 + 0.05 = 0.8.
- QUESTION: Consider the following experiment. You draw a square, of width 1 foot, on the floor. Inside the square, you inscribe a circle of diameter 1 foot. The circle will just fit inside the square. You then throw a dart at the square in such a way that it is equally likely to fall on any point of the
square. What is the probability that the dart falls inside the circle? (Think about area!) How might this process be used to estimate the value of π?
SOLUTION:
All points in the square are equally likely so that probability is the ratio of the area of the circle to the area of the square. The area of the square is 1 and the area of the circle is π/4 (since the radius is 1/2). If you don’t know π you can estimate it by repeating the experiment a very large number of times. Then π will be approximately the same as the proportion of times the dart fall in the circle multiplied by 4.
- QUESTION: I have in my pocket ten coins. Nine of them are ordinary coins with equal chances of coming up head and tail when tossed and the tenth has two heads.
(a) If I take one of the coins at random from my pocket, what is the probability that it is the coin with two heads? (b) If I toss the coin and it comes up heads, what is the probability that it is the coin with two heads ? (c) If I toss the coin one further time and it comes up tails, what is the probability that it is one of the nine ordinary coins?
SOLUTION:
Denote by D the event that the coin is the one with two heads.
(a) P(D) = 1/10.
18. QUESTION:
Two events A and B are such that P(A) = 0.5, P(B) = 0.3 and P(A ∩ B) = 0.1. Calculate
(a) P(A|B); (b) P(B|A); (c) P(A|A ∪ B); (d) P(A|A ∩ B); (e) P(A ∩ B|A ∪ B).
SOLUTION:
(Venn diagrams are helpful in understanding some of the events that arise below.)
(a) P(A|B) = P(A ∩ B)/P(B) = (^13) (b) P(B|A) = P(A ∩ B)/P(A) = (^15) (c) P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.7, and the event A ∩ (A ∪ B) = A, so
P(A|A ∪ B) = P(A)/P(A ∪ B) =
(d) P(A|A ∩ B) = P(A ∩ B)/P(A ∩ B) = 1, since A ∩ (A ∩ B) = A ∩ B. (e) P(A ∩ B|A ∪ B) = P(A ∩ B)/P(A ∪ B) = 17 , since A ∩ B ∩ (A ∪ B) = A ∩ B.
- QUESTION: An urn contains r red balls and b blue balls, r ≥ 1 , b ≥ 3. Three balls are selected, without replacement, from the urn. Using the notion of conditional probability to simplify the problem, find the probability of the sequence Blue, Red, Blue.
SOLUTION:
Let Bi be the event that a blue ball is drawn on the ith draw, and define Ri similarly. We require
P(B 1 R 2 B 3 ) = P(B 3 |R 2 B 1 )P(R 2 |B 1 )P(B 1 ) = ( b − 1 r + b − 2
r r + b − 1
b r + b
20. QUESTION:
Three babies are given a weekly health check at a clinic, and then returned randomly to their mothers. What is the probability that at least one baby goes to the right mother?
SOLUTION:
Let Ei be the event that baby i is reunited with its mother. We need P(E 1 ∪ E 2 ∪ E 3 ), where we can use the result
P r(A ∪ B ∪ C) = P r(A) + P r(B) + P r(C) − P r(A ∩ B) − P r(A ∩ C) − P r(B ∩ C) + P r(A ∩ B ∩ C).
for any A, B, C. The individual probabilities are P(E 1 ) = P(E 2 ) = P(E 3 ) = 13. The pairwise joint probabilities are equal to 16 , since P(E 1 E 2 ) = P(E 2 |E 1 )P(E 1 ) = ( 12 )( 13 ), and the triplet P(E 1 E 2 E 3 ) = (^16) similarly. Hence our final answer is 1 3
21. QUESTION:
In a certain town, 30% of the people are Conservatives; 50% Socialists; and 20% Liberals. In this town at the last election, 65% of Conservatives voted, as did 82% of the Socialists and 50% of the Liberals. A person from the town is selected at random, and states that she voted at the last election. What is the probability that she is a Socialist?
SOLUTION:
We organise the problem as follows: let C, S and L be the events that a person is Conservative, Socialist, or Liberal respectively. Let V be the event that a person voted in the last election. We require to find P(S|V ), where the information we are given can be summarised as:
P(C) = 0. 3 , P(S) = 0. 5 , P(L) = 0. 2 ,
P(V |C) = 0.65 P(V |S) = 0. 82 , P(V |L) = 0. 5. Now, by Bayes theorem, P(S|V ) =
P(V |S)P(S)
P(V )
Each term is known, excepting P(V ) which we calculate using the idea of a partition. We can calculate P(V ) by associating V with the certain partition C ∪ S ∪ L:
P(V ) = P(V ∩ (C ∪ S ∪ L)) = P(V C) + P(V S) + P(V L) = P(V |C)P(C) + P(V |S)P(S) + P(V |L)P(L) = (0.65)(0.3) + (0.82)(0.5) + (0.5)(0.2) = 0. 705.
Hence
P(S|V ) =
22. QUESTION:
Three prisoners, A, B, and C, are held in separate cells. Two are to be executed. The warder knows specifically who is to be executed, and who is to be freed, whereas the prisoners know only that two are to be executed. Prisoner A reasons as follows: my probability of being freed is clearly 13 until I receive further information. However, it is clear that at least one of B and C will be executed, so I will ask the warder to name one prisoner other than myself who is to be executed. Once I know which of B and C is to be executed, either I will go free or the other, unnamed, prisoner will go free, with equal probability. Hence, by asking the name of another prisoner to be executed, I raise my chances of survival from 13 to 12. Investigate A’s reasoning. [Hint: find the conditional probability that A is freed, given that the warder names B to be executed.]
SOLUTION:
A’s reasoning is unsound. It does not take into account the latitude that the warder has in naming another prisoner to be executed. To see this, let AF be the event that A goes free, and let WB be the event that the warder names B. We need to calculate
P(AF |WB ) =
P(WB |AF )P(AF )
P(WB )
We have
P(WB ) = P(WB |AF )P(AF ) + P(WB |BF )P(BF ) + P(WB |CF )P(CF ) = (
and we find thereby that P(AF |WB ) = 13. (This analysis supposes that the warder is equally likely to name either B or C in the situation that both are to be executed.)
k 0 1 2 3
P(X = k) 2764 2764 649 641 (b) Show that E(X) = 34 and that Var(X) = 169.
SOLUTION:
To answer this question, it is necessary both to think about how certain events can occur, and the probability that they occcur. Let H be the event that a child is healthy, and D be the event that a child has the disease. If there are three children, there are 8 possiblities (including different orderings) as shown in the next table, where by the sequence H, H, D we mean that the first two children were born healthy, and the third was born with the disease.
Sequence Probability X
H,H,H 343434 = 2764 0
H,H,D 343414 = 649 1
H,D,H 341434 = 649 1
D,H,H 143434 = 649 1
H,D,D 341414 = 643 2
D,H,D 143414 = 643 2
D,D,H 141434 = 643 2
D,D,D 141414 = 641 3
The probabilities for each sequence are shown in the second column; successive births are independent so that we can multiply probabilities. Notice that the sum of the probabilities is 1. The random variable X is the number of children having the disease. We see that only one sequence leads to X = 0, and this sequence has probability 2764. Hence P(X = 0) = 2764. There are three sequences leading to X = 1, each with probability 649. Hence P(X = 1) = 649 + 649 + 649 = 2764. The other probabilities are found similarly. It is easy to show that E(X) = 34 and that E(X^2 ) = 1816 , so that Var(X) = 169. Remark. This is an example of a binomial distribution with parameters n = 3 and p = 14. That is,
P(X = k) =
k!(3 − k)!
)k(
)^3 −k, k = 0, 1 , 2 , 3.
For such distributions it is well known that E(X) = np and that Var(X) = np(1 − p).
- QUESTION: A six-sided die has four green and two red faces and is balanced so that each face is equally likely to come up. The die will be rolled several times. Suppose that we score 4 if the die is rolled and comes up green, and 1 if it comes up red. Define the random variable X to be this score. Write down the distribution of probability for X and calculate the expectation and variance for X.
SOLUTION:
The distribution of X is as follows.
k 1 4
P(X = k) (^1323)
E(X) = (1)
E(X^2 ) = (1^2 )
+ (4^2 )
V ar(X) = E(X^2 ) − (E(X))^2 = 11 − 9 = 2.
26. QUESTION:
For two standard dice all 36 outcomes of a throw are equally likely. Find P (X 1 + X 2 = j) for all j and calculate E(X 1 + X 2 ). Confirm that E(X 1 ) + E(X 2 ) = E(X 1 + X 2 ).
SOLUTION:
The possible totals are j = 2, 3,... , 12 and P (X 1 + X 2 = j) = (j − 1)/36, j = 2,... , 7 and P (X 1 + X 2 = j) = (13 − j)/36, j = 8,... , 12. For each of the dice E(Xi) = 21/6 = 7/2 while for the total E(X 1 + X 2 ) = 361 (2 · 1 + 3 · 2 + · · · + 7 · 6 + · · · + 12 · 1) = 252/36 = 7
- QUESTION: X takes values 1, 2, 3, 4 each with probability 1/4 and Y takes values 1, 2, 4, 8 with probabilities 1 /2, 1/4, 1/8 and 1/8 respectively. Write out a table of probabilities for the 16 paired outcomes which is consistent with the distributions of X and Y. From this find the possible values and matching probabilities for the total X + Y and confirm that E(X + Y ) = E(X) + E(Y ).
SOLUTION:
There are 16 pairs and infinitely many ways to allocate the probabilities. Selecting one, say p(1, 4) = p(1, 8) = 1/8, p(2, 2) = 1/4, p(3, 1) = p(4, 1) = 1/4 we see this satisfies
∑^ j^ p(i, j) =^ P^ (X^ =^ i) and i(i, j) =^ P^ (Y^ =^ j). The possible values and probabilities are t 2 3 4 5 6 7 8 9 10 11 12 pt 0 0 1/2 3/8 0 0 0 1/8 0 0 0 where for instance P (X + Y = 4) = p(2, 2) + p(3, 1) = 1/2. From the table, E(X + Y ) = (4 · 4 + 5 · 3 + 9 · 1)/8 = 5. As E(X) = 5/2 and E(Y ) = 5/2 the required equality holds.
- QUESTION: Calculation practice for the binomial distribution. Find P (X = 2), P (X < 2), P (X > 2) when (a) n = 4, p = 0.2; (b) n = 8, p = 0.1; (c) n = 16, p = 0.05; (d) n = 64, p = 0.0125.
SOLUTION:
(a) P (X = 2) = 6 · 0. 22 · 0. 82 = 0.1536, P (X < 2) = P (X = 0) + P (X = 1) = 0.4096 + 0.4096 = 0.8192, P (X > 2) = 1 − P (X ≤ 2) = 1 − 0. 8192 − 0 .1536 = 0.0272. (b) P (X = 2) = 0.1488, P (X < 2) = 0.8131, P (X > 2) = 0.0381. (c) P (X = 2) = 0.1463, P (X < 2) = 0.8108, P (X > 2) = 0.0429. (d) P (X = 2) = 0.1444, P (X < 2) = 0.8093, P (X > 2) = 0.0463.
- QUESTION: A wholesaler supplies products to 10 retail stores, each of which will independently make an order on a given day with chance 0.35. What is the probability of getting exactly 2 orders? Find the most probable number of orders per day and the probability of this number of orders. Find the expected number of orders per day.
33. QUESTION:
Given that 0.04% of vehicles break down when driving through a certain tunnel find the probability of (a) no (b) at least two breakdowns in an hour when 2,000 vehicles enter the tunnel.
SOLUTION:
The number of breakdowns X has a binomial distribution which can be approximated by the Pn(λ) distribution with λ = 2000 × 0 .0004 = 0.8. Hence P (X = 0) ≈ 0 .4493 and P (X ≥ 2) = 1 − P (X ≤
- ≈ 1 − 0 .8088 = 0.1912.
- QUESTION: Experiments by Rutherford and Geiger in 1910 showed that the number of alpha particles emitted per unit time in a radioactive process is a random variable having a Poisson distribution. Let X denote the count over one second and suppose it has mean 5. What is the probability of observing fewer than two particles during any given second? What is the P (X ≥ 10)? Let Y denote the count over a separate period of 1.5 seconds. What is P (Y ≥ 10)? What is P (X + Y ≥ 10)?
SOLUTION:
P (X ≤ 1) = e−^5 (1 + 5) = 0.0404. P (X ≥ 10) = 0.0398. Y is Pn(7.5) and so P (Y ≥ 10) = 1 − P (Y ≤
- = 1 −
0 P^ (Y^ =^ i) = 0.5113.^ X^ +^ Y^ is Pn(12.5) and we find^ P^ (X^ +^ Y^ ≥^ 10) = 0.7986.
- QUESTION: A process for putting chocolate chips into cookies is random and the number of choc chips in a cookie has a Poisson distribution with mean λ. Find an expression for the probability that a cookie contains less than 3 choc chips.
SOLUTION:
The Poisson distribution gives probabilities for each possible number of choc chips but as a cookie can’t contain two different numbers simultaneously we add the probabilities for the possible values 0, 1 and 2. Hence P (X < 3) = e−λ
1 + λ +
λ^2 2
(if you wonder what happened to s, it equals 1 as we’re only looking at a single cookie – for 10 cookies we take s = 10 etc).
- QUESTION: Let X have the density f (x) = 2x if 0 ≤ x ≤ 1 and f (x) = 0 otherwise. Show that X has the mean 2 /3 and the variance 1/18. Find the mean and the variance of the random variable Y = − 2 X + 3.
SOLUTION:
To find expected values for continuous random variables we integrate e.g.
E(X) =
−∞
xf (x) dx =
0
x · 2 x dx = 2/ 3
and similarly Var(X) =
0
(x − 2 /3)^2 · 2 x dx = 1/ 2 − 8 /9 + 4/9 = 1/ 18
You could also use the formula Var(X) = E(X^2 ) − E(X)^2 where
E(X^2 ) =
0
x^2 · 2 x dx = 1/ 2
so that Var(X) = 1/ 2 − (2/3)^2 = 1/18. Use the linearity of expectation for the last bits to get
E(Y ) = − 2 × 2 /3 + 3 = 5/ 3 and Var(Y ) = (−2)^2 × 1 /18 = 2/ 9
37. QUESTION:
Let the random variable X have the density f (x) = kx if 0 ≤ x ≤ 3. Find k. Find x 1 and x 2 such that P (X ≤ x 1 ) = 0.1 and P (X ≤ x 2 ) = 0.95. Find P ( |X − 1. 8 | < 0 .6).
SOLUTION:
To make
0 f^ (x)dx^ = 1 we must take^ k^ = 2/9. For any^ c^ in (0,^ 3),^ P^ (X^ ≤^ c) =^ c
(^2) /9 so x 1 = √ 0 .9 = 0 .9487 and x 2 =
8 .55 = 2.9240. P ( |X − 1. 8 | < 0 .6) = P (1. 2 < X < 2 .4) = (2. 42 − 1. 22 )/9 = 0.48.
38. QUESTION:
A small petrol station is supplied with petrol once a week. Assume that its volume X of potential sales (in units of 10, 000 litres) has the probability density function f (x) = 6(x − 2)(3 − x) for 2 ≤ x ≤ 3 and f (x) = 0 otherwise. Determine the mean and the variance of this distribution. What capacity must the tank have for the probability that the tank will be emptied in a given week to be 5%?
SOLUTION:
Proceed as usual, E(X) = 5/2 and Var(X) = 1/20. Let T denote the capacity of the tank. We need to solve 0.05 =
T 6(x^ −^ 2)(3^ −^ x)dx^ =^
T − 2 6 y(1^ −^ y)^ dy^ = 1^ −^ 3(T^ −^ 2) (^2) + 2(T − 2) (^3) and doing this numerically (iterate the equation T − 2 =
0 .95(7 − 2 T )−^1 /^2 ) we find T ≈ 2 .86465 i.e. the tank should hold approximately 28,650 litres.
- QUESTION: Find the probability that none of the three bulbs in a set of traffic lights will have to be replaced during the first 1200 hours of operation if the lifetime X of a bulb (in thousands of hours) is a random variable with probability density function f (x) = 6[0. 25 − (x − 1 .5)^2 ] when 1 ≤ x ≤ 2 and f (x) = 0 otherwise. You should assume that the lifetimes of different bulbs are independent.
SOLUTION:
For a single bulb, P (X > 1 .2) = 6[ 32 x^2 − 13 x^3 − 2 x]^21. 2 = 0.8960. Hence P (no bulbs replaced) =
89603 = 0..
QUESTION: Suppose X is N(10, 1). Find (i) P [X > 10 .5], (ii) P [9. 5 < X < 11], (iii) x such that P [X < x] = 0.95. You will need to use Standard Normal tables.
SOLUTION:
Let Z denote a N (0, 1) random variable from now on and let Φ denote its cdf. (i) P [X > 10 .5] = P [X − 10 > 0 .5] = 1 − Φ(0.5) = 0.3085; (ii) P [9. 5 < X < 11] = Φ(1) − Φ(− 0 .5) = 0 .5328; (iii) P [X < x] = P [Z < x − 10] = 0.95 when x − 10 = 1.645 i.e. x = 11..
- QUESTION: Suppose X is N(− 1 , 4). Find (a) P (X < 0); (b) P (X > 1); (c) P (− 2 < X < 3); (d) P (|X + 1| < 1).
SOLUTION:
As X = 2Z + 1 we have (a) P (X < 0) = P (2Z + 1 < 0) = Φ(− 1 /2) = 1 − Φ(1/2) = 0.3085; (b) P (X > 1) = Φ(0) = 1/2; (c) P (− 2 < X < 3) = Φ(1) − Φ(− 3 /2) = 0.7745; (d) P (|X + 1| < 1) = P (− 2 < X < 0) = Φ(− 1 /2) − Φ(− 3 /2) = 0.2417.
- QUESTION: Suppose X is N(μ, σ^2 ). For a = 1, 2, 3 find P ( |X − μ| < aσ).
SOLUTION:
P ( |X − μ| < aσ) = P (−a < Z < a) = 2Φ(a) − 1 so the required values are approximately 0.682, 0. 954 and 0.998 respectively.
