2016-2018 – S3 – Mathematics – TEST 1 SOLUTIONS Page 1 / 3
IUT of Saint-Etienne – Sales and Marketing department
Mr. Ferraris Prom 2016-2018 26/10/2017
MATHEMATICS – 3
rdsemester, Test 1 length : 2 hours – coefficient 1/2
SOLUTIONS
Exercise 1 (4 points)
A bag contains 16 red coins and 4 green ones. A game consists in a simultaneous draw of ten coins, at random.
The player wins in case he gets at least three green coins.
1) After having determined the probability distribution of X, number of green coins among the ten, calculate
the probability to win. 2 pts
A coin may be green (success) or red. The ten coins draw is performed without putting back, and our variable X is the number of possible successes. X is then distributed by
H
(10, 4, 20).( ) (
= X = +) (
X =)
= 34×10167 + 44×10616 ≈ + ≈20 20
C C C C
p win p 3 p 4 0.2477 0.0433 0.2910
C C .
2) To be allowed to play once, you have to bet €2. The game owner settled the following gains:
€20 for 4 green coins, €4 for 3 green coins, and no gain in case of any other outcome. How much money
can the owner expect to win until 10,000 played games? 2 pts
By a table, let's sum up the "gain after bet" probability distribution:
gain Y (€) : -2 2 18
pi or p(Y = yi) : 0.7090 0.2477 0.0433
The expectation is (for a player) : E(Y) = €–0.1424, thus allowing the owner to forecast a 14.24 cents per game average income, on a large number of games, that is around €1424 on 10,000 games.
Exercise 2 (8.5 points)
In a supermarket, it has been stated that the pass-through of one customer in twenty causes a problem at the checkout (caisse) and blocks the line (file d'attente) for a while. At any given time, for a given checkout, the probability of occurrence of a problem is then 0.05. After having taken your products, you go choose a checkout and enter a line that consists of n customers in front of you. The variable X describes the number of problems that may occur and waste your time due to the pass-through of these n clients.
1) Give, and justify, the probability distribution of X. 1 pt
The success is here: "the pass-through of a customer causes a problem". The probability of success is given by the exercise's statement, invariable (p = 0.05). X is then distributed by
B
(n , 0.05)2) In this question, let's set n = 5.
a. What is the probability that at least one problem would occur before your turn? 1 pt
(
X ≥ = −) (
X =)
≈ − ≈p 1 1 p 0 1 0.7738 0.2262 .
b. Given that the supermarket opens 10 checkouts, and that there is an average of 5 clients per checkout, give an estimate of the number of checkouts blocked by a problem, at any time. 1 pt We can use our former result: 22.62% of the checkouts are blocked, so two checkouts out of the ten.
We can also consider the expectation of the number of problems for individual pass-throughs, that is, for 10 checkouts: E(X) = 10×0.05 = 0.5. Hence, 0.5 blocked checkout, on average, at any time. (this last answer is actually the right one, but the mention "at any time" isn't easy to translate in a right calculation, and then we will allow the point in case of any logical reasoning that doesn't give the right answer).
3) a. What is the maximum number of clients that have to be in front of you in the line, so that there are more than 90% chances that no problem would occur before your turn? 1 pt The probability of no success has to be more than 90%; so, we can check it for several values of n thanks to our calculator, or we can solve the inequality p
(
X =0)
>0.9 , that leads to 0.95n > 0.9 and thenn < ln(0.9)/ln(0.95) ≈ 2.054. You have to be behind a maximum of two people if you want the risk to be blocked to be under 10%.
How to check it for several values of n thanks to the calculator:
2016-2018 – S3 – Mathematics – TEST 1 SOLUTIONS Page 2 / 3 create a function representing p(X = 0):
Casio : Y1= binomialpd(0, X, 0.05) TI : Y1= binomial fdp((X, 0.05, 0)
where X will be a list of possible values of n.
The results for p(X = 0) are given on the right.
We can see that this probability is more than 90% provided that n ≤ 2.
b. What is the minimum number of clients that have to be in front of you in the line, so that there are more than 50% chances that at least one problem would occur
before your turn? 1.5 pt
The probability of no success has to be less than 50%; so, we can check it for several values of n thanks to our calculator, or we can solve the inequality p
(
X = <0)
0.5 , that leads to 0.95n < 0.5 and thenn < ln(0.5)/ln(0.95) ≈ 13.51. If you are behind at least 14 people in your line, the chance to be blocked at least once is more than one out of two.
Checking values thanks to the calculator: we can sue the same function as in the former question. The table (still on the right) shows that p(X = 0) becomes less than 50% since n values 14 or more.
4) We assume that, on average, one checkout registers 100 payments per day. In this question, we aim to use a Poisson distribution, and we will name Y the variable daily number of problems suffered by a checkout.
a. Show that Y can be distributed by a Poisson's law, whose parameter will be given. 1 pt n = 100 > 30, p = 0.05 < 0.1 and np = 5 < 10. Thus, Y can be distributed by
P
(5).b. What is the probability that, in a day, less than 3 problems occur at a checkout? 1 pt
(
Y≤ ≈)
p 2 0.12465 .
c. What is the minimum number of problems that have less than 1% chance to be exceeded? 1 pt
(
Y≤)
≈p 10 0.9863 and p
(
Y ≤11)
≈0.9945 . 11 problems, or more, have less than 1% chance to occur.Exercise 3 (7.5 points)
A raspberry's (framboise) mass forms a variable X, well-modeled by the law
N
(4 grams ; 0.5 g).1) Determine p(X < 3) ; p(3 < X < 5). 1 pt
p(X < 3) = 0.02275 and p(3 < X < 5) = 0.9545.
2) The raspberries whose mass is less than 3 grams might be removed by the producer, before the delivery. In a 40 kg batch, representing around 10,000 raspberries, how many of them would be removed? 1 pt 0.02275 = 2.275%, so 227 or 228 raspberries out of 10,000.
3) On a production line, it's planned to sort raspberries by mass: the lightest 20% will be packaged separately, sold at a reduced price and renamed "raspbabies". What is the maximum mass of a raspbaby? 2 pts We're looking for the real k such that p(X < k) = 0.2.
The tool "InvNorm" or "FracNormale" (inverse normal) gives immediately: p(X < 3.58) = 0.2.
Besides, the calculator (or the table), using the standard normal law, gives p(U < 0.84) = 0.8, that can be translated in p(U < -0.84) = 0.2, and finally transformed into our law by a variable change: X = µ + Uσ. k = 4 – 0,84×0,5 = 3,58. The raspbabies are all the raspberries weighting less than 3.58 grams.
4) From a production of 10,000, a sample of 300 raspberries has to be drawn. Y is the possible number of raspbabies among the 300.
a. Explain why a binomial distribution would correctly describe Y, and give its parameters. 1 pt The draw will be performed without putting back. But, as N > 20n (10000 > 6000), we can still use a binomial distribution instead of a hypergeometric one. Y is distributed by
B
(300 ; 0.2).b. Explain why a normal distribution could correctly replace this binomial one, and then give the
parameters of this new normal one. 1 pt
n = 300 > 30 and npq = 48 > 5, hence a possible use of
N
(60 ; 48≈6.9282 ).2016-2018 – S3 – Mathematics – TEST 1 SOLUTIONS Page 3 / 3 c. Using both distributions, give the probability of getting between 55 and 65 (included) raspbabies, and
then the probability of getting exactly 60 raspbabies. Comment the visible differences between the
results coming from different laws. 1.5 pt
binomial dist.: p(55 ≤ X ≤ 65) = p(X ≤ 65) – p(X ≤ 54) = 0.7879 – 0.2152 = 0.5727.
normal dist.: p(54.5 < X < 65.5) = 0.5727.
binomial dist.: p(X = 60) = 0.05750.
normal dist.: p(X = 60) = 0.05758 or p(59.5 < X < 60.5) = 0.05753, given the way you order.
The differences are visible, due to the fact that a normal law is an approximation of the binomial distribution (which is in itself an approximation of the Hypergeometric reality…). The differences would have been much bigger in case of misuse of your calculator… Common mistakes made with a discrete variable, or due to the change discrete/continuous:
* ask the calculator, with a binomial distribution, p(X ≤ 65) – p(X ≤ 55), which is not p(55 ≤ X ≤ 65), but is p(56 ≤ X ≤ 65) ;
* ask the calculator, with a normal distribution, p(X < 65) – p(X < 55),
which is not p(55 ≤ X ≤ 65), because 65 successes are modeled with the interval [64.5 ; 65.5] by a continuous variable, and same for 55 successes, associated with [54.5 ; 55.5].
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