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RRStatistics Questions
In a university, suppose that 6% of the students are in a program of joint honours statistics and economics, and that 30% are in honours statistics either by itself or in a joint program. What is the probability that a random honours statistics student is also taking honours economics?
Suppose that 60% of all business trips between Toronto and Montreal, using public transport, are by air, 30% by train, and 10% by bus. The probability of arriving late is 10% by air, 60% by train, and 5% by bus.
If a person on a business trip arrives late, what is the probability the person traveled by air? By train? And By bus?
A person always uses the same mode of transportation and arrives late every time on n independent trips. Find the probability that the person traveled by air. Evaluate the probability for n = 2,3, and 4.
Suppose that 0.2% of all commercial airline flights should have the takeoff aborted due to a malfunction. A warning light indicates danger 99% of the time when the takeoff should be aborted. However, the light can come on when there is no danger; suppose this occurs either a) 1%, b) 0.1%, c) 0.01% of the time. If the light comes on, what is the probability that the takeoff should be aborted in the above three cases? What do you conclude from these results about the reliability required in a warning system that is widely used to detect an important but rare fault.
Joe, Sue and Tom take a statistics course. Their probabilities of passing are 0.9, 0.8 and 0.6 respectively.
If only one of them passed the course, what is the probability it was Sue?
Suppose two of them passed the course. Find the probability that Sue failed the course
A doctor has just read of a new type of shock treatment which allegedly produces instantaneous recovery in 10% of schizophrenics upon whom it is tried. The doctor decides to check the validity of this information by trying the treatment on some randomly selected schizophrenic patients.
What is the smallest number of patients the doctor must plan to try it on so that there is a probability of at least 0.5 that at least one patient will instantaneously recover, if the information about the treatment is correct.?
If the information is correct, what is the probability that the first instantaneous recovery will occur on the seventh schizophrenic treated?
Suppose that the doctor decides that they must have two instantaneous recoveries to be convinced. If the information is correct, what is the probability that they will have to try the treatment on exactly 10 schizophrenics in order to obtain the two instantaneous recoveries?
A company recruiter is interviewing applicants for five identical jobs. The probability that any one applicant will be acceptable is 2/3. Find the probability that:
exactly eight applicants will have to be interviewed to fill all five positions.
eight or fewer interviews will have to be conducted to fill all five positions.
A telephone switchboard handles 600 calls an hour on average. The board can make a maximum of 20 connections per minute.
Use the Poisson distribution to evaluate the probability that the board will be over taxed during a given minute.
Find the probability that 10 seconds elapse with no calls.
Suppose that blackflies are randomly distributed throughout the air with an average concentration of 2 in 5 cubic feet.
Find the probability of getting exactly 2 blackflies in a tent of volume 40 cubic feet.
Find the probability of getting at least 1 blackfly in a box of volume 1 cubic foot.
For a campsite with three tents each of volume 40 cubic feet, find the probability that:
There are a total of 40 blackflies in the three tents.
Exactly two of the tents contain 2 blackflies each.
In the past, half the items received from a certain supplier by a retail merchant have been in good working order (G), a third have been defective but easily repairable (R), and the rest have been so badly defective that it is either impossible or uneconomical to repair them (U). If the retailer orders 6 more items from the same supplier, find the probability that:
three items will be G, 2 will be R, and 1 will be U
there will be at least one R and exactly one U
there will be at least one R
there will be at least one item which is either R or U
Referring to question #8 (about the blackflies), suppose that a campground contains 20 tents, each with a volume of 20 cubic feet.
Find the probability that 6 of the tents contain no blackflies, 4 contain one or two, 3 contain three, four or five blackflies, and 6 contain more than ten.
If 6 of the tents contain no blackflies, find the probability that 4 contain one or two blackflies, 3 contain three, four or five, and 6 contain more than ten.
Answers:
1)
Suppose S percent are taking honours statistics only, E honours economics only and SE joint program. We have:
SE = 6
S + SE = 30
The claimed probability is SE / (S+SE) = 6/30 = 1/5.
2)
Denote the events: A travelling was performed by air, T by train, B by bus. X the event that the person arrived late. We know:
P(A) = 0.6, P(T) = 0.3, P(B) = 0.1, P(X|A) = 0.1, P(X|T) = 0.6, P(X|B) = 0.05.
We use Bayes calculations:
P(X) = P(X|A)P(A) + P(X|T)P(T) + P(X|B)P(B) = 0.245
(a)
P(A|X) = P(X|A)P(A) / P(X) = 0.245 probability that person who was late traveled by air
P(T|X) = P(X|T)P(T) / P(X) = 0.735 by train
P(B|X) = P(X|B)P(B) / P(X) = 0.020 by bus
(b)
Denote the event that person arrived late on each trip X1, X2, , Xn. They are dependent until we know what transport is used by person. When we are sure about what transport is used, these events become independent under this condition. Denote Y = X1X2Xn.
P(Y) = P(Y|A)P(A) + P(Y|T)P(T) + P(Y|B)P(B) = P(X|A)nP(A) + P(X|T)nP(T) + P(X|B)nP(B)
P(A|Y) = P(Y|A)P(A) / P(Y) = P(X|A)nP(A) / P(Y)
n=2: P(Y) = 0.114, P(A|Y) = 0.053
n=3: P(Y) = 0.065, P(A|Y) = 0.009
n=4: P(Y) = 0.039, P(A|Y) = 0.0015
3)
Denote the events: A takeoff should be aborted, W warning light appears. Bayes formula:
P(A|W) = P(W|A)P(A) / ( P(W|A)P(A) + P(W|~A)P(~A) )
~A is the event not A; P(~A) = 1 P(A) = 0.998
We are given P(A) = 0.002, P(W|A) = 0.99, and
(a) P(W|~A) = 0.01, (b) P(W|~A) = 0.001, (c) P(W|~A) = 0.0001
Then, (a) P(A|W) = 0.166, (b) P(A|W) = 0.665, (c) P(A|W) = 0.952
I guess if the passengers are in danger, the warning system may be used even in case (a).
But it is more reliable in case (b) and (c).
4)
Denote the events: J Joe passed the course, S Sue passed the course, T Tom passed the course, X only one passed the course. X = (~J ~S T) ( (~J S ~T) ( (J ~S ~T).
P(X) = 0.1*0.2* 0.6 + 0.1*0.8*0.4 + 0.9*0.2*0.4 = 0.116
(a)
P(S|X) = P(~J S ~T) / P(X) = 0.1*0.8*0.4 / 0.116 = 0.276
(b)
Y two passed the course. Y = (~J S T) ( (J ~S T) ( (J S ~T).
P(Y) = 0.1*0.8*0.6 + 0.9*0.2*0.6 + 0.9*0.8*0.4 = 0.444
P(~S|Y) = P(J ~S T) / P(Y) = 0.243
5)
(a)
Suppose doctor tries n patients. Probability of recovering of a patient is p = 0.1, probability of fail is 0.9. Probability of recovering at least one of n is evidently 1 0.9n. We must solve the inequality 1 0.9n >= 0.5;
0.9n <= 0.5;
n >= log 0.9 (0.5) = ln(0.5) / ln(0.9) = 6.58.
So, n must be at least 7.
(b)
This means 6 fails and 7-th success, i.e. the probability is (1-p)6p = 0.053.
(c)
This means 1 success among first 9 patients and 10-th success.
The probability is 9*0.1*(0.9)8 * 0.1 = 0.039 (we have 9 ways to choose on which patient will be the first success).
6)
(a)
This means 4 accepts among first 7 applicants, and 8-th accept. C(7,4) ways to choose who was acceptable among first 7 applicants; the probability is C(7,4) * (2/3)5(1/3)3 = 0.171.
(b)
This means 5, 6, 7 or 8 interviews have to be conducted. For 5 interviews, the chance is (2/3)5; for 6 interviews, it is C(5,4)*(2/3)5(1/3); for 7 interviews, it is C(6,4)*(2/3)5(1/3)2; and for 8 interviews (a) it is C(7,4)*(2/3)5(1/3)3.
The sum is (2/3)5 ( 1 + 5/3 + C(6,4)/9 + C(7,4)/27 ) = 0.741.
7)
The Poisson distribution is P(X=k) = Lk / k! * exp(-L), with distribution mean M(X) = L.
600 calls per hour is 10 calls per minute => in our case L = 10.
(a)
The probability of overtaxing is
EMBED Equation.3 0.0016
(b)
The distribution for the number of calls in 10 seconds, Y, will be Poisson distribution with
L = 10/6 = 5/3. The probability of no calls during 10 seconds is P(Y=0) = exp(-5/3) = 0.189.
8)
(a)
Average number of blackflies in a tent of 40 cubic feet is 2/5*40 = 16. Hence, the number of blackflies in 40 cubic feet X has Poisson distribution with parameter L = 16.
P(X=2) = 162 / 2! * exp(-16) = 1.44 * 10-5
(b)
Average number L = 2/5 in 1 cubic foot. The answer is 1 P(X=0) = 1 exp(-2/5) = 0.33.
(c)
(i) Average number is L = 2/5*40*3 = 48 in three tents.
P(X=40) = 4840 / 40! * exp(-48) = 0.031
(ii) We know from (a) the probability that there are exactly 2 blackflies in 1 tent, P(X=2).
The claimed probability is 3 * P(X=2)2 * (1 P(X=2)) = 6.221 * 10-10
9)
(a) P(G) = , P(R) = 1/3, P(U) = 1/6.
We have a sequence of 6 items. There are C(6,3) ways to choose places for G-items, and C(3,2) to choose places for R-items in this sequence. Hence, the answer is
C(6,3)*C(3,2)*P(G)3P(R)2P(U) = 5/36
(b)
There are 6 possibilities to choose a place for U; then, we should have other 5 items such that there are no U and at least one R. The probability of no U among 5 is (1-P(U))5, and we must subtract the probability of no U and no R (i.e., only Gs) which is P(G)5. The answer is
6*P(U)*( (1-P(U))5 P(G)5 ) = 0.371
(c)
This means not all not-R: 1 (1-P(R))6 = 0.912
(d)
This means not all G: 1 P(G)6 = 0.984
10)
It is time to apply both strategies we used in #8 and #9. On the average, there are 2/5*20 = 8 blackflies in 1 tent. So, the number of blackflies in one tent X has a Poisson distribution with the parameter L = 8.
(a)
C(20,6) to choose tents with no blackflies, C(14,4) to choose those which contain 1 or 2, C(10,3) to choose those with 3, 4 or 5, and C(7,6)=7 to choose those which contain more than 10 (or one tent we dont know anything about). The answer will be
C(20,6)*C(14,4)*C(10,3)*7*P(X=0)6 (P(X=1) + P(X=2))4 (P(X=3)+P(X=4)+P(X=5))3 P(X>10)6
I dont want to calculate this. A very small number.
(b)
By the definition of conditional probability, the answer is the probability of (a) divided by the probability that 6 of the tents contain no blackflies. The latter probability is C(20,6)*P(X=0)6.
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