Question

(supposedly) Elementary probability:
A conservative design team, call it C, and an innovative design team, call it N, are asked to separately design a new product within a month. We know from past experience that:
(a) The probability that team C is successful is 2/3
(b) The probability that team N is successful is 1/2
(c) The probability that at least one team is successful is 3/4
Assume that exactly one successful design is produced. What is the probability that it was produced by team N?

Answer 1

I already have the answer form the book, I just want to see other approaches.

Answer 2

is answer 1/12 ?

Answer 3

no, should I say what it is?

Answer 4

gotta love how this is the first problem in the book... MIT lol

Answer 5

looks like i have to use conditional probability.

Answer 6

yeah

Answer 7

it's in the section on conditional probability, but their solution hardly seems to involve the formula they use

Answer 8

I mean, their solution doesn't involve the actual formula for conditional probability, which is how I tried (and failed) to solve it. Hence I want to see another approach.

Answer 9

by that formula ... i seem to get 6/4>1

Answer 10

I got something>1 when I tried to use the formula as well

Answer 11

last guess is it 1/24 ?

Answer 12

no

Answer 13

looks like i am out of guess. damn probability ... never been my forte.

Answer 14

should I write their solution, or just the answer, or neither and just try to remember how I tried to work it.
This is my first time playing with probability. I like it, but it sure is sneaky.

Answer 15

probability is indeed an interesting topic :p

Answer 16

(a) The probability that team C is successful is 2/3 = P(C)
(b) The probability that team N is successful is 1/2 = P(N)
(c) The probability that at least one team is successful is 3/4 = P(CUN)
using inclusion exclusion i seem to get \( P(C \cap N) = 5/12 \)

Answer 17

another guess ... is it 1/3 ?

Answer 18

...sorry, but still no

Answer 19

It frustrates me that in their solution they don't even speak of \(P(N\cap C),~P(N\cup C)\) or even P(N) or P(C)

Answer 20

woops!! 1/3 wouldn't work

Answer 21

If I were to try to model the problem using the formula I am setting up\[P(N|\{N^cC, C^cN\})\]

Answer 22

I got 2/9 this time.. which is still wrong.

Answer 23

what is the final answer?

Answer 24

actually using my method I got 5/6
the actual answer is 1/4

Answer 25

should I show their working?

Answer 26

no

Answer 27

The probability of at least one team being successful is 1 minus the probability that both teams fail, 1 - (1/3)(1/2)= 5/6.

Answer 28

you didn't make use of the probability of either being successful as 3/4

Answer 29

Use the odds ratio to get the probability that it was one team rather than the other.

Answer 30

I did not use that probability because I think it was mistaken. If the probabilities are independent, then
1 - both fail = at least one succeeds.

Answer 31

how can it be mistaken? it's a given in the problem.

Answer 32

even if I ignore that, I don't see how your method will give the right answer

Answer 33

How indeed? ! Sometimes the problems are wrong. How can it be correct? Are they partly collaborating How not independent? Perhaps we are not getting the correct answer because the statement was wrong but their solution was right.

Answer 34

well, keep in mind this is the first worked problem in a book that has been used for 40-ish years at MIT, so I highly doubt they are mistaken about anything.

Answer 35

If we have the probability that one team got it right, then the odds favor the team that is more often successful by the ration of the two success rates. "At least one" is the joker here, and my odds ratio may fail for this reason.

Answer 36

I grant you that it is unlikely that the error has remained for over 40 years.

Answer 37

1 - both fail = at least one succeeds=5/6=P(N C)+P(N ~C) +P(~N C)
how would you propose to wrestle P(N ~C), which is what we want, out of this?

Answer 38

Given that one was successful, the odds ratio of its being done by N seems to be (1/2)/(2/3) = 3/4 and: 1 and the probability is (3/4)/((1+3/4) = 0.43. x:y odds ratio is x/(x+y) probability.

Answer 39

I guess I just don't follow, but it still doesn't yield 1/4

Answer 40

I am probably wrong.

Answer 41

its bayes theorem

Answer 42

lemme post the question again so i dont hafta scroll up again

Answer 43

(supposedly) Elementary probability:
A conservative design team, call it C, and an innovative design team, call it N, are asked to separately design a new product within a month. We know from past experience that:
(a) The probability that team C is successful is 2/3
(b) The probability that team N is successful is 1/2
(c) The probability that at least one team is successful is 3/4
Assume that exactly one successful design is produced. What is the probability that it was produced by team N?

Answer 44

Bayes theorem sounds plausible, P(this|that).

Answer 45

we are nowhere near Baye's theorem in the book yet, but if it works I'd like to to see it.

Answer 46

lemme try

Answer 47

is the answer 3/7?

Answer 48

http://www.insidemathematics.org/problems-of-the-month/pom-growingstaircases.pdf

Answer 49

no, I wrote the answer above a few times if you want it

Answer 50

look at the link I posted

Answer 51

what about it?

Answer 52

It is just showing some ideas of how to arrive at the answer

Answer 53

explain

Answer 54

seems quite irrelevant to me, what part specifically helps in this problem?

Answer 55

that's my book ....

Answer 56

so that's the solution I *don't* want

Answer 57

lol..it shows how to get the answer

Answer 58

yes, I'm quite aware of that

Answer 59

sorry...I tried

Answer 60

sorry I wanted to delete that so others don't read it... which I'm sure a few have by now

Answer 61

my bad....I didn't know

Answer 62

I know, it's cool
I think I'll repost this Q, this thread is too long anyway

Answer 63

Afterthought: I see I mistakenly equated the idea that the teams worked independently to the probabilities' being independent, no
overlap P(N v C)=0, when their talents and training could/did mean significant overlap.