Independent/dependent probability summation

Hi all,
This problem is an example in the book "Introduction To Probability", and I am having trouble understanding a specific part in it. The problems is as follows: two design teams called A and B are given a design task; the probability of success for A is 2/3, and for B it's 1/2. The probability that at least one team will succeed is 3/4. If both teams are successful, the design of B is adopted, assuming that exactly 1 successful design is produced, what is the probability that is was designed by team N?

The given solution in the book: There are four possible outcomes - SS, SF, FS, and FF (the first letter corresponds to team A, the latter for team B; S stands for success, F stands for failure).
The given probabilities are as follows(SS) + P(SF) = 2/3, P(SS) + P(FS) = 1/2, P(SS) + P(SF) + P(FS) = 3/4
From these relations, together with the normalization equation P(SS) + P(SF) + P(FS) + P(FF) = 1, we can obtain the probabilities of all the outcomes.

This is the part I can't figure out. How are the above relations used to work out the probabilites for the individual events? The book goes on to specify that P(SS) = 5/12, P(SF) = 1/4, P(FS) = 1/12, P(FF) = 1/4. I'm not really sure how these were worked out, having spent quite some time trying to work it out but to no avail. Any thoughts or help would be most appreciated.

tytytyty said:

Hi all,
This problem is an example in the book "Introduction To Probability", and I am having trouble understanding a specific part in it. The problems is as follows: two design teams called A and B are given a design task; the probability of success for A is 2/3, and for B it's 1/2. The probability that at least one team will succeed is 3/4. If both teams are successful, the design of B is adopted, assuming that exactly 1 successful design is produced, what is the probability that is was designed by team N?

The given solution in the book: There are four possible outcomes - SS, SF, FS, and FF (the first letter corresponds to team A, the latter for team B; S stands for success, F stands for failure).
The given probabilities are as follows(SS) + P(SF) = 2/3, P(SS) + P(FS) = 1/2, P(SS) + P(SF) + P(FS) = 3/4
From these relations, together with the normalization equation P(SS) + P(SF) + P(FS) + P(FF) = 1, we can obtain the probabilities of all the outcomes.

This is the part I can't figure out. How are the above relations used to work out the probabilites for the individual events? The book goes on to specify that P(SS) = 5/12, P(SF) = 1/4, P(FS) = 1/12, P(FF) = 1/4. I'm not really sure how these were worked out, having spent quite some time trying to work it out but to no avail. Any thoughts or help would be most appreciated.

Click to expand...

I always draw a Venn diagram for this type of question - two overlapping circles labeled A and B. P(SS) is the intersection, and P(FF) is the exterior, not in either circle. That area is 1/4. P(A)=2/3, P(B)=1/2, and the union is P(A) + P(B) - P(SS) = 3/4.

The region corresponding to "exactly one" is the sum [P(A) & notB] + [P(B & notA)]
The conditional probability you are looking for is P(B | "exactly one")

Does that help?

tytytyty said:

Hi all,
This problem is an example in the book "Introduction To Probability", and I am having trouble understanding a specific part in it. The problems is as follows: two design teams called A and B are given a design task; the probability of success for A is 2/3, and for B it's 1/2. The probability that at least one team will succeed is 3/4. If both teams are successful, the design of B is adopted, assuming that exactly 1 successful design is produced, what is the probability that is was designed by team N?

The given solution in the book: There are four possible outcomes - SS, SF, FS, and FF (the first letter corresponds to team A, the latter for team B; S stands for success, F stands for failure).
The given probabilities are as follows(SS) + P(SF) = 2/3, P(SS) + P(FS) = 1/2, P(SS) + P(SF) + P(FS) = 3/4
From these relations, together with the normalization equation P(SS) + P(SF) + P(FS) + P(FF) = 1, we can obtain the probabilities of all the outcomes.

This is the part I can't figure out. How are the above relations used to work out the probabilites for the individual events? The book goes on to specify that P(SS) = 5/12, P(SF) = 1/4, P(FS) = 1/12, P(FF) = 1/4. I'm not really sure how these were worked out, having spent quite some time trying to work it out but to no avail. Any thoughts or help would be most appreciated.

Click to expand...

Build a table. Leave some room at the top. On the left, write SX. Next line down, write FX. Third line down, write SX or FX. Beside each set of words, draw three adjacent boxes. On top, where you left room, write YS above the leftmost set of boxes, YF above the middle set of boxes, and YS or YF above the rightmost set of boxes. You now have nine boxes. The first row of boxes running across gives you places to put the probabilities for SS, SF, and SS or SF when you know them. The second row gives you places to put the probabilities for FS, FF, and FS or FF when you know them. The first two boxes in the bottom row give you places to put the probabilities for SS or FS and SF or FF respectively. The last box in the lowest row gives you a place to put the probability of SS or SF or FS or FF. But those four possibilities are mutually exclusive and exhaustive so they add up to what? Now look at the last box in the first row. You are supposed to put the probability of SS or SF in that box, and the problem gives you the value to put in there. Now look at the box below that. You are supposed to put the probability of FS or FF in that box. Well you know that the box above holds the probability of SS or SF and the box below holds the probability of SS or SF or FS or FF so can you figure out what goes in the middle box in the rightmost column. Now deal with the bottom row in the same way. You now have filled in the rightmost column and bottom-most row. If you knew just ONE probability to put in one of the remaining boxes, you could figure out all the remaining boxes. The problem gives you the information needed to fill in one of the boxes.

DrPhil said:

I always draw a Venn diagram for this type of question - two overlapping circles labeled A and B. P(SS) is the intersection, and P(FF) is the exterior, not in either circle. That area is 1/4. P(A)=2/3, P(B)=1/2, and the union is P(A) + P(B) - P(SS) = 3/4.

The region corresponding to "exactly one" is the sum [P(A) & notB] + [P(B & notA)]
The conditional probability you are looking for is P(B | "exactly one")

Does that help?

Click to expand...

Thanks very much, the Venn diagram explained everything. I shouldn't have tried to do it in my head, cheers.