Math Is Fun Forum

this is the root cause for my last thread.

In genetics, as some of you may know, we often inherit recessive traits from our parents, but they are not manifested if we get a dominant trait from our other parent. However, the recessive traits that are not manifested are still contained in us and can be passed on to our children, thus given them characteristics which we don't have.

Basically the way it works is this. For blue eyes and brown eyes, brown eyes is a dominant trait (B) and blue eyes are a recessive trait (b).

Now there are 3 possible combinations for blue and brown traits within a person:

BB, Bb, bb (note, order is not considered)

if a person has BB or Bb, they posess the dominant brown eyed trait B. The only way a person can be blue eyed is if they are both blue eyed traits (bb).

Now here's where the probabity kicks in. If a person with one Brown eyed trait and one blue eyed trait (Bb, which means he/she has brown eyes) mates with a person who is also Bb, then what happens is one random gene from the first parent is paired with one random gene fom the second parent, and this pair is given to the offspring. To work out the probability and possibility for certain combinations, they usually draw what they call a punett square
_B_b
B
b
then consider the intersections of each row and collumn as a possible combination. From this we see the possible combinations are
BB, Bb, Bb, and bb. There is therefore a 75% chance that the offspring will be brown eyed, and 25% that they will be blue eyed. Moreover, there is a 25% chance they will be BB specifically, a 50% chance they will be Bb, and a 25% chance they will be bb.

Its all rather simple really. Hope that makes sense. They've been discussing this for like 2 weeks in my biology class but it seems there is little more to it than this.

What i'm wondering is what do you do if you have a problem like this:
suppose a brown eyed man and a blue eyed woman have a child.  What is the probability that the child will be BB, bb, or Bb?

This problem is a bit difficult because while we know the mother is bb, (else she wouldn't have blue eyes) the father, being brown eyed, could be either BB, or Bb. So what is the probability?

If the father is BB, then we end up with a 100% chance of Bb, (BB crossed with bb, take one from each pair and you always get bb)

If the father is Bb, then we end up with a 50% chance of bb, and a 50% chance of Bb, (Bb crossed with bb, take one from each pair and you either get Bb or bb)

so the question is, 100% chance of Bb if the father is BB, and 50/50 chance of Bb or bb if the father is Bb. Note, there is a 50/50 chance of the father being BB or BB.

so now what do we do? We have (Bb) or (bb) evenly distributed on one half of the scale, and only Bb on the other half. But both halfs can't happen at once. So what do we do now?

Moreover, suppose it says both parents are brown eyed. Well that could mean parent 1 is Bb while parent two is Bb, or it could mean parent 1 is BB and parent 2 is BB OR it could mean parent 1 is BB or parent 2 is Bb! Each of these scenarios brings with them their own set of probabilities for their respective offspring. So what is the probability all four considered? That is, what is the probability that a given combination (such as Bb) will occur?

there will probably not be questions of this calibur on the test but, i want to be ready for it just in case.

suppose you flip a coin, if its heads, i have to guess a number between 1 and 6 and roll a die to see if i'm right. If its tails, i have to guess a number between 1 and 12 and role a 12 sided die.

What are the odds that i will guess correct?

I was working on some recreational maths, and I was faced with a problem, a problem which required a proof. I tried several times to disprove it without luck so.. i then decided to try to prove it, and...  I think i did it. But i'm not certain. Hopefully one of you can verify if its valid and if not, where the error in my logic is.

show that there is some solution, y = y1, to the inequality

(1): (m + b)y <= (q + d)

such that (y = y1, x = x1) is a solution to the inequalities

a(x) + b(y) <= d AND
-a(x) + m(y) <= q

for some value x1 (which must be shown to exist) and for any non zero real number a.

suppose now, that  m + b != 0, then it just so happens that the system of equations

a(x) + b(y) = d and
-a(x) + m(y) = q has a solution, which also happens to give you y = (q + d)/(b+m) which satisfies (1), and x = (d - by)/a. (recall a != 0)

thus the three inequalities

(m + b)y <= (q+d)
a(x1) + b(y1) <= d
-a(x1) + m(y1) <= q

have a solution where m + b != 0, namely, the solution to the two equations listed above.

Now suppose that (m+b) = 0, we have then that (m+b)y <= q + d,  and it must be so that q + d >= 0. Moreover, y may be anything at all and will still satisfy (1).

we must now show that

a(x) + b(y) <= d AND
-a(x) + m(y) <= q

has a solution, and since m = -b, we need only show that

a(x) + b(y) <= d AND
-a(x) -b(y) <= q     

for some two values (x,y) with no restrictions placed on their values. Equivilently, this can be written as

-q <= a(x1) + b(y1) <= d,

note that 0 <= q + d so -q <= d, in otherwords, there must lie some number s between -q and d, if we therefore chose any values of x and y such that a(x)+ b(y) = s (trivial) all inequalities are thereby satisfied and the proof is complete. ∈

that worked out a little too nicely, especially with the q + p >= 0 part. Is this proof legit? or do i need more sleep?