1) Consider the genetic markers Kell and Duffy. For both these markers a person may be either positive (+) or negative (-).\nIt is known that P(Kell+) = 0.6 and P(Duffy+ | Kell+) = 17/30.\n\na)What is the probability a person chosen at random is both Kell+  and Duffy+ ?\nb)What is P(Duffy- | Kell+) ?
215829
a) P(D+|K+) = P(D+ and K+)/P(K+)\n17/30 = P(D+ and K+)/.6\nP(D+ and K+) = .34\n\nb) P(D- | K+) = P(D- and K+)/P(K+)\n=P(D- and K+)/.6\n=(P(K+) - P(D+ and K+))/.6\n=.26/.6\n= 13/30
35666
F- even and odds.  There are 3 even sides and 3 odd sides. So the likelihood is the same for both.
299433
This problem is solvable.\nProbability of event A is written P(A).\n\nP(A)= Total # of outcomes that correspond to event A (divided by -or- over) Total # of outcomes in sample\n\nAnswer:\nP(A) = 12/20 = 3/5
110612
P(Yes OR Female) = P(Yes) + P(Female) - P(Yes AND Female)\n                 = (100/167) +  (147/167)  - (86/167)\n                 = 0.964
616006
the others are correct, it is 1/120.\n\nIf you are not sure how to do the calculation, it is 3,000,000 divided by 25,000.\n\nOr you can cut it down to:\n\n  25  \n------\n3000\n\n  5\n-----\n600\n\n\nThis cancels down to 1/120.\n\nThat means that the probability is 1 in 120, or 1:120  (one person out of every 120 people has VD) or 8.33% of the population has VD.   (The 3 is recurring - you show that by putting a dot above the last 3)\n\nWhy did you need to ask us what the answer was when you could work it out yourself?
339424
First you have to make out your alleles.  Your first parent PpLl will produce spermatozoids that can be either PL Pl pL pl.  As for the mother, she is also heterozygotous for both traits, so she is also PpLl and will produce eggs that are either PL Pl pL or pl.  The you make the Punnett square with all those four colomns and four rows.  You should get something like\n___|_PL__|_Pl_|_pL_|_pl__\nPL_|_PPLL|_PPLl_|_PpLL_|_PpLl_\nPl_|PPLl|_PPll_|_PpLl_|_Ppll_\npL_|PpLL_|_PpLl_|_ppLL_|_ppLl_\npl_|PpLl_|_Ppll_|_ppLl_|_ppll_\n\nOKAY NOW...\nPPLL means healthy all the way\nppll means has all the diseases\nIn between they are heterozygotous for one or both disease.\n\nSince you have 16 possibilities, then yes your 1/16 answer is correct for the probabilities of having both diseases\n\ncan you do the square for your second problem :-)
540991
This probability problem can be used using Binomial Distribution.  \n\nGiven a number of 'x' success, the probability in a binomial experiment is:\n\nP(x) = [n choose x]*p^x*q^(n-x)\n\nwhere:\n\nn choose x is a combination problem: n!/[(n-x)!x!]\np = probability of success\nq = probability of failure\nn = total selection\nx = successes\n\nIn this problem, we are looking for at least one case has a success (which is one of the highest 9 numbers).  In essence, we are solving for:\n\nP(x=1) + P(x=2) + P(x=3) + P(x=4) + P(X=5).  \n\nHowever it would be easier to find the probability of the failures and subtract from the total probability because we only have to calculate 1 probability.  This would be:\n\n1 - P(x=0) is what we are looking for.  Now all we need is to calculate the probabilty.\n\nFor P(x=0), we have the following:\n\nn = 5\nx = 0\np = (9/26) = probability of success\nq = (17/26) = probability of failure\n\nPlugging into P(x), we get:\n\nP(x=0) = (5!/5!)(9/26)^0(17/26)^5\n\nSolving, you get P(x=0) = 0.12\n\nTherefore, \n\n1 - P(x=0) is ------>\n\n1- 0.12 = 0.88\n\nBelow is a website devoted to Binomial Distribution exercises\n---------\n\nHope this helps
549080
b) and c) are pretty straightforward.  \n\nb) is "not excellent" (.80 chance) on 1, and excellent (.15 chance) on 2.  .80 * .15 = 0.12, or 12% chance.\n\nc) is "not excellent" on either, so .80 * .85 = 0.68, or 68%.\n\na) is requires a more complicated formula, and I haven't taken "sadistics" in about 6 years, so I can't help you there.
541604
First the probability that a family of 4 is all girls is:\n1/16 (and the chance that they are not is 15/16).\n\nRather than figuring the chance of at least one family having this arrangement, figure the probability that *none* of the families do.  Then subtract from 1.\n\nThe answer will be 1 - (15/16)^10\nor about 47.6%\n\nIt's not a simple multiplication by n... if it were, then with 16 families you should have a 100% chance.  And with 32 families you would have a 200% chance.  That logic would be correct if you couldn't repeat a combination, but you can.  Just because family A has four boys, for example, doesn't mean family B can't also have four boys.\n\nLabel the 10 families as A, B, C, D, ... , I, J.  The probability that A has four girls is 1/16.  The probability that B has four girls is 1/16, etc.\n\nThe chance that *none* of them have four girls is 15/16 x 15/16 x 15/16 x ... x 15/16.  This is (15/16)^10 or about 52.4%.  Therefore the chance that at least one of them has four girls is 100 - 52.4 or 47.6%
504113
When you are computing the probability of a given event such as 2 heads, 2 tails, or one of each, you are essentially counting the number of ways a given event can occur and dividing this number by the number of all possible outcomes.\n\nIf you toss 2 coins once ( or one coin twice) there are 4 possible outcomes:\n\nHH = both heads\nHT = 1st heads, 2nd tails\nTH = 1st tails, 2nd heads\nTT = both tails\n\nAssuming the coins are "fair" each outcome is equally likely.\n\nThere is one way in four that two heads (or 2 tails) can happen so the probability for each of these outcomes is 1/4\nBut there are two ways you can have one head and one tail and so the probability of this happening is 2/4 =1/2. The sum of all the probabilities of the distinct outcomes will always equal 1. \n1/4 +1/4 + 1/2 =1\n\nHow about if you have 3 tosses? The 8 possible outcomes are as follows:\n\nHHH \nHHT\nHTH\nTHH\nHTT\nTHT\nTTH\nTTT\nWe see there is only one way to have 3 heads (or 3 tails) and so the probability of 3 heads (or 3 tails) = 1/8\nHowever there are 3 ways you can have 2 heads and one tail, so the probability of this happening is 3/8. And there are also 3 ways you can have one head and 2 tails, so 3/8 is the probability for this too.
347949
When it states the frequency of the dominant allele in a population, think of p as one allele that 36% of the population is carrying. so, we can set p=36%, or .36\n\nThe next part of the problem asks for the frequency not of the recessive allele, but of the recessive trait. In order to carry the recessive trait, the person must have both alleles. The probability that both parents passed down the allele for the person to have the trait, using the multiplicative rule of probability would be q*q. We are solving for q^2 since we are looking for the trait not the allele. \n\n1-p=q\n1-.36=.64\nq*q=.64*.64=0.4096\n0.4096*100%=40.96%
473901
For the first one:\n\nz = (2725-2500)/150 = 1.5\n\nThe area would be 0.0668.\n\nFor the second one: You are given that X (the tensile strength) is greater than 2775, and you want to find the probability that X is greater than 2725.  This probability can be written this way.\n\nP(X > 2775 | X > 2725)\n\nRemember how to compute a conditional probability.\n\nP(A|B) = P(A and B)/P(B)\n\nso...\n\nP(X > 2775 | X > 2725) = P(X>2775 and X>2725)/P(X>2725)\n= P(X>2775)/P(X>2725) (The only way X>2775 and X>2725 is if X>2775.)\n\nConverting 2775 to standard units...\nz = (2775-2500)/150 = 1.83\n\nSo 0.0336/0.0668 = 0.503 approximately\n\nThe third one is zero.  Since a normal is continuous, the probability of any particular number is zero.  Think of trying to find the area of a line; it has length but no width.\n\nedit: jojo, you did not read very carefully.  This is not refering to a hypothesis test, and you did not write the second problem correctly.  It is a conditional probability.
152857
This is easy, just say there are 100 shaft holders.  So of A, 50 of them are processed there, and 3% don't work, so 1.5 out of 50.  For B, it is 30 x 4%=1.2, and C is 20 x 5%=1.  \n\nSo out of a total 100, 3.7 don't work, the the probability that this one chosen at random wont work is 3.7%.  \n\nHopefully the way i calculated the answer helps you out in the future.
622872
0.8037 or 80.37%\n\nThis is why:\n\nP(E)=1-P(not E); so the probability of at least 2 people having the same number is 1- the probability that each person has a unique number.\n\nThe probability that each person has a unique number:\n\nFirst person has 100 choices out of 100 numbers, so 100/100.\nSecond person can't choose the first person's number, so only 99 choices out of 100, so 99/100;\nThird only 98/100;\n. . .\n18th person, has (100-17)/100=84/100\n\nTherefore the probability that none of these people have the same number is (100/100)•(99/100)•(98/100)•(97/100)•. . .•(84/100) ≈19.63%\n\nSo the probability that at least two people have the same number is 100%-19.63%=80.37%.
633273
25% chance of being O\n75% chance of being A\nonly if both parents are heterozygous\n\neven if one is homozygous no chance of getting "O"
545904
Okay, keep in mind that this problem is asking the probablility that AT LEAST 8 were married - that means the probability that 8 were married plus the probability that 9 were married plus the probablility that 10 were married.\n\nSo let's start with the probability that 8 were married:\n\nIf the probability that someone was never married is .15, that means the probability that they were married is 1 minus .15, or .85. And if 8 people in the sample were married, that means that 2 were not. So what you're going to do is multiply .85^8 (for married people) x .15^2 (for unmarried). \n\nThe principle is the same for 9 married people and ten married people: \n\nFor 9 married, you multiply .85^9 x .15^1\n\nFor 10 married, you multiply .85^10 x .15^0\n\nFinally, add all three answers together, which gives you the probability that either 8, 9, or 10 people in the sample were married, which is what the question was asking! Hope that helped!
504520
c is the right answer.\n\nA makes no sense in relation to the question\nB natural selection has a bit of randomness in it of course but is basically not random. I mean, the fastest critter in the herd may have had an unlucky day and so he will not procreate, oops. But otherwise, the slowest are intentionally selected by the lion and that is never random.\nD just isn't true, plenty of proof there.