Probability (iii)Questions:pls solve:?
415287
Although this seems to be a problem about Gregor Mendel and his tall and dwarf pea plants, once it is realised that the probability of any child being short is 1/2, it simplifies to being the simpler and more familiar one of tossing a coin, heads or tails, n times\n\nIt would have produced more interesting results perhaps if the parents were Aa and Aa and three children out of four were therefore tall and we had looked at grandchildren (offspring of children who are one quarter AA, one half Aa and one quarter aa, eg if they intermarry with another such distribution among their partners.) However ...\n\n(i) is imprecisely worded. How big is the family?\n\nIf there is one child, the chance of that child being short is  1/2\n\nIf there are two children, the chance that both are tall is 1/4 and the chance that at least one child is short is 3/4. (Of one being short is 1/2 and of both being short is a quarter.)\n\nIf there are three children, the chance that all three are tall is 1/8 and the chance that at least one child is short is 7/8 (of one being short: 3/8, of two being short also 3/8, of all three being short: 1/8).\n\nWith 4 children, the chance of at least 1 being short is 15/16, of 1  being short is 4/16 or 1/4; of 2 being short is 6/16 or 3/8; of 3   being short is 4/16 or 1/4; of all 4 being short is 1/16;\n\nOne just reads across successive lines of Pascal's Triangle.\n\n(ii) the answer to this is buried in the above: 3/8ths\n\n(iii) again, imprecisely phrased. Does "such families" refer back to (i) or to (ii)? Taking it to mean families of four children, there is one chance in 16 of 4 tall children, 4 chances in 16 of 3 tall children, 6 chances in 16 of 2 tall children, 4 chances in 16 of 1 tall child and 1 chance in 16 of 0 tall children.\n\nIn 16 such families, there are 64 children and unsurprisingly 32 are tall\n(1 x 4) + (4 x 3) + (6 x 2) + (4 x 1) + 0\n\nwith an average number of 2 tall children per family, and as you would expect an average number of 2 short children per family, too.\n\n(iv) 256 = 16 squared and so the expectation is 16 families out of 256 will have 4 tall children, given 1 family in 16 has this particular expectation.
1113295
(1, 2, 3,4) or (-1, -2, -3, -4) or (1, 2, -3, -4) or (1, -2, -3, 4)\nYou also have\n(-1, -2, 3, 4), (-1, 2, -3, 4), (-1, 2, 3, -4), and (1, -2, 3, -4)\nSo there are 8 possibilities.\nEach having a probability of 1/8 or occurring.\nFor (1,2,3,4) and (-1,2,-3,4), the prob. of an even natural number is 2/4 = 1/2\nMultiply this by 1/8 to get the conditional prob.\n(1/2)(1/8) = 1/16\nFor each of the following: (1,2,-3,-4), (1,-2,-3,4), (-1, -2, 3, 4), and (-1, 2, 3, -4), the prob. of an even natural number is 1/4\n(1/4)(1/8) = 1/32\nFor (-1,-2,-3,-4) and (1,-2,3,-4), the prob. of an even natural number is 0.\n0(1/8) = 0\nAdd all these together.\n2(1/16) + 4(1/32) + 2(0)\n= 1/8 + 1/8\n= 2/8\n= 1/4\nThis doesn't match your solution, but you can see if this makes any sense to you.
821305
a) effective or not - 2 choices - binomial\nb) smoke or note - 2 choices -binomial\nc) answers could be 1,2,3,4,5,6 . . .therefore NOT binomial
955894
x1*p(x1) + x2*p(x2)\n\nx1 = 1 for  event dots> 3,\nx2 = 4 for event  dots<=3.  \nThese are equally probable because there are 6 dots on the die.  \n\n1*(1/2) + 4*(1/2)
867162
5| 2 -9 -7 13\n..... 10 ..5 -10\n-----------------\n....2 1 ..-2 3\n\nSo P(5) = 3
672123
The probability of A is 26%.  Probability of B is 42%.  Both A and B is 7%.  Probability of neither = 100% - [(26%+42%)-7%] = 39%.  Or in your format, 0.39.
631161
This is a binomial probability.\nP(success) = 1/5, P(failure) = 4/5.  There are 6 events here, so P(2 successs) = (1/5)(1/5)(4/5)(4/5)(4/5)(4/5).\nP(at least 2 successes) = P(2 successes) + P93 successes) + . . . . P(6 successes) or 1 - P(O successes) + P(1 success) + P(2 successes) which you already computed above.\nLet me know if you don't understand.
1155743
Mutually exclusive - They both cannot occur at the same time.\nTherefore P(A and B) = 0\nP(A or B) = .5
733920
a) There are 16 combinations and 4 of them include exactly 3 heads (HHHT, HHTH, HTHH, THHH).  So the probability is 4/16 or 1/4.\n\nb) This question results in the same probability (4/16 = 1/4), except it asks for it as 'odds' which is the probability of success divided by the probability of failure.  That is 1 to 3 odds.\n\nc) The odds against heads are just the reverse.  Probability of failure divided by probability of success.  That is 3 to 1 odds against.
1169384
The previous responders are incorrect. \n\nLet the expected length of the game (in tosses) be L. We can classify each game into one of three starting sequences (in the following table P=probability,E=expectation value for total length of game):\n\nH --> P = 1/2 , E = L+1\nTH --> P = 1/4 , E = L+2\nTT --> P = 1/4 , E = 2\n\nThe first two cases are equivalent to starting the game over after 1 and 2 tosses respectively.\n\nSumming P*E for these three events gives L:\n\nL =(1/2)*(L+1) + (1/4)*(L+2) + (1/4)*(2)\n= (3/4)L + 3/2\n\n--> L = 6
686917
As it is certain that the first one is red and the second one is blue, we can ignore the fact that they were even mentioned. (It would be different if you phrased it as find the probability of the first ball being red, the second being blue and the third being yellow...)\n\nP(Yellow) = 1/5
194085
In the experiment of choosing a committee of 6 from a group of 8 women and 7 men,\n\nTotal no. of possible choices \n= no. of ways in which the 6 members of the committee can be chosen from the total 15\n==> n = C(15,6)\n==> n = 15*14*13*12*11*10/6*5*4*3*2*1\n==> n = 7*13*11*5\n==> n = 5005\n\nLet "A" be the event of choosing the committee with 3 women and 3 men\n\nFor Event "A"\n\n---------- Men----Women---Total\nAvailable---8-------7------15\nTo Choose---3-------3-------6\nChoices----C(8,3)-C(7,3)--C(15,6)\n\nNo. of favorable choices\n= (no. of ways in which the 3 men can be chosen from the total 8) * (no. of ways in which the 3 wommen can be chosen from the total 7) \n==> m(a) = C(8,3) * C(7,3)\n==> m(a) = 8*7*6/3*2*1 * 7*6*5/3*2*1\n==> m(a) = 8*7 * 7*5\n==> m(a) = 1960\n\nProbabilty of choosing the committe with 3 women and 3 men \n==> Probability of occurance of event "A" = m(a)/n\n==> P(A) = 1960/5005\n==> P(A) = 392/1001\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/
395430
Assuming that the chance of having a girl is 50%, the chance would be 3/8.  This answer can be most simply reached a couple different ways.  First, you could list the possibilities:\n\nBBB\nBBG\nBGB\nGBB\nGGB\nGBG\nBGG\nGGG\n\n3 of the 8 possibilities fit the condition (exactly one girl), so the probability is 3/8.\n\nAnother way, better for larger numbers where the above method is impractical, is to use permutations.  The above works fine for this problem, though.
67102
43 out of 400.\nThere is 120*120 pairs of minuites within which in which they can arrive resturent. And to be within 3 minuite, the diffrence of this of this  two pair have to be 3 or less. If the first one of this is between 4 to 117, there will be 7 possibility of  second one.\nAnd teher will be 6 for 3 and 118, 5 for 2 and 119,and 4 for 1 and 120.\nso total possibolity is 114*8+6*2+5*2+4*2 that is 828\nSo the probability is 828/14400=43/200.
906595
C.\nThis is a binomial probability problem.\nThe binomial formula says \nP(r out of n) = (nCr)(p^r)(q^(n-r)), where n is the number of trials (here 6 rolls), r is the number of successes (here 0, 1, 2, and 3 twos), p is the probability of success (here 1/6), and q is the probability of failure (here 5/6)\nYou need to plug in 0, 1, 2, and 3 for "r" into this formula.  If you do, you get:\nP(0 twos) = .3348979767\nP(1 two) = .401877572\nP(2 twos) = .200938786\nP(4 twos) = .05358367627\nYou add up these answers to get your final answer, about .9913
799156
Use Binomial distribution:\n\n(n nCR x) p^x (1-p)^(n-x)\n\nn nCR x = n!/(x!(n-x)!)\np=probabiliy (.10)\nn=patients (10)\nx= desired\n\na) P(X<=2) = P(X=0) + P(X=1) + P (X=2)\nb) 1- P (X<=2)... 1-prevoius answer
95999
Don't you think you could have written this more clearly? \n\nIf there is a uniform distribution of the weight w being between 5 and 10, then each of these probabilities is equal:\n\nP5_6=P{5<=w<6}\nP6_7=P{6<= w<7}\nP7_8=P{7<=w<8}\nP8_9=P{8<=w<9}\nP9_10=P{9<=w<10} \n\nThe sum of these is 1, so \n\n1=P5_6+P6_7+P7_8+P8_9+P9_10, and\nP5_6=P6_7=P7_8-P8_9=P9_10\n\nso 1=5xP9_10, so P9_10=0.2\n\nThe probability is 0.2.