first methoduse binomial distribution to solve this.. here we have 4 independent throws. in each throw the sample space is (1,2,3,4,5,6) so the prob of getting a 6 is 1/6.so our p=1/6 i.e prob of coming 6 and q= 1-p = 1-(1/6)=5/6 i.e prob of not coming 6 is 5/6p= prob of success. here our success event is getting 6 q= prob of failure. here our failure event is not getting 6no of trials n=4now using the binomial distribution formula. (C signifies combination)c(n,r)*p^r(p raised to power r)*q^n-r(q raised to power (n-r))here we want at least two 6. i.e. desirable outcomes are two 6 or three 6 or four 6.because at least two 6 doesn't mean that we can't get more than two 6 . so r in these cases take the value of 2,3,4 respectively.so for r=2c(4,2)*(1/6)^2*(5/6)^2= upon solving we get 150/1296for r=3c(4,3)*(1/6)^3*(5/6)^1 = 20/1296for r = 4c(4,4)*(1/6)^4*(5/6)^0 = 1/1296here we will add these prob ..Getting two 6 fulfills our condition so we will stop and calculate the prob and keep it asidenow we got another way of fulfilling the condition by getting three 6 again our work is done so we will stop and calculate the prob and keep it asideagain we got another way of fulfilling the condition by getting four 6 again our work is done so we will stop and calculate the prob and keep it aside.if our condition is not fulfilled then we multiply the probabilities and if our condition is fulfilled then we add the probabilities so here we will add all these probabilitiesnow adding these we get (150/216) +(20/216) + (1/216) = 171/12962nd methodwe can calculate the probability of not getting at least two 6 i.e. getting at most one 6 i.e getting either no six or getting one 6here every parameter defined above will be same only r will take 0 and 1 as now we want only at most one 6.so for r=0c(4,0)*(1/6)^0*(5/6)^4= 625/1296for r=1 c(4,1)*(1/6)^1*(5/6)^3= 500/1296adding the probabilities we get (625/1296) + (500/1296) = 1125/1296now in question it is given that we want at least two 6 so we subtract the prob of getting at most one 6 from 1 we will get our answer1-(prob of getting at most one 6) = prob of getting at least two 61-(1125/1296)= 171/1296hope both methods are simple to understand.
This is basically a binomial probability distribution where the probability of throwing a six is 1/6 and not a six is 5/6. Therefore to get 6,6',6' where 6' is mathematically anything that isn't a six is 1/6 x (5/6)^2. Still there are three ways this sequence can occur, you throw the six first and the rest are not six or you throw the six second or you throw the six third. Thus P(a)=3C1 x 1/6 x (5/6)^2=75/216=0.3472222...
Find the probability that in 200 tosses of a fair six-sided die, a five will be obtained at least 40 times?
If you have a computer program that can compute binomial probabilities, you can use that. If you are doing this by hand, however, I would suggest using the normal approximation to the binomial here. n = number of trials = 200 p = prob. of success = 1/6 if we consider getting a five to be a success. Then np = 200(1/6) = 33.33 and n(1-p) = 166.67 are both >= 5 or 10 (the cutoff is different for different classes) then the sample size is large enough to use the normal. This binomial is approximately normal with mean = np = 33.33 and SD = sqrt(np(1-p)) = sqrt(200(1/6)(5/6)) = 5.27. Now for the continuity correction. The rectangle over 40 goes from 39.5 to 40.5. Since we are finding the probability that the number of 5's is at least 40, then we should start at 39.5 so that all of 40 is included. z = (39.5-33.33)/5.27 = 1.17 Area to the right of 1.17 on the standard normal is 0.121.
Let's find our sample space first. There are 36 equally likely outcomes:(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)Now we find all the outcomes that match the desired result:(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)That's [math]\frac{20}{36}=\frac{5}{9}[/math].
In English language NATIVE Speakers and teachers HELP PLEASE!! I study English and need to check?
PLEASE PUT THE VERBS IN BRACETS INTO THE MOST SUITABLE FORMS. 1) Farmers, as we all ...............(know), ..................(have) a hard time of it lately, and ........(turn) to new ways of earning income from their land. This .........(involve) not only planting new kinds of their crops, but some strange ways of making money, the most unusual of which has got to be sheep racing . Yes , you ........(hear) me correctly! A farmer in the West of England now...........(hold)sheep races on regular basis, and during the past year over 100 000 people ............(turn up) to watch the proceedings. 2) Ask hundreds of people what they ..................(do) on a certain day in August next year, or the year after and there ................(be) only reply. Provided of course that the people you..........(ask)...........(belong) to the Elvis Presley Fan Club. Although the King of the Rock and Roll.......(die) nearly decade ago, his fans .........(meet) every year since then outside his home in Memphis to show respect for the singer they .......(love) so much. 2) Employees protesting at the planned closure of the Magnat electronics factory have begun a protest outside the factory in Brook Road. It .......(reveal) last week that production at the factory, where over 3000 local people ........(employ), ......(transfer) to the Magnat plant in Luton next month.''Why ........(we not inform)about this earlier? We........(only tell)about this two days ago', said Marjory Calder, representing the workforce.' It's about time companies such as this ......(start) thinking abut how local communities ....(affect)by their policies.
Here's a general way how to think about problems like this one:At any moment, the game will be in one of finitely many states. In general, the state can be the entire history of the game -- in this case, the sequence of coin throws we already made. But, in order to simplify our calculations, we want the number of states to be as small as possible. Hence we are looking for ways to merge multiple states that share the same answer into one new state.For example, suppose that you are playing our game, and you just made the following sequence of tosses: HTTHTHTT. What is the probability that if we play on, HH will appear before TTT does? We don't know. But we do know it is the same as for the sequence HTHTT. Or for the sequence TT. All we care about are the last two coin tosses, the previous ones don't influence the answer in any way.In our game, probably the simplest possible description involves five states:A: HH appeared, game overB: TTT appeared, game overC: game in progress, last two tosses were HTD: game in progress, last two tosses were THE: game in progress, last two tosses were TTWe are now interested in the following questions: What is the probability P(X) that my game will end in state A, given that I'm currently in state X?Obviously, P(A)=1 and P(B)=0. We cannot compute any single of the other three answers separately, but we can compute all of them at the same time.The trick is to write a system of equations that describe what happens in the next coin toss:If I'm in state C, the next coin toss will get me either to state D or to state E with equal probability. Hence, P(C) = 0.5*P(D) + 0.5*P(E).If I'm in state D, the next coin toss will get me either to state A or to state C with equal probability. Hence, P(D) = 0.5*P(A) + 0.5*P(C) = 0.5 + 0.5*P(C).If I'm in state E, the next coin toss will get me either to state D or to state B with equal probability. Hence, P(E) = 0.5*P(D) + 0.5*P(E) = 0.5*P(D).This is a system of linear equations with a unique solution: P(C)=3/5, P(D)=4/5, and P(E)=2/5.Now we can answer the original question. After the very first two coin tosses of the game, we will be in one of the states A, C, D, and E with equal probability. Hence, the probability of seeing HH before TTT is 0.25*P(A) + 0.25*P(C) + 0.25*P(D) + 0.25*P(E) = 7/10.
Construct a table of the possible outcomes. So this would have 6 rows and 6 columns. Columns are headed “first throw and numbered 1 to 6. Rows are headed “2nd throw” and numbered 1 to 6. Then fill in all the dice scores from the different outcomes - it doesn’t take long. If you are new to probability questions, this is a good way to help get a feel for what’s going on. You will see you have 36 squares (6x6) ie 36 possible outcomes. The probability of each outcome is 1/36 (because with fair dice they are all equal and also must add up to one).In row 6, column 6, your score is 12. This is the only square with 12 in and thus the probability of scoring 12 is 1/36. 2 squares have 11 in - from throws 5,6 and 6,5. 3 squares have 10 in (5,5, 4,6 , 6,4) and 4 squares have 9. So your total probability is 1/36 + 2/36 + 3/36 + 4/36 = 10/36 =5/18.
If a fair coin is tossed 3 times what is the probability of getting at least 2 heads?
If this is a truthful coin, then each and every time there's a million/2 possibility each and each and each and each of heads and tails. hence, the probabilities of any throw being a tail is a million/2. the probability of three throws being tails is a million/2*a million/2*a million/2=a million/8. Now, for 2 heads there are 3 possibilities, ie the tail could be first, 2d or 0.33. Any specific such probability has a probability of a million/2*a million/2*a million/2=a million/8. thinking the certainty that there are 3 such possibilities, this is a finished of three/8. Now upload to the previous a million/8, and you have a finished of four/8 or a million/2. greater substantially, there are 3 possibilities as quickly as you throw a coin thrice. 0 heads, a million head, 2 heads or 3 heads, it particularly is such as 3 tails, 2 tails, a million tail, or 0 tail. thoroughly, those have a probability of a million. 0 heads would have a matching probability as 0 tail (3 heads). 2 heads would have a matching probablity as 2 tails (a million head). you will see that that those cover each and each and each and all of the recommendations. hence No heads or 2 heads are precisely 0.5 the probability ie a million/2.