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A coin is tossed, landing on "HEADS" with an unknown probability p . You have the option to either observe the outcome of the toss without betting or place a fixed bet on an interval where you believe p lies. Betting involves selecting a confidence interval where you believe the probability p exists. The bet is fixed, meaning you bet the same ...
In this coin toss example, when we say "3 choose 2", it helps to think of the three events as not yet having a result. Only after we choose two of the three events will we give the events their heads/tails results. So doing 3C2 we get {(Event 1, Event 2), (Event 1, Event 3), (Event 2, Event 3)}. Then we assign heads to the chosen events.
Question: Two players A and B, alternatively toss a fair coin (A tosses the coin first, then B, than A again, etc.). The sequence of heads and tails is recorded and if there is head followed by a tail (HT subsequence), the game ends and the person who tosses the tail wins.
Note that if we define the random variable C = 0, 1, 2 C = 0, 1, 2 with probability 1 3 each (for the coin toss with 0 0 representing the coin falling on an edge, 1 1 representing tail and 2 2 representing head), then. Now we can split into cases. Suppose that 1 ≤ k <2 1 ≤ k <2. Then, note that the first and last term above would be zero ...
Sigma Algebra on Coin Toss. I am trying to study probability space, and so far I have come to point that probability space is defined as (Ω, F, P) where F is the σ − algebra. I know what σ − algebra is, but I am confused that the σ − algebra can be easily obtained by the power set of Ω .i.e. 2Ω. Since a set can have many σ − ...
1. We toss n = 200 n = 200 Euro coins on the table. I want to calculate, using the central limit theorem, the probability that at least 110 110 coins have tutned on the same side. Xi ={1, 0, coin i shows head coin i does not show head X i = {1, coin i shows head 0, coin i does not show head. Or do we not have to consider one specific side of ...
The last case is if you toss HH, you get two more tosses which gives you a favorable outcome of (HHTH, HHHT, HHTT, HHHH} out of $2^4=16$ ways it can play out. Thus it is for a prob of $\frac{1}{16}$ you get two heads (HHTT), and for a prob of $\frac{2}{16}$ you get three heads and for a prob of $\frac{1}{16}$ , you get four heads.
Coin toss problem, get exactly 2 heads in 5 tosses. 2. Coin toss bias estimation problem. 1. Probability ...
For a single coin toss, you will also need to indicate what are you counting as a success, ...
where in the middle we relied on the fact that since Z Z only can take values either 0 0 or 1 1, we have Z2 = Z Z 2 = Z. Takeaway: this problem can be modeled by a Binomial (n, p) (n, p) distribution, where n = 3600 n = 3600 and p = 1/2 p = 1 / 2: since you sum 3600 independent coin tosses with same probability of Heads (here, 1/2).