The AVERAGE of the VALUES of a set of events = the sum of (PROBABILITY of each e
ID: 3206665 • Letter: T
Question
The AVERAGE of the VALUES of a set of events = the sum of (PROBABILITY of each event) x (the VALUE associated with that event). For example, if I pay you $1 if you flip a coin and it comes up HEADS and I pay you NOTHING if it comes up TAILS, then the AVERAGE amount I pay you = ($1)x(1/2) + ($0)x(1/2) = $0.50, because there is a PROBABILITY = 1/2 that each time the coin is either HEADS or TAILS. OK, let's play a different game, you keep flipping a coin UNTIL it comes up HEADS. You have to keep playing UNTIL you get a HEAD. If you get a HEADS on the first toss, I give you $2. The probability of this happening is 1/2. If you first get a TAIL and then a HEAD, I give you $4, that is you got TAIL-HEAD. If you got TAIL-TAIL-HEAD, I give you $8. And so on. That is, if you get a HEAD on the n-th toss, I give you $2^n. What is the PROBABILITY of getting that first HEAD? What is the PROBABILITY of getting TAIL-HEAD? What is the PROBABILITY of getting TAIL-TAIL-HEAD? What is the PROBABILITY of getting (n-1) TAILS and then a HEAD? What is the AVERAGE value that I give you for playing just ONE coin toss? Are you sure about your answer to e)? Use the internet to find REAL examples of physical, biological, or social process that are similar to this problem.Explanation / Answer
a) Coin is tossed one time, therefore probability of getting a head is 1/2
b) Coin is tossed two times.
Possible outcomes are {H,T}X{H,T} = {HH,TH,HT, TT} = 4 = 22
therefore probabiility of getting TH is 1/4
c) coin is tossed 3 times
Possible outcomes are {H,T}X{H,T}x{H,T} = {HHH, HTH,HHT,HTT, THH, TTH, THT,TTT} = 8= 23
Probability of getting TTH = 1/8
d) Coin is tossed n times
Possible number of outcomes will be 2n
Probability of getting (n-1) Tails and then a Head is 1/2n
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