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Answered 2015-05-04 20:59:12

There are 4 events: 3 heads, 2 heads 1 tail, 1 head 2 tails, and 3 tails.

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Two coins tossed sample space is (H=Heads, T = Tails) as follows. HH, HT, TH, TT is the sample space.


Sample space for two coins tossed is: HH HT TH TT Therefore at most one head is HT TH TT or 3/4 or 0.75.



Assuming the variable of interest is the face on top: H (= heads) or T (= tails), then they are the four possible outcomes: HH, HT, TH and TT.



The probability of tossing two heads in two coins is 0.25.



The answer depends on how many coins were tossed.


The sample space is 23 or 8; which can be listed out as: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. There are 2 of the 8 that have exactly 2 heads; so the probability of exactly two coins landing on heads is 2/8 or 1/4.


The total number of outcomes is 2^5 = 32.


Coins do not have numbers, there is only the probability of heads or tails.


The possible outcomes areHH, HT, TH and TT.


Coins, for wealth and prosperity in the new year.


The sample space for tossing 2 coins is (H = Heads & T = Tails): HH, HT, TH, TT


It is the event that one of the two coins lands showing tails and the other shows heads.


The sample space when tossing 3 coins is [HHH, HHT, HTH, HTT, THH, THT, TTH, TTT]


You find the sample space by enumerating all of the possible outcomes. The sample space for three coins is [TTT, TTH, THT, THH, HTT, HTH, HHT, HHH].


I believe there would be 11 possible outcomes!


They are:HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.




Heads ( H ) Tails ( T ) HH , TT , HT , TH


This question can be rather easily answered, as soon as outcomes 'a' and 'b' are defined.


The answer depends on the experiment: how many coins are tossed, how often, how many dice are rolled, how often.


Simple answer: All coins are graded by the same scale. Circulated coins by how much wear the coin has. Uncirculated coins by how well the coins are struck.



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