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anonymous
 3 years ago
If I flip four coins what is the probability that at least two of the coins are heads?
anonymous
 3 years ago
If I flip four coins what is the probability that at least two of the coins are heads?

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amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1its the opposite of P(at most 1 is heads)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1there are 16 outcomes tttt ttth ttht thtt httt ... 1 4 6 4 1 = 16 0 heads is 1 out of 16 1 head is 4 out of 16

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What about at least 2 heads

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1at least 2 heads means that you have 2, 3, or 4 heads P(0) + P(1) + P(2) + P(3) + P(4) = 1 P(2) + P(3) + P(4) = 1  P(0)  P(1)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1since P(0) = 1/16 and P(1) = 4/16 the solution is determined

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So what would the answer be written as?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1im not going to do the subtraction for you ....

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1you can tell me what you get ... and i can verify if you did it correctly

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0bro your even confusing me.... amistre64

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1im not a mind reader, so youll have to be more detailed in what is confusing :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0nevermind .. the probability would be 50% John

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1the probability is not 50%; for starters, a probability is a number between 0 and 1 ... not a percentage.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1and for another, 1  P(0)  P(1) does not equal .5000 or 1/2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0he asked for a probability. if you flip 4 coins and the odds of 2 of the comming out heads is 2/4 and tht equals 50% and wtf is p

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0its not rocket science he just wants a simple answer.. idk whats wrong with tht

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1\[\binom{4}{0}.5^0.5^4+\binom{4}{1}.5^1.5^3+\binom{4}{2}.5^2.5^2+\binom{4}{3}.5^3.5^1+\binom{4}{4}.5^n.4^0=1\] \[(1).5^0.5^4+(4).5^1.5^3+(6).5^2.5^2+(4).5^3.5^1+(1).5^4.5^0=1\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1P(E) is notation for "probability of an event"

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1the last 3 terms are 4 choose 2, 4 choose 3, and 4 choose 4 which is what we want if we are to determine "at least 2" out of 4

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1subtracting the first 2 terms from each side gives us the solution

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0your making this way more complex than it already isnt lol if you know the damn answer then tell him and quit arguing with a degenerate!

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1i suggest you tone down your language. and, we are not providing a free answering service, we are offering a way to learn the material.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1my first attempt was to simplify it .... since that did not work i resorted to the longer, more detailed version of it all.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So the answer is .6875?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.111/16 is correct either way is fine, but the format is up to whoever is doing the grading.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks I just forgot how to do this from the beginning of the year

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1you did fine :) good luck
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