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Johnmathstudent
If I flip four coins what is the probability that at least two of the coins are heads?
its the opposite of P(at most 1 is heads)
there 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
What about at least 2 heads
at 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)
since P(0) = 1/16 and P(1) = 4/16 the solution is determined
So what would the answer be written as?
im not going to do the subtraction for you ....
you can tell me what you get ... and i can verify if you did it correctly
bro your even confusing me.... amistre64
im not a mind reader, so youll have to be more detailed in what is confusing :)
nevermind .. the probability would be 50% John
the probability is not 50%; for starters, a probability is a number between 0 and 1 ... not a percentage.
and for another, 1 - P(0) - P(1) does not equal .5000 or 1/2
he 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
its not rocket science he just wants a simple answer.. idk whats wrong with tht
\[\binom{4}{0}.5^0.5^4+\binom{4}{1}.5^1.5^3+\binom{4}{2}.5^2.5^2+\binom{4}{3}.5^3.5^1+\binom{4}{4}.5^n.4^0=1\] \[(1).5^0.5^4+(4).5^1.5^3+(6).5^2.5^2+(4).5^3.5^1+(1).5^4.5^0=1\]
P(E) is notation for "probability of an event"
the 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
subtracting the first 2 terms from each side gives us the solution
your 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!
i suggest you tone down your language. and, we are not providing a free answering service, we are offering a way to learn the material.
my first attempt was to simplify it .... since that did not work i resorted to the longer, more detailed version of it all.
So the answer is .6875?
11/16 is correct either way is fine, but the format is up to whoever is doing the grading.
Thanks I just forgot how to do this from the beginning of the year
you did fine :) good luck