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If I flip four coins what is the probability that at least two of the coins are heads?
 10 months ago
 10 months ago
If I flip four coins what is the probability that at least two of the coins are heads?
 10 months ago
 10 months ago

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amistre64Best ResponseYou've already chosen the best response.1
its the opposite of P(at most 1 is heads)
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
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
 10 months ago

JohnmathstudentBest ResponseYou've already chosen the best response.0
What about at least 2 heads
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
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)
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
since P(0) = 1/16 and P(1) = 4/16 the solution is determined
 10 months ago

JohnmathstudentBest ResponseYou've already chosen the best response.0
So what would the answer be written as?
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
im not going to do the subtraction for you ....
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
you can tell me what you get ... and i can verify if you did it correctly
 10 months ago

MAYSTEPHBest ResponseYou've already chosen the best response.0
bro your even confusing me.... amistre64
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
im not a mind reader, so youll have to be more detailed in what is confusing :)
 10 months ago

MAYSTEPHBest ResponseYou've already chosen the best response.0
nevermind .. the probability would be 50% John
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
the probability is not 50%; for starters, a probability is a number between 0 and 1 ... not a percentage.
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
and for another, 1  P(0)  P(1) does not equal .5000 or 1/2
 10 months ago

MAYSTEPHBest ResponseYou've already chosen the best response.0
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
 10 months ago

MAYSTEPHBest ResponseYou've already chosen the best response.0
its not rocket science he just wants a simple answer.. idk whats wrong with tht
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
\[\binom{4}{0}.5^0.5^4+\binom{4}{1}.5^1.5^3+\binom{4}{2}.5^2.5^2+\binom{4}{3}.5^3.5^1+\binom{4}{4}.5^n.4^0=1\] \[(1).5^0.5^4+(4).5^1.5^3+(6).5^2.5^2+(4).5^3.5^1+(1).5^4.5^0=1\]
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
P(E) is notation for "probability of an event"
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
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
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
subtracting the first 2 terms from each side gives us the solution
 10 months ago

MAYSTEPHBest ResponseYou've already chosen the best response.0
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!
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
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.
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
my first attempt was to simplify it .... since that did not work i resorted to the longer, more detailed version of it all.
 10 months ago

JohnmathstudentBest ResponseYou've already chosen the best response.0
So the answer is .6875?
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
11/16 is correct either way is fine, but the format is up to whoever is doing the grading.
 10 months ago

JohnmathstudentBest ResponseYou've already chosen the best response.0
Thanks I just forgot how to do this from the beginning of the year
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
you did fine :) good luck
 10 months ago
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