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Johnmathstudent
 one year ago
If I flip four coins what is the probability that at least two of the coins are heads?
Johnmathstudent
 one year ago
If I flip four coins what is the probability that at least two of the coins are heads?

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

amistre64
 one year 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

Johnmathstudent
 one year ago
Best ResponseYou've already chosen the best response.0What about at least 2 heads

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

Johnmathstudent
 one year ago
Best ResponseYou've already chosen the best response.0So what would the answer be written as?

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

amistre64
 one year 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

MAYSTEPH
 one year ago
Best ResponseYou've already chosen the best response.0bro your even confusing me.... amistre64

amistre64
 one year 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 :)

MAYSTEPH
 one year ago
Best ResponseYou've already chosen the best response.0nevermind .. the probability would be 50% John

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

MAYSTEPH
 one year 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

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

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

MAYSTEPH
 one year 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
 one year 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
 one year 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.

Johnmathstudent
 one year ago
Best ResponseYou've already chosen the best response.0So the answer is .6875?

amistre64
 one year 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.

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

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