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Never saw a reply to this one, so just for kicks...Solution: 1/2*[1-ln(1/2)] or .846574 and confirmed by monte carloMethod:xy < .5y<.5/x *now integrate the area which the condition is true within the event space x{0,1} & y{0,1}Y = 1/2*ln(x) *from x{0,.5} the area under the curve is the same as event space'sProbability = Area of event space from x{0,.5} + Area under curve 1/2*ln(x) from x{.5,1}Probability = 1/2 + [0-1/2*ln(1/2)]Probability= 1/2*[1-ln(1/2)]
Never saw a reply to this one, so just for kicks...
Solution: 1/2*[1-ln(1/2)] or .846574 and confirmed by monte carlo
Method:
xy < .5
y<.5/x *now integrate the area which the condition is true within the event space x{0,1} & y{0,1}
Y = 1/2*ln(x) *from x{0,.5} the area under the curve is the same as event space's
Probability = Area of event space from x{0,.5} + Area under curve 1/2*ln(x) from x{.5,1}
Probability = 1/2 + [0-1/2*ln(1/2)]
Probability= 1/2*[1-ln(1/2)]
