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If you throw pair of dice once P of not getting double 4 is 5^2/6^2
So if you throw N times then not getting even one double 4 = 5^2n/6^2n
So P of getting atleast one double 4 = (6^2n-5^2n)/6^2n
If we equate it to 1/2
6^2n = 5^2n Only for N = 0 this equation is satisfied.
This is how I approached it.
Probability of getting two 4s: 1/6*1/6 = 1/36
When n>18, then probability of getting two 4s will be greater than 1/2. Therefore, 19/36>1/2 and n would then be equal to 19... Btw, there is no OA on this one so it's open to debate
Computing probability for one dice is itself very complicated. Two dice ! Hope we won't see these problems in real GMAT
I've used calculator to compute the answers and it is 24.
Method will fallow, if it is correct.
The probabilty of getting this question in GMAT is less than the probabilty of a nuclear war in a year. (Two years back, Warren E Buffett predicted that the world will see a nuclear war or catastrophe in 20 years)
So, We may have other issues to worry about
campared to this problem.
1 - (35/36)^n > 1/2
N = 25.
Qus: Does anyone know a method to solve the above equation
with out a calculator.