Posted by **Koia** on Thursday, May 10, 2012 at 3:23am.

A fair die is tossed and the number facing up is noted. If the probability of getting at least one ‘six’ is to exceed 0.9, how many times should the die be tossed.

- Math -
**MathMate**, Thursday, May 10, 2012 at 7:34am
The geometric distribution gives the probability of getting a success at the xth trial:

P(X=x)=Px(x)=(1-p)^(x-1)p

The probability of getting the first success in n trials is therefore the sum of the above, or

P(n)=Σ(1-p)^(n-1)p

=pΣ(1-p)^(n-1) (geom seq.)

=p(1-(1-p)^n)/p

=1-(1-p)^n

So

P(1)=1/6

P(2)=11/36

...

P(5)=4651/7776=0.598

P(10)=50700551/60466176=0.838

P(13)=11839990891/13060694016=0.906

P(15)=439667406451/470184984576=0.939

...

- Math -
**Koia**, Sunday, May 13, 2012 at 12:06am
it really make me confusing what you are mention, how could i describe what you did

- Math -
**MathMate**, Sunday, May 13, 2012 at 10:17pm
If you have not done the geometric distribution at school, you need to read up about it before. I agree that it is not obvious if you have not done the distribution before, or if you have not done summation of geometric series before.

In my calculations above,

P(1) is the probability of obtaining at least one favourable outcome with one trial (throw of die)

P(2) is the probability of obtaining at least one favourable outcome (throwing a six) with 2 trials.

...

P(n) is the probability of obtaining at least one favourable outcome with n throws of the die.

It turns out that with 13 throws, the probability of getting at least "six" once is 0.9, as shown above.

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