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.
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
...
it really make me confusing what you are mention, how could i describe what you did
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.
To find out how many times the die should be tossed, we need to calculate the probability of not getting a 'six' on a single toss and then calculate how many tosses would exceed a probability of 0.9 for at least one 'six'. Let's break down the problem step by step:
Step 1: Calculate the probability of not getting a 'six' on a single toss.
A fair die has six equally likely outcomes: numbers 1, 2, 3, 4, 5, and 6. The probability of not getting a 'six' on a single toss is given by:
P(not 'six') = 5/6
Step 2: Calculate the probability of not getting a 'six' in multiple tosses.
If we toss the die multiple times, the probability of not getting a 'six' in all the tosses is given by multiplying the probabilities of not getting a 'six' in each individual toss. So, for n tosses, the probability of not getting a 'six' in any of them is:
P(not getting a 'six' in n tosses) = (5/6)^n
Step 3: Calculate the probability of getting at least one 'six' in multiple tosses.
Since we want the probability of getting at least one 'six' to exceed 0.9, we can subtract the probability of not getting any 'six' from 1:
P(at least one 'six' in n tosses) = 1 - P(not getting a 'six' in n tosses)
Step 4: Set up an inequality and solve for n.
We want the probability of getting at least one 'six' to exceed 0.9. So, we have:
1 - P(not getting a 'six' in n tosses) > 0.9
Simplifying, we get:
P(not getting a 'six' in n tosses) < 0.1
Substituting the formula from Step 2, we have:
(5/6)^n < 0.1
Now, to solve for n, we can take the logarithm of both sides. Assuming base 10 logarithm:
log((5/6)^n) < log(0.1)
n * log(5/6) < log(0.1)
n > log(0.1) / log(5/6)
Using a calculator, we can evaluate the right-hand side of the inequality to get the minimum value for n:
n > log(0.1) / log(5/6)
n > -1 / log(5/6)
n > approximately 12.95
Therefore, the die should be tossed at least 13 times to ensure that the probability of getting at least one 'six' exceeds 0.9.