Binomial Distribution Question?
The variable T~B(n, 0.96) and it is given that P(T=n) > 0.5. Find the greatest possible value of n.
- Ian HLv 71 month ago
A binomial distribution refers to situation that has only two possible outcomes such as taking a test or flipping a coin, so it is about the probability of some sort of success or failure but repeated multiple times.
I think the question means this. If 0.96 is the probability of a single successful event, what is the greatest possible value of the number of events n, such that the combined probability of n good events remains above 5?
If we directly solve 0.96^n = 0.5 we get n ~ 16.97
n must be an integer, and 0.96^17 ~ 0.499587 which is just below 0.5
So, the greatest value of n is 16.
- PuzzlingLv 71 month ago
T ~ B(n,0.96) represents a binomial distribution with n independent trials and probability of success being 0.96.
You are looking for all n trials to be successful, hence:
P(T=n) = 0.96^n
And you want the probability of all n trials being successful to be higher than 0.5:
0.96^n > 0.5
One method would be to just try values of n until the result is below 0.5 and take the value just before that point. But that could be time-consuming.
A better approach is to just solve mathematically for n.
Start by taking the log of both sides:
log(0.96^n) > log(0.5)
n log(0.96) > log(0.5)
log(0.96) is negative, so we change the direction of the inequality symbol when we divide:
n < log(0.5)/log(0.96)
n < 16.9797480183
n represents the number of trials so must be whole number. The largest whole number less than 16.9797480183 is just 16 (round down).
0.96^16 ≈ 0.5204 <--
For completeness, we can show that the probability drops below 0.5 when you have 17 trials:
0.96^17 ≈ 0.4996
n = 16