The question asks to determine the smallest number of trials n such that a certain probability (typically P(X ≥ 1) or a complement probability) meets or exceeds a given threshold.
On a given day, an automatic checkout triggers 15 checks. The probability that a check reveals an error is $p = 0.165$. The detection of an error during a check is independent of other checks. We denote $X$ the random variable equal to the number of errors detected during the checks on this day.
We admit that the random variable $X$ follows a binomial distribution. Specify its parameters.
Determine the probability that exactly 5 errors are detected. The answer will be given rounded to the nearest hundredth.
Determine the probability that at least one error is detected. The answer will be given rounded to the nearest hundredth.
We wish to modify the number of checks triggered by the checkout so that the probability that at least one error is detected each day is greater than $99\%$. Determine the number of checks that the checkout must trigger each day for this constraint to be satisfied.
The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least $90 \%$ is: (1) 2 (2) 4 (3) 5 (4) 3
In a bombing attack, there is $50\%$ chance that a bomb will hit the target. At least two independent hits are required to destroy the target completely. Then the minimum number of bombs, that must be dropped to ensure that the probability of the target being destroyed is at least $0.99$, is ...