Q.How many times must Sumit toss a fair coin so that the probability of getting at least one head is more than 90%?

Binomial Probability Minimum Trials – A First Look

Imagine you're a quality inspector at a factory that makes light bulbs. You know from past data that 10% of the bulbs are defective. You want to be at least 95% sure that you'll find at least one defective bulb in your sample. How many bulbs must you test?

This is the classic "minimum trials" problem. You're not asked for the probability of a specific number of successes. Instead, you're asked: How many times must I repeat the experiment so that a certain event (like "at least one success") happens with a given high probability?

The Intuition

The key trick is to flip the question around. Instead of directly calculating the probability of "at least one success" (which gets messy for large numbers), you calculate the probability of the opposite event — "zero successes" — which is much simpler.

If you want "at least one success" to be very likely, then "zero successes" must be very unlikely. So you set up an inequality:

is equivalent to

which rearranges to

For a binomial experiment with $n$ trials, probability of success $p$ on each trial, the probability of zero successes is $(1-p)^n$. So the inequality becomes:

Now you solve for the smallest integer $n$ that satisfies this.

The Precise Statement

Problem: In a binomial experiment with success probability $p$ (where $0 < p < 1$), find the smallest number of trials $n$ such that the probability of obtaining at least one success is at least a given threshold $P_0$ (usually close to 1, like 0.95 or 0.99).

Solution: Solve for the smallest integer $n$ satisfying

Taking natural logs (or any base) on both sides:

Since $\log(1-p)$ is negative (because $1-p < 1$), dividing flips the inequality:

The answer is the ceiling (smallest integer greater than or equal to) of that fraction.

Worked Example

Back to the light bulbs: $p = 0.10$ (defective rate), target $P_0 = 0.95$.