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This might be a simple question, but can anyone tell me how to solve it.

The probability of success of a particular type of experiment is 0.004. How many times the same experiment is to be repeated so that the probability of success becomes more than 0.7??

Is it simply, 0.004 * x = 0.7 and then find x which is 175


(0.996)^x * (0.004) implies where x = 731.

I did not understand how to approach this problem...

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I suppose the word 'success' has two different meanings in your question. Do you mean: "How many times the same experiment has to be repeated so that the probability of at least one success occurring becomes more than 0.7?" Edit: if so, see Gerry Myerson's answer. – Kuba Helsztyński Sep 9 '12 at 12:21

If you do the experiment $n$ times, the probability of no success is $(.996)^n$, so the probability of at least one success is $1-(.996)^n$. So you want $$1-(.996)^n\gt0.7$$

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Assume you are tossing a coin and the probability that a Head comes is $0.004$ which is your success probability. If you repeat this experiment $N$ times and expecting at least one of the experiments would yield a Head according to your first conclusion there should be $175$ experiments. However if this conclusion would be correct, then $250$ experiments would be enough to get a probability $1$. Is it possible to get a Head for sure in $N$ experiment? It is not because the experiments are random and there is always a possibility that no Head comes unless you repeat the experiments infinitely many times.

We expect that in $N$ experiments when $N\rightarrow \infty$ the probability of at least one success should go to $1$ and the probability of failure should go to $0$. Somewhere in between we get $0.7$ probability. The probability of at least one success for $N$ independent bernoilli trials is $p=1-\prod_i^N q_i$, where $q_i$ is the failure probability. Es we expect when $N\rightarrow \infty$ failure probability goes to $0$ and the probability of at least one time success goes to $1$

In general we have $$\sum_{k=1}^{n}\binom{n}{k}p^k(1-p)^{n-k}=1-(1-p)^n$$

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