What is the probability that the second stage of the test will not be required?

The second stage will not be required if more than one IC fails.

So the test will be required if either 5 ICs pass or 6 ICs pass.

The chances of 6 ICs passing:

The first passes: probability p.

The second passes: probability p.

etc..

The probability of them ALL passing is p*p*p*p*p*p = p^6.

OR you could have 5 that pass and one fail.

So you'd have five passing similarly to above with probability p^5 and then the probability of one failing is 1-p.

So multiply these and you get (p^5)*(1-p) and then you multiply this by 6 to take into account that any one of the ICs could fail giving: (p^5)*(1-p)*6.

So you add together these two possible ways of getting onto the next stage: p^6 + (p^5)*(1-p)*6. This is your probability of getting onto the next stage, so the probability of NOT getting into the next stage, which is what the question asks - is 1 - p^6 + (p^5)*(1-p)*6.

For the second part - is it given whether there actually IS a second test? Which would mean your answer would be between 5 and 6 - or is the need for the second test still an uncertainty?