Original post by mqb2766

Use independence to get a quadratic in p repreenting the joint probability, then get the corresponding q(s) to sum to 1.

Thanks, i get it now

Original post by Dell PC 2

I also need help with this:

A batch contains n components which have been randomly selected.

The probability that no components are faulty in batches of this size is 0.2957 to 4

decimal places.

(iii) Find the value of n.

A batch contains n components which have been randomly selected.

The probability that no components are faulty in batches of this size is 0.2957 to 4

decimal places.

(iii) Find the value of n.

I need help as well, as youve missed out important info. 3% are faulty, so p of being not faulty is 0.97, so using the usual binomial or independent components or ...

0.97^n = 0.2957

Original post by mqb2766

I need help as well, as youve missed out important info. 3% are faulty, so p of being not faulty is 0.97, so using the usual binomial or independent components or ...

0.97^n = 0.2957

0.97^n = 0.2957

Thank you, should the answer be 40?

Original post by Dell PC 2

A die is biased so that the probability of obtaining a 3 is p. When the die is thrown 4 times the

probability that there is at least one three is 0.9375. Find the value of p.

i keep getting p=0.5 but the ans is p=0.825

probability that there is at least one three is 0.9375. Find the value of p.

i keep getting p=0.5 but the ans is p=0.825

Spoiler

I agree that corresponds to p=1/2 as 1/2^4=1/16=0.0625

Original post by mqb2766

I agree that corresponds to p=1/2 as 1/2^4=1/16=0.0625

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