box 1 contains 5000 bulbs of which 5% are defective. box 2 contains 4000 bulbs of which 3% are defective. two bulbs are draw without replacement from

Question

Q.box 1 contains 5000 bulbs of which 5% are defective. box 2 contains 4000 bulbs of which 3% are defective. two bulbs are draw without replacement from a randomly selected box. find the probability that both bulbs are defective.

Well I only know Bayes theorm and find it hard even to put in the formula p(x=r)=nCrp^rq^n-r

Answer 1

If you want the exact answer (and presuming that you will choose each box with equal probability, you need to apply the hypergeometric distribution and then apply the law of total probability, viz.

• Box $1$ has $5$% of $5000 = 250$ defectives,
  P($2$ defectives from box $1$) $= \frac{250}{5000}\cdot\frac{249}{4999} = A,\; (say)$

• Box $2$ has $3$% of $4000 = 120$ defectives,
  P($2$ defectives from box $2) = \frac{120}{4000}\cdot\frac{119}{3999}= B,\; (say)$

• Applying the law of total probability,
  required probability $= (1/2)(A+B)$

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Note

The answer I get is $\large\frac{1127233}{666366700}, \approx 0.1691$% to $4$ dp

But in practical terms, an accuracy of this order is not needed, and we could well have simply computed it as $(\frac12)[(\frac5{100})^2+(\frac3{100})^2] = 0.17$%, had not the question specifically mentioned without replacement