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Example question

It is known that 30% of all laptops of a certain brand experience hard-drive failure within 3 years of purchase. Suppose that 20 laptops are selected at random. Let the random variable $X$ denote the number of laptops which have experienced hard-drive failure within 3 years of purchase.

If it is known that at least 3 laptops experience hard-drive failure, what is the probability that no more than 6 laptops will experience hard-drive failure?

I know $X\sim\operatorname{Bin}(20,0.3)$.

I know how to calculate probabilities like $P(X = x), P(X \gt x), P(X \ge x),$ and $P(X \le x).$

Relevant formulae:

$$P(X = x) = {n \choose p}p^x(1-p)^{n-x}$$ $$P(X \gt x) = 1 - P(X \leq x)$$

and if $X$ is discrete,

$$P(X \ge x) = 1 - P(X \leq (x-1))$$

Additionally, what if the question was:

If it is known that at least 3 laptops experience hard-drive failure, what is the probability that at least 6 laptops will experience hard-drive failure?

or

If it is known that no more than 6 laptops experience hard-drive failure, what is the probability that at least 3 laptops will experience hard-drive failure?

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$$ \Pr(X\le 6 \mid X\ge 3) = \frac{\Pr(X\le6\ \&\ X\ge 3)}{\Pr(X\ge 3)} = \frac{\Pr(X=3\text{ or } X=4 \text{ or } X=5 \text{ or }X=6)}{1 - \Pr(X=0\text{ or }X=1\text{ or } X=2) } $$

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  • $\begingroup$ Awesome. I'm reviewing for an exam and forgot that conditional probability came before this section, so I didn't think about using the conditional probability formula. Is your answer the same as $\dfrac{P(X \le 6) - P(X \le 2)}{1-P(X \le 2)}$? $\endgroup$
    – Shea
    Nov 6, 2017 at 0:30
  • $\begingroup$ Yes. $\qquad\qquad$ $\endgroup$ Nov 6, 2017 at 0:40

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