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Been working on this problem for quite some time, it has to do with binomial distribution but i'm lost from there. The answers to the questions were given as well but I have yet to recreate their results. Answers will be shown in bold, any help is appreciated, would appreciate if you could show formulas used in getting numbers as well.

  1. Suppose that the probability that any Toyota has faulty brakes when tested is P = 0.10.

(a) If a garage tests n = 4 cars what is the probability they find at least 2 with faulty brakes? 5.23%

(b) If a second garage tests n = 10 cars what is the probability they find at least two with faulty brakes?

With n = 10 and P = 0.10 the probability is 1 − 0.3487 − 0.3874 = 0.2639 or 26.39%

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With homework problems it's customary to explain what you've tried to solve a problem (and to use the homework "tag"). –  hardmath Oct 15 '12 at 2:34
    
it's not homework. Its midterms that have been posted for review, so i'm trying to use their answers to work backwards. In this case i have no idea how they get the value of 5.23% to start with, i have tried using the binomial distribution formula but can only get 4.86% through that. –  Alex Oct 15 '12 at 2:40

1 Answer 1

up vote 1 down vote accepted

What you need is called cumulative distribution function, or CDF:$ P(X \leq k)=\sum_{j=0}^{k}f(k)$. In your case you need $$ P(2 \leq X \leq 4)=P(X\geq 2) = 1-P(X=0)-P(X=1) = 1-\binom{4}{0} .1^0 \cdot 0.9^4-\binom{4}{1} \cdot 0.1 \cdot 0.9^3 $$ In the second case the idea is the same: $$ P(X \geq 2) = 1-P(X=0)-P(X=1) $$

Can you handle from here?

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