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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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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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