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The Long Shot Golf Ball Company determined that, on average, one percent of the balls they produce are defective. (a) Find the probability that, out of 100 balls, at least one ball is defective. (b) Find the probability that, out of 100 balls, at most one ball is defective.

I am confused of what method to use?

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a. The probability that at least one is defective is equal to: $1 -Pr$(none are defective). Probability that none are defective=$(0.99)^{100}$

So the answer is $1-(0.99)^{100}$

b. This is straight binomial probability, so the answer should be: $\binom{100}{1}(0.99)^{99}(0.01)+(0.99)^{100}$. If you don't understand that let's approach it combinatorially.

First we let 1 ball to be defective. There are 100 balls, to arrange, 99 of which are not defective, and 1 of which is defective which accounts for the factor of $\binom{100}{1}$, as we do not know whether the defective ball is first or last ball inspected.Then we simply multiply probabilities to get $\binom{100}{1}(0.99)^{99}(0.01)$.

We then add the probability that no balls are defective- $0.99^{100}$, as the question asks for AT MOST 1 defective ball- which includes having no defective balls.

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For a), \begin{gather*} P(\{\text{at least one ball defective}\}) = 1 - P(\{\text{no balls defective}\}), \end{gather*} while $$P(\text{{no balls defective}}) = (1-0.01)^{100}.$$ For b), \begin{gather*} P(\text{{at most one ball defective}}) = P(\text{{exactly 1 ball defective}} \cup \text{{exactly 0 balls defective}}). \end{gather*} You already have $P( \text{{exactly 0 balls defective}})$, and the other probability is distributed binomially (why?). To conclude, use the fact that the two events on the right-hand side above are disjoint: $P(A\cup B) = P(A) + P(B)$ if $A \cap B = \emptyset$.

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