Question

2. (15 pts.) Bronze, the first man-made metal alloy, is formed by combining copper with tin. Cast bronze bells are manufactured by the Verdin Bells & Clocks company (htplwww.verdin.com). but even with all their experience, about 25% of bells have to be scrapped. If too much tin is combined with copper, for example, then the bell will be too brittle and break when struck For what follows, suppose bells are scrapped independently a Comsider the next thirteen bronze bells cast. What is the chance that exactly 3 must be scrapped? Give a numerical (decimal) answer b. Again, consider the next thirten bronze bells cast. What is the chance that ar least 3 must be scrapped? Again, give a numerical (decimal) answer.

Answer 1

The probability of scrapping a bell = p = 25% = 0.25

The probability of accepting a bell = (1-p) = (1-0.25) = 0.75

a)

For scrapping exactly 3 out of 13 bells. we need to choose 3 bells out of 13 bells. This can be done in $\binom{13}{3}$ ways. So, the probability of scrapping exactly 3 bells is:

$\binom{13}{3}p^3(1-p)^{13-3}=\frac{13!}{3!10!}\cdot 0.25^3\cdot (1-0.25)^{10}=\frac{13\cdot 12\cdot 11}{3\cdot 2\cdot 1}\cdot 0.25^3\cdot 0.75^{10}\\ = 286\cdot 0.25^3\cdot 0.75^{10}\approx 0.2517$

So, the probability of scrapping exactly 3 bells is 0.2517.

b)

To find the probability of scrapping at least 3 bells, we first find the probability of scrapping at most 2 bells and then subtract it from 1 to get the desired answer.

P(scrapping at most 2 bells)

= P(scrapping exactly 0 bells) + P(scrapping exactly 1 bell) + P(scrapping exactly 2 bells)

$= \binom{13}{0}p^0(1-p)^{13-0}+\binom{13}{1}p^1(1-p)^{13-1}+\binom{13}{2}p^2(1-p)^{13-2}\\ =1\cdot 1\cdot (1-0.25)^{13}+13\cdot 0.25\cdot (1-0.25)^{12}+\frac{13\cdot 12}{2\cdot 1}\cdot 0.25^2\cdot (1-0.25)^{11}\\ =0.75^{13}+13\cdot 0.25\cdot 0.75^{12}+78\cdot 0.25^2\cdot 0.75^{11}\approx 0.3326$

So, probability of scrapping at most 2 bells = 0.3326.

So, probability of scrapping at least 3 bells = 1 - 0.3326 = 0.6674.