Question

Suppose you have just received a shipment of 16 modems. Although you​ don't know​ this, 4 of the modems are defective. To determine whether you will accept the​ shipment, you randomly select 5 modems and test them. If all 5 modems​ work, you accept the shipment.​ Otherwise, the shipment is rejected. What is the probability of accepting the​ shipment?

Answer 1

$|\Omega|={_{16}C_5}=\dfrac{16!}{5!11!}=\dfrac{12\cdot13\cdot14\cdot15\cdot16}{120}=4368\\|A|={_{12}C_5}=\dfrac{12!}{5!7!}=\dfrac{8\cdot9\cdot10\cdot11\cdot12}{120}=792\\\\P(A)=\dfrac{792}{4368}=\dfrac{33}{182}\approx18\%$

Answer 2

The probability of the event is defined as the ratio of the number of cases favourable to an occurrence, and the further calculation can be defined as follows:

4 of the 16 modems are defective, while the remaining 12 are not.

P(accepting shipment) = P (all 5 modems work)

$\bold{^{12}C_{5}}$ methods could be used to choose 5 non-defective modems from a pool of 12 non-defective modems.

$\to \bold{^{12}C_{5} = \frac{12!}{ (12 -5)! \times 5! }}$

             $\bold{ = \frac{12!}{ 7! \times 5!}}\\\\\bold{ = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{ 7! \times 5!}}\\\\\bold{ = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1}}\\\\\bold{ =11 \times 9 \times 8}\\\\\bold{=792}$

The total number of methods to choose 5 modems from a pool of 16 modems is $\bold{^{16}C_{5}}$.

$\to \bold{^{16}C_{5} = \frac{16!}{ (16 -5)! \times 5! }}$

             $\bold{ = \frac{16!}{ 11! \times 5!}}\\\\\bold{ = \frac{16 \times 15 \times 14 \times 13 \times 12 \times 11 !}{ 11! \times 5!}}\\\\\bold{ = \frac{16 \times 15 \times 14 \times 13 \times 12}{ 5!}}\\\\\bold{ = \frac{16 \times 15 \times 14 \times 13 \times 12}{ 5\times 4\times 3 \times 2 \times 1}}\\\\\bold{ = 8 \times 3 \times 14 \times 13 }\\\\\bold{ = 4368 }$

P(accepting shipment) = P(all 5 modems work):

$= \bold{\frac{^{12}C_5}{ ^{16}C_{5}}}$

$\bold{=\frac{792}{4368}}\\\\\bold{=0.18131}\\\\\bold{=0.18131 \times 100= 18.131 \approx 18.131\%}$

Therefore, the final answer is "18.131%".

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