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Question 1:

A box contains 12 light bulbs of which 5 are defective. All the bulbs look alike and have equal probability of being chosen. Three bulbs are picked up at random. What is the probability that at least 2 are defective?

using combinations:

2 defective: $ {5 \choose 2}{7 \choose 1} $

3 defective: $ {5 \choose 3}{7 \choose 0} $

so:

Probability = $ \Large\frac{{5 \choose 2}{7 \choose 1} + {5 \choose 3}{7 \choose 0}}{12 \choose 3} =\frac {4}{11}$


Then I used binomial probability:

2 defective: $ \Large{3 \choose 2} \left(\frac{5}{12}\right)^2\left(\frac{7}{12}\right)^1 = \frac{175}{576}$

3 defective: $ \Large{3 \choose 3} \left(\frac{5}{12}\right)^3\left(\frac{7}{12}\right)^0 = \frac{125}{1728}$

Probability $=\large\frac{175}{576} + \frac{125}{1728} = \frac{325}{864}$


Here is another similar problem. Binomial probability is used in the book to solve this.

Question 2

A package contains 50 similar components and inspections shows that four have been damaged during transit. If six components are drawn at random from the contents of the package determine the probabilities that in this sample (a) one and (b) less than three are damaged.

a.) P(1 damaged) = ?

b.) P(<3 damaged) = ?

using combinations:

a.)

1 damaged: $ \Large\frac{{4 \choose 1}{46 \choose 5}}{50 \choose 6} = 0.3450$

b.)

0 damaged: $ {4 \choose 0}{46 \choose 6} $

1 damaged: $ {4 \choose 1}{46 \choose 5} $

2 damaged: $ {4 \choose 2}{46 \choose 4} $

so:

Probability = $ \Large\frac{{4 \choose 0}{46 \choose 6} + {4 \choose 1}{46 \choose 5} + {4 \choose 2}{46 \choose 4} }{{50 \choose 6}} =0.9961$


using binomial probability:

a.)

1 damaged: $ \Large{6 \choose 1} \left(\frac{4}{50}\right)^1\left(\frac{46}{50}\right)^5 = 0.3164$

b.)

0 damaged: $ \Large{6 \choose 0} \left(\frac{4}{50}\right)^0\left(\frac{46}{50}\right)^6 = 0.6064$

1 damaged: $ \Large{6 \choose 1} \left(\frac{4}{50}\right)^1\left(\frac{46}{50}\right)^5 = 0.3164$

1 damaged: $ \Large{6 \choose 2} \left(\frac{4}{50}\right)^2\left(\frac{46}{50}\right)^4 = 0.0688$

Probability $= 0.6064 + 0.3164 + 0.0688 = 0.9916$

Why am I getting different answers?

What did I do wrong here?

When should I use binomial probability?

How should I choose on what procedure I should do in these type of problems?

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The binomial distribution is used to model the number of successful outcomes when a sample of size $n$ is drawn with replacement from a population of size $N$. The hypergeometric distribution is used to model the number of successful outcomes when a sample of size $n$ is drawn without replacement from a population of size $N$.

Thus, in each of the examples you considered, it makes sense to use the hypergeometric distribution since sampling is done without replacement. That said, when $n << N$, the binomial distribution can be used as a good approximation to the hypergeometric distribution. This is why the author of your text used the binomial distribution in the second example.

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  • $\begingroup$ Just to clarify. All the answers I obtained are valid, but using combinations (hypergeometric distribution) is more accurate because in this case the sample size is drawn without replacement. Binomial distribution could also be used even if sampling is done without replacement because n << N. But the value you will get will just be an approximation. Did I say something wrong here? $\endgroup$ – user41567 Jul 3 '17 at 1:57
  • $\begingroup$ That is correct. $\endgroup$ – N. F. Taussig Jul 3 '17 at 2:24

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