a lot of 100 bulbs from manufacturing process is known to contain 10 defective and 90 nondefective bulbs. if a sample of 8 bulbs is selected at random what is the probability that the sample has 3 defective and 5 nondefective bulbs

Dear Student,

Please find below the solution to the asked query:

Given : Total number of bulbs in the lot    =  10 Defective +  90 Non-Defective =  100

Number of bulbs selected = 8

We know formula for probability :

Probability P ( E )  = $\frac{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{desired} \mathrm{events} \mathrm{n} ( \mathrm{E} )}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{events} \mathrm{n} ( \mathrm{S} )}$

And Selecting 8 bulbs from the lot

Total No. of Possible Choices =  Number of ways in which the 8 bulbs can be selected from the total 100 bulbs = n ( S ) = ¹⁰⁰C₈
So,
n ( S ) = $\frac{100 !}{8 !\hspace{0.17em}\times \left( 100 - 8 \right)!} = \frac{100 \times 99 \times 98 \times 97 \times 96 \times 95 \times 94 \times 93 \times 92 !}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\hspace{0.17em}\times 92 !}= \frac{100 \times 99 \times 98 \times 97 \times 96 \times 95 \times 94 \times 93 \times 92 !}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\hspace{0.17em}\times 92 !}= 10 \times 33 \times 7 \times 97 \times 95 \times 94 \times 93$

And 
Sample of 8 bulbs is selected at random and the sample has 3 defective and 5 non defective bulbs , So 

n ( E ) = ¹⁰C₃ $\times$ ⁹⁰C₅  =
$$\begin{gathered} \frac{10 !}{3 !\hspace{0.17em}\times \left( 10 - 3 \right)!} \times \frac{90 !}{5 !\hspace{0.17em}\times \left( 90 - 5 \right)!} \\ \Rightarrow \frac{10 \times 9 \times 8\times 7 !}{3 \times 2 \times 1\hspace{0.17em}\times 7!} \times \frac{90 \times 89 \times 88 \times 87 \times 86 \times 85 !}{5 \times 4 \times 3 \times 2 \times 1\hspace{0.17em}\times 85!} \\ \Rightarrow 5 \times 3 \times 8 \times 3 \times 89 \times 22 \times 87 \times 86 \end{gathered}$$
Therefore,


Probability that the sample has 3 defective and 5 non defective bulbs = $\frac{5 \times 3 \times 8 \times 3 \times 89 \times 22 \times 87 \times 86}{10 \times 33 \times 7 \times 97 \times 95 \times 94 \times 93 } = \frac{\mathbf{4}\mathbf{ }\mathbf{\times }\mathbf{ }\mathbf{89}\mathbf{ }\mathbf{ }\mathbf{\times }\mathbf{ }\mathbf{87}\mathbf{ }\mathbf{\times }\mathbf{ }\mathbf{86}}{\mathbf{ }\mathbf{7}\mathbf{ }\mathbf{\times }\mathbf{ }\mathbf{97}\mathbf{ }\mathbf{\times }\mathbf{ }\mathbf{95}\mathbf{ }\mathbf{\times }\mathbf{ }\mathbf{47}\mathbf{ }\mathbf{\times }\mathbf{ }\mathbf{31}\mathbf{ }}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)}$

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