# Probability of selecting defective light bulbs

1. Jan 29, 2009

### mamma_mia66

1. The problem statement, all variables and given/known data

A box contains 24 light bulbs of which 4 are defective. If one person selects 10 bulbs from the box in a random manner, and a second person then takes the remaining 14 bulbs, what is the probability that all 4 detective bulbs will be obtained by the same person.?

2. Relevant equations

I don't know how to show the solution because I can't find the big parentheses and numbers are one above the other (24 10) 10 from 24 Combinatorial Methods

3. The attempt at a solution

Here we 24 light bulbs in that 4 aare defectives.

First person takes 10 bulbs from the box that can be done in the (10 form 24) ways.

In that 4 are defective items remeaning are not defective items.

probability that all 4 detective bulbs will be obtained by the same person
Required probability

Last edited: Jan 29, 2009
2. Jan 29, 2009

### mamma_mia66

Re: probability

probability (4 from 4)*(6 from 20)/(10 from 24) + (10 from 20)/(10 from 24).

Does someone know how to write this in form for combinatorial methods. I want to verify my answer.

Please help.

3. Jan 29, 2009

### kasse

Re: probability

Looks correct to me. To calculate (24 10) for example; use the $$_n C_r$$ on your calculator: $$24 \ _nC_r \ 10$$.

Probability that the first person selects all the four defective ones:

(4 4)(20 6)/(24 10) = 0.01976

Probability that the second person selects all the four defective ones:

(4 0)(20 10)/(24 10) = 0.09420

Then you can add the two probabilities because they are independent occurences:

P = 0.1140

Last edited: Jan 29, 2009
4. Jan 29, 2009

### mamma_mia66

Re: probability

Thank you. I forgot to check the the answer in the textbook, it is given only :

(20 6) + (20 10) / (24 10).

Thank you again.

5. Jan 29, 2009

### mamma_mia66

Re: probability

where is the remaining 14. In that 4 are defective items remaining are not defective items.
then:

(4 4)(20 6) / (24 10) + (20 10) /(24 14)=0.114

that way answer is the same but not the same in the book.

6. Jan 29, 2009

### kasse

Re: probability

Just two different ways of regarding the same problem. The probability of choosing 14 (including all the 4 deficient ones) out of 24, is the same as choosing 10 normal bulbs out of 24.

7. Jan 29, 2009

### mamma_mia66

Re: probability

Thanks.

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