Probability of selecting defective light bulbs

[mamma_mia66]

Homework Statement

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.?

Homework 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

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

 

[mamma_mia66]

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.

 

[kasse]

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

 

[mamma_mia66]

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.

 

[mamma_mia66]

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.

 

[kasse]

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.

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.

 

[mamma_mia66]

Thanks.