I have a probability and statistic question that is listed in the description area.

Question

A tray of electronic components contains 15 components, 4 of which are defective. If 4 components are selected, what is the possibility for the following

All four are defective.
Three are defective, and one is good.
None are defective.
Explain the differences in the solution process for each of the scenarios with the same basic information as given above. Provide lots of detail regarding formulas and rationale, as the explanation is what is important.

Answer 1

15 items, 4 defective, therefore 15-4=11 non-defective. --------------------------------------------------------------------------------- 1st time we pick a component the probability to be defective is 4/15 because we have 15 component total and 4 are defective. 2nd time the probability to be defective is 3/14. Since we have 14 component left and 3 are defective. 3rd time the probability to be defective is 2/13. Since we have 13 component left and 2 are defective.

4th time the probability to be defective is 1/12. Since we have 12 component left and 1 is defective.

 

P ( All 4 defective) = 4/15 * 3/14 * 2/13 * 1/12 = 1/1365

 

General formula for this kind of questions is: n1Cr1* n2Cr2 * .... / nCr and nCr = n!/r!(n-r)! is the combination formula (order does not matter) We are choosing r1= 4 item from n1= 4 defective 4!/ [(4-4)! * 4!] = 1 and r2=0 item from n2=11 non-defective 11!/[ (11-11)! * 0!} =1 All or nothing combination is always 1. when r=n or r=0 . We choose r=4 out of n=15 in total. 15!/ [(15-4)! * 4!] = 15!/ [11! *4!] = (12*13*14*15)/(1*2*3*4) = 1365 after simplifying. P ( 4 defective) =( 1 * 1)/1365 = 1/1365

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P ( 3 defective and 1 non-defective) n1Cr1* n2Cr2 * .... / nCr and nCr = n!/r!(n-r)! is the combination formula (order does not matter) We are choosing r1= 3 item from n1= 4 defective 4!/ [(4-3)! * 3!] = 4 and r2=1 item from n2=11 non-defective 11!/[ (11-1)! * 1!} =11 We choose r=4 out of n=15 in total. 15!/ [(15-4)! * 4!] = 15!/ [11! *4!] = (12*13*14*15)/(1*2*3*4) = 1365 after simplifying. P ( 3 defective and 1 non-defective) = (4*11)/1365=44/1365

 

 

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None are defective

 

Like the first senario.

 

1st time we pick a component the probability not defective is 11/15 because we have 15 component total and 11 are not defective.
2nd time the probability not to be defective is 10/14. Since we have 14 component left and 10 are not defective.
3rd time the probability not to be defective is 9/13. Since we have 13 component left and 9 are not defective.
4th time the probability not to be defective is 8/12. Since we have 12 component left and 8 are not defective.

 

P ( All 4 not defective) = 11/15 * 10/14 * 9/13 * 8/12 = 22/91

n1Cr1* n2Cr2 * .... / nCr and nCr = n!/r!(n-r)! is the combination formula (order does not matter) We are choosing r1= 4 item from n1= 11 non-defective 11!/ [11-4)! * 4!] = 11!/ [7! * 4!] = (8*9*10*11)/(1*2*3*4) = 330 and r2=0 item from n2=4 defective 4!/[ (4-4)! * 0!} =1 All or nothing combination is always 1. when r=n or r=0 . We choose r=4 out of n=15 in total. 15!/ [(15-4)! * 4!] = 15!/ [11! *4!] = (12*13*14*15)/(1*2*3*4) = 1365 after simplifying. P ( 4 non-defective) =( 330 * 1)/1365 = 330/1365= 22/91