What property does this binomial probability calculation use? I have an equation sum_{k=2}^{7} ((7),(k)) 0.01^k(1-0.01)^{7-k}=1-(0.99)^7-7(0.01)(0.99)^6 approx 0.002031

Gaige Haynes

Gaige Haynes

Answered question

2022-09-09

What property does this binomial probability calculation use?
I have an equation
k = 2 7 ( 7 k ) 0.01 k ( 1 0.01 ) 7 k = 1 ( 0.99 ) 7 7 ( 0.01 ) ( 0.99 ) 6 0.002031
I don't know what property the teacher used to quickly transform the summation to two simple equations. Can someone please give me a hint?

Answer & Explanation

cerfweddrq

cerfweddrq

Beginner2022-09-10Added 15 answers

Step 1
I'm going to call the random variable of interest X and that it follows a Bin(7,.01). Then the probability of interest appears to be
P ( X 2 ) = k = 2 7 ( 7 k ) ( .01 ) k ( .99 ) 7 k .
Step 2
However, using complements results in an easier calculation:
P ( X 2 ) = 1 ( X 1 ) = 1 k = 0 1 ( 7 k ) ( .01 ) k ( .99 ) 7 k ,
which is what you have.
ridge041h

ridge041h

Beginner2022-09-11Added 2 answers

Explanation:
The property is the Law of Total Probability, but you can also do it this way.
k = 2 7 7 C k p k ( 1 p ) 7 k = ( k = 0 7 7 C k p k ( 1 p ) 7 k ) ( k = 0 1 7 C k p k ( 1 p ) 7 k ) = ( p + ( 1 p ) ) 7 ( k = 0 1 7 C k p k ( 1 p ) 7 k ) Binomial Expansion = 1 ( k = 0 1 7 C k p k ( 1 p ) 7 k ) = 1 7 C 0 p 0 ( 1 p ) 7 0 7 C 1 p 1 ( 1 p ) 7 1 = 1 ( 1 p ) 7 7 p ( 1 p ) 6 k = 2 7 7 C k ( 0.01 ) k ( 0.99 ) 7 k = 1 ( 0.99 ) 7 7 ( 0.01 ) ( 0.99 ) 6 = 0.002 031 041 634 94

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