Question

I am still struggling to understand the logic of th Ti-inspire CX calculator for inverse binomial N

Sample Problem
Monique is practicing goal shots for netball. She knows from experience that her chance of making any one shot is70%. She plans to practice until she scores 50 goals. How many shots must she attempt to ensure that the probability of making at least 50 goals is more than 0.99?

What do I need to input in the Ti nspire under:

--> probability  --> distributions --> Inverse binomial N

Ti-nspire CX Input

Cummulative Prob: 0.01
Prob of Success: 0.7
Successes, x: 49
Result min 86 trials.

What i still don't understand is, why I need to input the probability for the distribution between equal or less than 49 goals which is P=0.01 and then the successes x = 49 while I am looking for the probability for 50 or above goals with a probability of more then 0.99. Hence I am inputting the values for the goals 0=49 while the result reflects the goals from 50 and above ....this is confusing to understand.

can you please explain...?

Thanks

Answer 1

See the binomial distribution is a discrete distribution and X takes values as 0, 1, 2, ...,n

By using the law of total probability, we have

P(X = 0) + P( X = 1) + P(X= 2) + .... +P( X = n-1) + P( X = n) = 1 ....( 1 )

Here we want to find "n", such that P(X 50) 0.99

That is P(X = 50) + P(X = 51) + .... +P( X = n-1) + P(X = n) 0.99

Let's break equation ( 1 ), in two parts:

[ P(X = 0) + P( X = 1) + P(X= 2) + .... + P( X = 49) ] + [ P(X = 50) + P(X = 51) + .... +P( X = n-1) + P(X = n) ]= 1

P(X = 0) + P( X = 1) + P(X= 2) + .... + P( X = 49) = 1 - [ P(X = 50) + P(X = 51) + .... +P( X = n-1) + P(X = n) ] ....( 2 )

Here we want P(X = 50) + P(X = 51) + .... +P( X = n-1) + P(X = n) 0.99

So just put 0.99 in equation ( 2 ) instead of [P(X = 50) + P(X = 51) + .... +P( X = n-1) + P(X = n) ] ,

then we get the following equation:

P(X = 0) + P( X = 1) + P(X= 2) + .... + P( X = 49) 1 - 0.99 = 0.01  

P(X = 0) + P( X = 1) + P(X= 2) + .... + P( X = 49) can be written as P(X 49)

So finally, we get P(X 49) 0.01

Also it is given that the probability of success is 0.70

Note that: Here we convert greater than or equal to probability ( ) into less than or equal probability ( ) , by using the law of tot, because this calculator find "n" only for less than or equal to probability ().

Therefore, our inputs are as:

Cumulative Prob: 0.01

Prob of Success: 0.7

So we put successes , x: 49