# 4 part ap stats question, auto 5 stars?

I'm really having trouble with this example and I would appreciate someone doing the steps so I can see the work and how you did it.

Glenn likes the game at the state fair where you toss a coin into a saucer. You win if the coin comes to rest in the saucer without sliding off. There is 8% chance of winning and each toss is independent.a. How many tosses should Glenn expect to take before he wins? With what standard deviation?b. What is the probability that glenn will have his first win on his 9th toss? 17th?c. What is the probability that he sees a win within his first 10 tosses?d. What is the probability that it takes more than 24 tosses to win a stuffed animal?

### 1 Answer

- AlanLv 72 months agoFavorite Answer
Standard format for the Negative Binomial Distribution

gives us the number of successes before the rth failure.

However, wikipedia gives an alternative format for

the equation of the number of failures before the kth success.

I used this formula from this wikipedia article for answers b,c, d .

You are looking for the numbers of rolls, but in your case, the number of rolls

is just the number of failures + 1

https://en.wikipedia.org/wiki/Negative_binomial_di...

so b,c, and d are just plug and chug into a formula mostly.

a.

Is a little harder.

so for the standard Negative Binomial Distribution

E(number of failures ) = pr / (1-p)

We can treat our case as reverse success for failure and r for k

so k = 1

new p = 1- original_p

r=k = 1

new 1-p = p

E(number of failures) = (1-original_p) / p = 0.84/0.16 = 5.25

but we want the number of rolls

so number of rolls = one success + number of failures

E(number of rolls ) = 5.25 + 1 = 6.25

so remembering that when you add a constant to variable it doesn't change

its standard deviation.

Variance from wikipedia for standard form = pr/(1-p)^2

again we can reverse success for failure and

k for r

we get

Variance = (1-original)1/ (p)^2 = 0.84/ (0.16)^2 =

Variance = 32.8125

Standard deviation = sqrt( 32.8125) = 5.728219619

Alternate formula shown for this case.

Negative binomial probability function is

P( X=r) = ((k+r -1) (r) )(p)^k (1-p)^r

where r = number of failures before the kth success

where p = probability of success

b. What is the probability that glenn will have his first win on his 9th toss? 17th?

Alternate formula shown for this case.

Negative binomial probability function is

P( X=r) = ((k+r -1) (r) )(p)^k (1-p)^r

This version gives the number of failures before the 1st success.

but we are using special case or k = 1 (first success)

P(x =r) = ( (1+r -1) r) (p)^1 (1-p)^r

P(x =r) = ( r r ) p*(1-p)^r= p*(1-p)^r

or in terms of n

but we are looking for

n = number of rolls before 1st success

k = 1

n = r+1

r = n-1

so we need to equal the number of rolls

P(X =n )= P(r = (n-1) ) =

probability on 9th toss

P(x_rolls=9) = P(r = 8 ) =1*(0.08)*(0.92)^8

P(X_rolls = 9) = 0.04105751

P(X_rolls= 17)= P(r = 16) = (0.08)*(0.92)^16 = 0.021071489

c.

P( X_rolls<= 10) = P(X_rolls =1) + P(X_rolls =2 ) + P(X_rolls=3 )

+P(X_rolls= 4) + P(x_rolls = 5) + P(x_rolls = 6) +P(X_rolls = 7)

P(X_rolls =8) + P(x_rolls = 9) + P(x_rolls = 10)

P(X_rolls<=10)= P(r = 0) + P(r=1 ) +P(r= 2) + P(r =4)

P(r =5) + P(r = 6 ) + P(r=7) +P(r =8) +P(r=9)

P(x_rolls <=10) = 0.08*(1 + 0.92^1 + 0.92^2 + ... + 0.92^9)

since 2nd half is geometric series with a = 1 , n = 9 , r = 0.92

P(x_rolls < = 10) = 0.08 *( a (1-r^n) / (1-r) ) =

P(x_rolls < = 10) = 0.08 *((1 - 0.92^9) / (1-0.92)) = (1- 0.92^9)

P(x_rolls < = 10) = 0.527838637

d.

P(X_rolls > 24 ) = 1- P(x<=24)

using the example from c

P(X_rolls<= 24) = (1- 0.92^23)

P(X_rolls <= 24) = 1 - (1 -0.92^23) = 0.92^23 = 0.146933231

I appreciate you!