Bayesian Probability

[AvgStudent]

Randall uses a random number generator to generate a number q that is uniformly distributed between 0 and 1. Randall splits a wheel into two segments with relative areas q and 1−q, respectively.

Then, Randall spins the wheel repeatedly. If the spin lands on the segment that corresponds to q, Randall gets a piece of candy. Randall makes ten spins and gets seven pieces of candy.

Calculate the probability that Randall will get a piece of candy on his 11th spin.

Answer Choices: [imath]\dfrac{1}{3}, \dfrac{1}{2},\dfrac{3}{5},\dfrac{2}{3}, \dfrac{7}{10}[/imath]

I don't understand the question, so I can't set up the problem...

 

[BigBeachBanana]

Randall uses a random number generator to generate a number q that is uniformly distributed between 0 and 1. Randall splits a wheel into two segments with relative areas q and 1−q, respectively.

Then, Randall spins the wheel repeatedly. If the spin lands on the segment that corresponds to q, Randall gets a piece of candy. Randall makes ten spins and gets seven pieces of candy.

Calculate the probability that Randall will get a piece of candy on his 11th spin.

Answer Choices: [imath]\dfrac{1}{3}, \dfrac{1}{2},\dfrac{3}{5},\dfrac{2}{3}, \dfrac{7}{10}[/imath]

I don't understand the question, so I can't set up the problem...

As I understand the question...
Let [imath]N_i[/imath] be a trial that Randall lands on q. Since q varies with the random generator, then we have: [imath]N_i|q \sim\text{Bernoulli}(q)[/imath].
Also, you were given that [imath]q\sim\text{Uniform}(0,1)[/imath]
Furthermore, after 10 trials, Randall landed on q 7 times and did not land on q 3 times, so [imath]\Pr(N_1,...,N_{10}|q)=q^7(1-q)^3[/imath]
By Bayesian probability, [imath]\Pr(q|N_1,...,N_{10})=\Pr(N_1,...N_{10}|q)\cdot\Pr(q)[/imath]
The question is asking for [imath]\Pr(N_{11}|N_1,...,N_{10})=E[\Pr(N_{11}=1|q)|N_1,...,N_{10}][/imath]

 

[AvgStudent]

As I understand the question...
Let [imath]N_i[/imath] be a trial that Randall lands on q. Since q varies with the random generator, then we have: [imath]N_i|q \sim\text{Bernoulli}(q)[/imath].
Also, you were given that [imath]q\sim\text{Uniform}(0,1)[/imath]
Furthermore, after 10 trials, Randall landed on q 7 times and did not land on q 3 times, so [imath]\Pr(N_1,...,N_{10}|q)=q^7(1-q)^3[/imath]
By Bayesian probability, [imath]\Pr(q|N_1,...,N_{10})=\Pr(N_1,...N_{10}|q)\cdot\Pr(q)[/imath]
The question is asking for [imath]\Pr(N_{11}|N_1,...,N_{10})=E[\Pr(N_{11}=1|q)|N_1,...,N_{10}][/imath]

Thank you! I got the right answer 2/3.