Conditional probability problem

[MathMan2022]

Homework Statement

45% of the population is said to for a certain political block A at an election. 10 people are sampled.

a) Whats the prob that 5 of them vote block A?

b) What the prob that none of them vote block A?

Relevant Equations

P(A and B) = P(A) * P(B)
P(Not B) = 1 - P(B)

A) P(A and B) = 0.45 * 5/10
B P(Not B) = 1 - ( 0.45 * 5/10)

Is it like this?

 

[PeroK]

Homework Statement:: 45% of the population is said to for a certain political block A at an election. 10 people are sampled.

a) Whats the prob that 5 of them vote block A?

b) What the prob that none of them vote block A?
Relevant Equations:: P(A and B) = P(A) * P(B)
P(Not B) = 1 - P(B)

A) P(A and B) = 0.45 * 5/10
B P(Not B) = 1 - ( 0.45 * 5/10)

Is it like this?

It's nothing like that. What have you learned about probability theory so far? Have you heard the term binomial coefficients?

 

[MathMan2022]

It's nothing like that. What have you learned about probability theory so far? Have you heard the term binomial coefficients?

Oh its like that? Yes I have heard of that.

P(X = r) = K(n,r)*p^r*(1-p)^(n-r) right?

 

[MathMan2022]

So
a) P(X = 5) = K(10, 5)*0.45^5*(1 - 0.45)^(10 - 5)= 0.23
b) P(X = no votes) = 1- P(X=5) = 1-0.23

 

[PeroK]

So
a) P(X = 5) = K(10, 5)*0.45^5*(1 - 0.45)^(10 - 5)= 0.23

That looks a lot better.

b) P(X = no votes) = 1- P(X=5) = 1-0.23

Why would no votes be the complement of 5 votes? That's the probability of any number of votes except 5.

 

[MathMan2022]

If 45% is sampled to vote for block A. Then 55 % must non voters for block A?

 

[PeroK]

If 45% is sampled to vote for block A. Then 55 % must non voters for block A?

No, it means that 55% don't vote for block A.

 

[PeroK]

If 45% is sampled to vote for block A. Then 55 % must non voters for block A?

Look at it this way, suppose you change part a) to calculate $P(X = 4)$. Would your answer to part b) change to $P(X = 0) = 1 - P(X = 4)$?

 

[MathMan2022]

Look at it this way, suppose you change part a) to calculate $P(X = 4)$. Would your answer to part b) change to $1 - P(X = 4)$?

That would that is the prob that 4 people or less voted for block A?

 

[PeroK]

That would that is the prob that 4 people or less voted for block A?

No, that would be the probablity that 0, 1, 2, 3, 5, 6, 7, 8, 9 or 10 people vote for block A. Any number but $4$. Note that we have:
$$\sum_{n = 0}^{10} P(X = n) = 1$$

 

[MathMan2022]

No, that would be the probablity that 0, 1, 2, 3, 5, 6, 7, 8, 9 or 10 people vote for block A. Any number but $4$. Note that we have:
$$\sum_{n = 0}^{10} P(X = n) = 1$$

Then that would be P(X=0) I am searching for? Because that would be the prob that non of the 10 voted for block A.

 

[PeroK]

Then that would be P(X=0) I am searching for? Because that would be the prob that non of the 10 voted for block A.

Yes, exactly.

 

[PeroK]

Here's a tip. The Excel spreadsheet has a binomial distribution function (and other useful statistical things). For example, if you type:

=BINOMDIST(5, 10, 0.45, FALSE)

Then, you'll get the answer $0.23$.

See the Excel help pages for more information.

 

[Office_Shredder]

You can do the exact same thing for 0 people that you did for 5 people.

 

[Hornbein]

This is not conditional probability. It's plain ordinary probability.

 

[BWV]

Although you don’t need binomial for zero, it’s analogous to flipping heads ten times in a row (as there is only one combination)