# Probability Question - colored balls in 2 bowls - Baye's formula?

• GreenPrint
In summary, Bayes' formula is a mathematical equation used in probability to calculate the likelihood of an event occurring based on prior knowledge or information. It can be applied to various situations, such as the colored balls in 2 bowls problem, to determine the probability of a specific outcome. However, it has limitations, including the need for accurate prior probabilities and the assumption of independent events.
GreenPrint
Probability Question -- colored balls in 2 bowls -- Baye's formula?

Bowl #1 contains 2 red balls and 2 white balls
Bowl #2 contains 3 red balls and 2 white ball

one bowl is chosen at random (each is equally likely)

(A) What is the probability of choosing a red ball?

Let R stand for choosing a red ball
$B_{#}$ = Bowel #

$P(R) = P(B_{1})P(R|B_{1})+P(B_{2})P(R|B_{2})$
$P(R) = \frac{1}{2} \frac{1}{2} + \frac{1}{2} + \frac{3}{4} = .625$

(B)
$P(B_{1}) = P(R)P(B_{1}|R)$

I don't know how to calculate $P(B_{1}|R)$. Do I need to use Baye's formula?

Thanks for any help! Hopefully I'm not making this to complicated again lol.

Last edited by a moderator:

GreenPrint said:
Bowl #1 contains 2 red balls and 2 white balls
Bowl #2 contains 3 red balls and 2 white ball

one bowl is chosen at random (each is equally likely)

(A) What is the probability of choosing a red ball?

Let R stand for choosing a red ball
$B_{#}$ = Bowel #

$P(R) = P(B_{1})P(R|B_{1})+P(B_{2})P(R|B_{2})$
$P(R) = \frac{1}{2} \frac{1}{2} + \frac{1}{2} + \frac{3}{4} = .625$

(B)
$P(B_{1}) = P(R)P(B_{1}|R)$

I don't know how to calculate $P(B_{1}|R)$. Do I need to use Baye's formula?

Thanks for any help! Hopefully I'm not making this to complicated again lol.

I'm not an expert in these things but you got part (A) wrong. Try it again. And for part (B), now it sounds like a good time to use Baye's formula.

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Is it just (1/2)(3/4) for part A?

GreenPrint said:
Is it just (1/2)(3/4) for part A?

No, no. It's given by exactly the formula you wrote except you've got P(R|B_2) wrong. What is that? It's not 3/4.

GreenPrint said:
(B)
What is the actual question for B?
$P(B_{1}) = P(R)P(B_{1}|R)$
That would not be true in general. $P(B_{1} \wedge R) = P(R)P(B_{1}|R)$, $P(B_{1}) = P(R)P(B_{1}|R) + P(\neg R)P(B_{1}|(\neg R)$

Ok so I figured out my mathematical miscalculation in A and was able to figure out the answer to B (I realized I forgot to post it).

What is that symbol mean?
¬?

GreenPrint said:
Bowl #1 contains 2 red balls and 2 white balls
Bowl #2 contains 3 red balls and 2 white ball

one bowl is chosen at random (each is equally likely)

(A) What is the probability of choosing a red ball?

Let R stand for choosing a red ball
$B_{#}$ = Bowel #

$P(R) = P(B_{1})P(R|B_{1})+P(B_{2})P(R|B_{2})$
$P(R) = \frac{1}{2} \frac{1}{2} + \frac{1}{2} + \frac{3}{4} = .625$
Surely you don't mean that second "+"?
$$\frac{1}{2}\frac{1}{4}+ \frac{1}{2}\frac{3}{5}= \frac{1}{8}+ \frac{3}{10}$$
Ignoring your typo with the "+", the probability that, drawing from the second bowl, you will draw a red ball, is 3/(3+ 2)= 3/5, not 3/4.

(B)
$P(B_{1}) = P(R)P(B_{1}|R)$
What, exactly, is the question? This is simply a "formula". What are you to do with it?
(And, by the way, an incorrect formula. After having picked a single ball, the probability that it was picked from bowl 1 is $P(B_1)= P(R)P(B_1|R)+ P(W)P(B_1|W)$)

I don't know how to calculate $P(B_{1}|R)$. Do I need to use Baye's formula?
That at least is easy to answer. There are a total of 5 red balls, 2 of them from bowl 1. Given that you picked a red ball, the probability it came from bowl 1 is 2/5.

Thanks for any help! Hopefully I'm not making this to complicated again lol.

GreenPrint said:
What is that symbol mean?
¬?

It's a logical 'not'.

## What is Bayes' formula and how is it used in probability?

Bayes' formula is a mathematical equation used in probability to calculate the likelihood of an event occurring based on prior knowledge or information. It is often used in situations where there are multiple possible outcomes and some uncertainty about the true probability of each outcome. The formula is P(A|B) = (P(B|A) * P(A)) / P(B), where P(A) represents the prior probability of event A, P(B) represents the prior probability of event B, and P(A|B) represents the probability of event A occurring given that event B has occurred.

## What is the difference between a prior probability and a posterior probability?

A prior probability is the initial probability assigned to an event before any new information or data is taken into account. A posterior probability is the updated probability of an event after new information or data is considered. Bayes' formula allows for the calculation of posterior probabilities by incorporating prior probabilities and new information.

## How can Bayes' formula be applied to the colored balls in 2 bowls problem?

In the colored balls in 2 bowls problem, Bayes' formula can be used to calculate the probability of drawing a certain color ball from a specific bowl, given that a ball of that color has been drawn. The prior probabilities in this case would be the initial number of balls of each color in each bowl, while the posterior probabilities would be the updated number of balls after each draw. By inputting these values into the formula, the probability of drawing a certain color ball from a specific bowl can be calculated.

## How does the number of balls and colors in each bowl affect the outcome in the colored balls in 2 bowls problem?

The number of balls and colors in each bowl have a direct impact on the probability of drawing a specific color ball from a specific bowl. The more balls of a certain color in a bowl, the higher the probability of drawing that color. Additionally, the total number of balls in each bowl will also affect the overall probability of drawing a specific color, as it will impact the denominator in Bayes' formula.

## Are there any limitations to using Bayes' formula in probability calculations?

While Bayes' formula is a useful tool for calculating probabilities, it does have its limitations. One limitation is that it relies on accurate prior probabilities, which may not always be available or easy to determine. Additionally, the formula assumes that events are independent, which may not always be the case in real-world scenarios. Furthermore, the formula can become more complex and difficult to use when there are multiple events and factors at play.

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