Exercise in Probability - balls drawn from a box

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Discussion Overview

The discussion revolves around a probability exercise involving drawing balls from a box, where 10% of the balls are red. Participants explore the probability of drawing more than 3 red balls when 20 balls are drawn with replacement. The focus is on understanding the binomial distribution and calculating probabilities related to this scenario.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • Some participants identify the problem as involving a binomial distribution and inquire about the probability of drawing exactly 0 red balls.
  • One participant suggests calculating the probability of drawing more than 3 red balls by using the complement rule: \( P[X > 3] = 1 - P[X \leq 3] \).
  • Another participant confirms the approach and provides the formula for calculating \( P(X > 3) \) in terms of \( P(X = i) \) for \( i = 0, 1, 2, 3 \).
  • Participants express familiarity with the binomial distribution, indicating that it may simplify the calculations involved.

Areas of Agreement / Disagreement

Participants generally agree on the use of the binomial distribution and the approach to calculating the desired probabilities, though specific calculations and interpretations are still being discussed.

Contextual Notes

Some assumptions about the understanding of binomial distribution and probability calculations are present, but not all participants have confirmed their familiarity with the necessary concepts.

evinda
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Hey! I need some help at the following exercise...

We have a box with balls and 10% of them are red. If we choose at random 20 balls with replacement, which is the probability to pick more than 3 red balls?

Thanks in advance!
 
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Re: Exercise in Probability

evinda said:
Hey! I need some help at the following exercise...

We have a box with balls and 10% of them are red. If we choose at random 20 balls with replacement, which is the probability to pick more than 3 red balls?

Thanks in advance!

Welcome to MHB, evinda! :)

This is about a binomial distribution.
Do you have notes on that?

Can you say for starters what the probability on exactly 0 red balls is?
 
Re: Exercise in Probability

evinda said:
Hey! I need some help at the following exercise...

We have a box with balls and 10% of them are red. If we choose at random 20 balls with replacement, which is the probability to pick more than 3 red balls?

Thanks in advance!

Hi evinda,

Welcome to MHB! It seems to me this is the binomial distribution, but maybe it's not necessary to worry about that if you haven't been introduced to this distribution. How would you find the probability that all twenty balls are red?

I see that I like Serena has beaten me to a reply but I still want to say hello and welcome. :)

Jameson
 
Re: Exercise in Probability

I like Serena said:
Welcome to MHB, evinda! :)

This is about a binomial distribution.
Do you have notes on that?

Can you say for starters what the probability on exactly 0 red balls is?

The probability on exactly 0 re balls is P(X=0)={20 choose 0}(0.1)^0*(0.9)^(20-0)=(0.9)^20...

- - - Updated - - -

Jameson said:
Hi evinda,

Welcome to MHB! It seems to me this is the binomial distribution, but maybe it's not necessary to worry about that if you haven't been introduced to this distribution. How would you find the probability that all twenty balls are red?

I see that I like Serena has beaten me to a reply but I still want to say hello and welcome. :)

Jameson

The probability that all 20 balls are red is {20 choose 20}*(0.1)^20*(0.9)^(20-20)=0.1^20...Thank you very much! ;)
 
Re: Exercise in Probability

So you are familiar with binomial distribution, great! That will make this much easier to do. There is one "trick" you can use here to make this calculation much easier. Let $X$ be a random variable which represents the number of red balls drawn. $$P[X >3]=1-P[X \le 3]$$. So instead of over 15 probabilities to calculate now you should be able to solve this through 4 calculations. Do you see how?
 
Re: Exercise in Probability

P(X>3)=1-P(X<=3)=1-(P(X=0)+P(X=1)+P(X=2)+P(X=3)), where P(X=i)={20 choose i}(0.1)^i*(0.9)^(20-i), i=0,1,2,3...Right?
 
Re: Exercise in Probability

Right! ;)
 
Re: Exercise in Probability

Ok,thanks! :p
 

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