Sample space and probability measure.

  • #1
If we have two red balls and we are to choose one it is true that the probability measure of picking a red ball is 1. In this case it is understood that the sample space only contains a red ball and this is because events in the sample space are to be disjoint.

But how come this is not the case when we try to calculate the probability of picking a red ball out of a basket with 49 yellow balls and 1red ball.

If I write down the "sample space" such that the outcomes are disjoint I will only have 2 outcomes (namely the red and yellow ball) in which case the probability measure of picking a red or yellow is 0. 5. I got this from dividing the size of the event ( red) by the size sample space.

Ofcourse this is not the case because red and yellow balls are not equally likely. But by writing down the sample space such that outcomes are disjoint it appears as if red and yellow are equally likely.

What is the reason or flaw in my reasoning? Do you we only use the formula sizeof(event)/sizeof(sample space) when the sample space is understood ?
 

Answers and Replies

  • #2
statdad
Homework Helper
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First sample space: [itex] \{R1, R2\} [/tex] (red one and red two), each with probability 1/2. Probability of red is 1/2 + 1/2 = 1

Second sample space: [itex] \{Y1, \dots, Y49, R\} [/itex]

Probability of yellow:

[tex]
\underbrace{\frac 1 {50} + \frac 1 {50} + \cdots + \frac 1 {50}}_{49 \text{ terms}} = \frac{49}{50}
[/tex]
 
  • #3
Okay. So basic assumption is that the sample space is understood in a meaning way.

Thanks, that is all I needed.
 

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