I Probability Spaces | What You Need to Know

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The probability that all five shots land in the outer part of the disk is calculated as (3/4)^5. The probability of at least one shot landing in the inner disk is the complement of this value. The figures provided illustrate the concentric circles with radii of 3/4, 1/2, and 1/4, although they are not to scale. The final probability for at least one shot hitting the inner disk is 1 - (3/4)^5. Understanding these calculations is essential for grasping probability spaces.
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Probability for firing shots independently and at random into the circular target with unit radius.
1664284103995.png
 
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The book answer to b) is correct. The probability that all five shots land in the outer part of the disk is ##(\frac 3 4)^5##. And the probability that at least one lands in the inner disk is the complement of this.
 
PeroK said:
The book answer to b) is correct. The probability that all five shots land in the outer part of the disk is ##(\frac 3 4)^5##. And the probability that at least one lands in the inner disk is the complement of this.
Please study the figure given below:

1664287749370.png
1, 3/4 1/2, 1/4 are the radii of the concerned concentric circles.
 
Yes, I know. It's not to scale, but the answer remains ##1- (\frac 3 4)^5##.
 
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There is a nice little variation of the problem. The host says, after you have chosen the door, that you can change your guess, but to sweeten the deal, he says you can choose the two other doors, if you wish. This proposition is a no brainer, however before you are quick enough to accept it, the host opens one of the two doors and it is empty. In this version you really want to change your pick, but at the same time ask yourself is the host impartial and does that change anything. The host...

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