For Paul to select 1 of his 4 children, the simulation tool must have the possibility of giving a result of 1/4 (i.e. 1 in 4 children). The simulation tools capable of these results are: (1), (3) and (4)
To determine which simulation tool is correct for the situation, we simply put each option to test.
Option (1)
[tex]Sections = 12[/tex]
[tex]Sub= 3[/tex]
Here, the simulation is done by calculating the probability of each subsection.
The probability is:
[tex]Pr = \frac{Sub}{Sections}[/tex]
[tex]Pr = \frac{3}{12}[/tex]
Simplify
[tex]Pr = \frac{1}{4}[/tex]
Option 2
[tex]Chips =12[/tex]
[tex]Sections = 2[/tex]
Here, the simulation is done by calculating the probability of each section.
The probability is:
[tex]Pr = \frac{Sections}{Chips}[/tex]
[tex]Pr = \frac{2}{12}[/tex]
Simplify
[tex]Pr = \frac{1}{6}[/tex]
Option 3
A standard deck has:
[tex]Cards = 52[/tex]
[tex]Suit = 13[/tex]
Here, the simulation is done by calculating the probability of each suit.
The probability is:
[tex]Pr = \frac{Suits}{Cards}[/tex]
[tex]Pr = \frac{13}{52}[/tex]
Simplify
[tex]Pr = \frac{1}{4}[/tex]
Option 4
[tex]Coins = 2[/tex]
Here, he needs to toss the coin 2 times. So, the possible sample space are:
[tex]S = \{HH, HT, TH, TT\}[/tex]
The probability of one of the outcomes is:
[tex]Pr = \frac{1}{n(S)}[/tex]
Where:
[tex]n(S) = 4[/tex] ---- number of outcomes
So:
[tex]Pr =\frac{1}{4}[/tex]
Option 5:
[tex]Sides =6[/tex]
The probability of getting 1 of the 6 faces is:
[tex]Pr = \frac{1}{6}[/tex]
Paul can only use simulations that give 1/4. Hence, (1), (3) and (4) are correct because they all equal 1/4.
Read more about probability simulation at:
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