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Sagot :
The given problem can be modeled as a Binomial distribution since the following conditions are satisfied.
There are only two possible outcomes either you get a 2 or you dont.
The number of trials is fixed. (n = 4)
The probability of success is fixed.
The probability of success is the probability of getting a 2 that is 1/6
The Binomial distribution is given by
[tex]P(x)=^nC_x\cdot p^x\cdot(1-p)^{n-x}[/tex]Where n is the number of trials that is 4
x is the outcome of interest which means getting a 2.
At least three 2's means three or more than three.
So, x = 3, 4
nCx is the number of possible combinations
So, this means that we have to find
[tex]P(x\ge3)=P(x=3)+P(x=4)_{}[/tex]P(x = 3):
[tex]\begin{gathered} P(x=3)=^4C_3\cdot(\frac{1}{6})^3\cdot(1-\frac{1}{6})^{4-3} \\ P(x=3)=4\cdot(\frac{1}{6})^3\cdot(\frac{5}{6})^1 \\ P(x=3)=0.0154 \end{gathered}[/tex]P(x = 4):
[tex]\begin{gathered} P(x=4)=^4C_4\cdot(\frac{1}{6})^4\cdot(1-\frac{1}{6})^{4-4} \\ P(x=4)=1\cdot(\frac{1}{6})^4\cdot(\frac{5}{6})^0 \\ P(x=4)=0.00077 \end{gathered}[/tex]So, the probability of obtaining at least three 2's is
[tex]\begin{gathered} P(x\ge3)=P(x=3)+P(x=4)_{} \\ P(x\ge3)=0.0154+0.00077 \\ P(x\ge3)=0.01617 \end{gathered}[/tex]Therefore, the probability of obtaining at least three 2's is 0.01617
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