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Sagot :
The probability that there is at least two 4s in the three throws is [tex]\frac{2}{27}[/tex]
Permutations and combinations
permutations and combinations, the various ways in which objects from a set may be selected, generally without replacement, to form subsets. This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor
given that
we throw a fair die three times
Due to the fact that each die has a 4 on it, the equation would be [tex]\frac{1}{6}[/tex]+ [tex]\frac{1}{6}[/tex]+ [tex]\frac{1}{6}[/tex]=[tex]\frac{1}{2}[/tex] if all the dice were added together.
(There are fifteen ways to get 2 fours: 441,414,144,442,424,244.. and so on until 446,464,644; another way to get the solution is using permutations and combinations:
=(1[tex]C_{1}[/tex])(1[tex]C_{1}[/tex])(5[tex]C_{1}[/tex])+1[tex]C_{1}[/tex])(5[tex]C_{1}[/tex])(1[tex]C_{1}[/tex])+(5[tex]C_{1}[/tex](1[tex]C_{1}[/tex])(1[tex]C_{1}[/tex])/[tex](6C_{1} )^{3}[/tex]
= [tex]\frac{1+15}{216}[/tex]
=[tex]\frac{2}{27}[/tex]
The probability that there is at least two 4s in the three throws is [tex]\frac{2}{27}[/tex]
To learn more about combinations:
https://brainly.com/question/19692242
#SPJ4
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