# Dice

We throw 10 times a play dice, what is the probability that the six will fall exactly 4 times?

Result

p =  0.054

#### Solution:

$q = 1/6 = \dfrac{ 1 }{ 6 } \doteq 0.1667 \ \\ C_{{ 4}}(10) = \dbinom{ 10}{ 4} = \dfrac{ 10! }{ 4!(10-4)!} = \dfrac{ 10 \cdot 9 \cdot 8 \cdot 7 } { 4 \cdot 3 \cdot 2 \cdot 1 } = 210 \ \\ \ \\ p = { { 10 } \choose 4 } \cdot \ q^4 \cdot \ (1-q)^6 = 210 \cdot \ 0.1667^4 \cdot \ (1-0.1667)^6 \doteq 0.0543 = 0.054$

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