Three reds

What is the probability that when choosing 3 carats from seven carats, all 3 reds will be red?

Correct result:

p =  1.694 %

Solution:

C3(9)=(93)=9!3!(93)!=987321=84 C3(32)=(323)=32!3!(323)!=323130321=4960 p1=100 (9 8 7)/(32 31 30)=105621.6935 % p=100 (39)/(332)=10562=1.694%C_{{ 3}}(9) = \dbinom{ 9}{ 3} = \dfrac{ 9! }{ 3!(9-3)!} = \dfrac{ 9 \cdot 8 \cdot 7 } { 3 \cdot 2 \cdot 1 } = 84 \ \\ C_{{ 3}}(32) = \dbinom{ 32}{ 3} = \dfrac{ 32! }{ 3!(32-3)!} = \dfrac{ 32 \cdot 31 \cdot 30 } { 3 \cdot 2 \cdot 1 } = 4960 \ \\ p_{1}=100 \cdot \ (9 \cdot \ 8 \cdot \ 7) / (32 \cdot \ 31 \cdot \ 30)=\dfrac{ 105 }{ 62 } \doteq 1.6935 \ \% \ \\ p=100 \cdot \ { { 3 } \choose 9 } / { { 3 } \choose 32 }=\dfrac{ 105 }{ 62 }=1.694 \%



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