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Sagot :
[tex]P=\dfrac{\binom{4}{1}\binom{6}{2}}{\binom{6+4}{3}}=\dfrac{4*(6*5/2)}{10*9*8/(3*2)}=1/2[/tex]
The number of all possible 3-person groups
[tex]{10 \choose 3}=\dfrac{10!}{3!7!}=\dfrac{8\cdot9\cdot10}{6}=120[/tex]
The number of all possible groups consisting of 1 girl and 2 boys
[tex]4\cdot{6\choose 2}=4\cdot\dfrac{6!}{2!4!}=4\cdot\dfrac{5\cdot6}{2}=60[/tex]
The probability is [tex]\dfrac{60}{120}=\dfrac{1}{2}[/tex]
[tex]{10 \choose 3}=\dfrac{10!}{3!7!}=\dfrac{8\cdot9\cdot10}{6}=120[/tex]
The number of all possible groups consisting of 1 girl and 2 boys
[tex]4\cdot{6\choose 2}=4\cdot\dfrac{6!}{2!4!}=4\cdot\dfrac{5\cdot6}{2}=60[/tex]
The probability is [tex]\dfrac{60}{120}=\dfrac{1}{2}[/tex]
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