Answer:
C
Step-by-step explanation:
There are 6 girls and 7 boys in the class. 10 players can be selected in
[tex]C^{13}_{10}=\dfrac{13!}{10!(13-10)!}=\dfrac{11\cdot 12\cdot 13}{2\cdot 3}=286[/tex]
different ways.
1. A team of 10 players with 4 boys includes 6 girls.
[tex]C^7_4=\dfrac{7!}{4!(7-4)!}=\dfrac{5\cdot 6\cdot 7}{2\cdot 3}=35[/tex] - possible ways to choose 4 boys.
[tex]C^6_6=\dfrac{6!}{6!(6-6)!}=1[/tex] - possible way to choose 6 girls.
2. A team of 10 players with 6 boys includes 4 girls.
[tex]C^7_6=\dfrac{7!}{6!(7-6)!}=\dfrac{7}{1}=7[/tex] - possible ways to choose 6 boys.
[tex]C^6_4=\dfrac{6!}{4!(6-4)!}=\dfrac{5\cdot 6}{2}=15[/tex] - possible ways to choose 4 girls.
3. The probability that a randomly chosen team includes 4 or 6 boys is
[tex]Pr=\dfrac{35\cdot 1+7\cdot 15}{286}=\dfrac{140}{286}=\dfrac{70}{143}\approx 0.49.[/tex]
