Respuesta :
Answer:4
Step-by-step explanation:
Let a denotes the no of odd ball selected and b denotes the no of even ball selected
then [tex]X=3 a+4 b[/tex]
for [tex]a=0,b=3, X=12[/tex]
for [tex]a=1,b=2, X=11[/tex]
for [tex]a=2,b=1, X=10[/tex]
for [tex]a=3,b=0, X=9[/tex]
thus X can take 4 values i.e. 9,10,11 and 12
[tex]P(X=9)\ i.e. 3\ odd\ balls\ and\ 0\ even\ balls=\frac{^{14}C_{3}}{^{28}C_3}=\frac{1}{9}[/tex]
[tex]P(X=10)\ i.e. 2\ odd\ balls\ and\ 1\ even\ balls=\frac{^{14}C_{2}\times ^{14}C_1}{^{28}C_3}=\frac{7}{18}[/tex]
[tex]P(X=11)\ i.e. 1\ odd\ balls\ and\ 2\ even\ balls=\frac{^{14}C_{1}\times ^{14}C_2}{^{28}C_3}=\frac{7}{18}[/tex]
[tex]P(X=12)\ i.e. 0\ odd\ balls\ and\ 3\ even\ balls=\frac{^{14}C_{3}}{^{28}C_3}=\frac{1}{9}[/tex]
