Answer: [tex]\dfrac{665}{1496}[/tex]
Step-by-step explanation:
Given : In a group , there are 20 boys and 14 girls.
Total people = 20+14=34
Number of ways to choose 3 people out of 34 = [tex]^{34}C_3[/tex]
Number of ways to choose 1 girl and 2 boys = [tex]^{14}C_1\times^{20}C_2[/tex]
The probability that 1 girl and 2 boys are selected = [tex]\dfrac{^{14}C_1\times ^{20}C_2}{^{34}C_3}[/tex]
[tex]\dfrac{14\times\dfrac{20!}{2!18!}}{\dfrac{34!}{3!31!}}\ \ \ [^nC_r=\dfrac{n!}{r!(n-r)!}]\\\\=\dfrac{14\times\dfrac{20\times19}{2}}{\dfrac{34\times33\times32\times31!}{3\times2\tims3!}}\\\\=\dfrac{665}{1496}[/tex]
Hence, required probability = [tex]\dfrac{665}{1496}[/tex]
