Given:
The number of blue sweets is 9.
The number of green sweets is 12.
Two sweets are picked at random.
To find:
The probability that both the sweets are chosen is blue in colour.
Explanation:
Using the combination rule,
[tex]\begin{gathered} P(Both\text{ blue\rparen= }\frac{Number\text{ }of\text{ }favourable\text{ }outcomes}{Total\text{ n}umber\text{ }of\text{ }outcomes} \\ =\frac{^9C_2}{^{21}C_2} \\ =\frac{\frac{9!}{(9-2)!2!}}{\frac{21!}{(21-2)!2!}} \\ =\frac{\frac{9!}{7!2!}}{\frac{21!}{19!2!}} \\ =\frac{8\times9}{20\times21} \\ =\frac{72}{420} \\ =\frac{6}{35} \end{gathered}[/tex]Convert into a percentage,
[tex]\begin{gathered} \frac{6}{35}\times100\text{ \%}=17.14\text{ \%} \\ \approx17\text{ \%} \end{gathered}[/tex]Final answer:
The probability that both the sweets are chosen is blue in colour in percentage is 17%.