Answer:
![(1)/(30)](https://img.qammunity.org/2021/formulas/mathematics/high-school/rn17jsfxvoyidob9l68ku89ak8kbkumfkt.png)
Explanation:
6 symbols and 4-symbol code so number of combinations wich doesn't contain the same symbol twice would be:
6•5•4•3 = 360
If code starts with a star followed by triangle that means only one posibility for first two symbols and code doesn't containing the same symbol twice means that left 4 symbols for next two.
So number of combinations of code starting with a star followed by triangle and doesn't containing the same symbol twice will be:
1•1•4•3 = 12
The probability that a randomly chosen code starts with a star followed by triangle and doesn't contain the same symbol twice:
![(1\cdot1\cdot4\cdot3)/(6\cdot5\cdot4\cdot3)=(1\cdot1)/(6\cdot5)=(1)/(30)](https://img.qammunity.org/2021/formulas/mathematics/high-school/mj3evjpl9w6fv9pgkvm6xyb6iiqkewldwy.png)