Aiko is a keen basketball player and knows that her chance of scoring a basket on any one throw is 0.61. If Aiko takes 5 shots for a goal, find the probability, correct to 3 decimal places (and put a 0 in front of decimal point), that she scores a basket at least once.
My attempt: $$1*(\frac{39}{100})^5$$
However, the answer is 0.991.
There are a few problems with your calculation.
First, if Aiko's probability of scoring a basket on any one throw is $0.61$ (or $\frac{61}{100}$), then her probability of missing on any one throw is $1 - 0.61 = 0.39$, or $\frac{39}{100}$. So, $\frac{35}{100}$ should never come into the equation.
You definitely seemed to have grasped the idea of calculating the probability of $0$ baskets, and then subtracting the resulting probability from $1$, but you didn't go about it right. You should be calculating as follows:
$$1-(\frac{39}{100})^{5}$$