Aiko is a keen basketball player and knows that her chance of scoring a basket on any one throw is 0.65. If Aiko takes 6 shots for a goal, find the probability, (correct to 4 decimal places and put a 0 in front of the decimal point), that she scores a goal five times, given that she scored a goal at least three times.
Taken $$binomPcf(6,0.65,3) = 0.2355$$ and $$binomPcf(6,0.65,3) = 0.2437$$
$$\frac{0.2355}{0.2437}=0.9664$$
But the answer is 0.2761...
On
The probabilities that Aika gets exactly $0$, $1$, and $2$ successes are
$$0.35^6\,\,\,$$
$$0.35^5 \cdot 0.65 \cdot 6\,\,\,$$
$$ 0.35^4 \cdot 0.65^2 \cdot 15\,\,$$
respectively. The sum of these is $\approx 0.1174\,\,\,$. So the probability that she gets at least $3$ goals is
$$1-0.1174 \approx 0.8826\,\,\,\,$$
The probablity that she gets exactly $5$ goals is
$$ 0.65^5 \cdot 0.35 \cdot 6 \approx 0.2437 \,$$
So the searched probability is
$$\frac{0.2437}{0.8826}\approx 0.2761$$
The answer 0.2761 is correct (I approximated along the way, so there might be errors in the last digit or so); the question should have specified EXACTLY 5 times.
So, you need to find the probability of the event that she hits at least 3 (so, she hits either 3, 4, 5, or 6 (the way I did it was that she doesn't hit 0 or 1 or 2, which is also valid)) and find the probability that she hits exactly 5. It works; just try it a couple more times.
(Also, I would have put this as a comment, but I don't have commenting privileges so if someone wants to move this to the comments, they can)