I was attempting a problem that involved conditional probability, and I admit, I am completely lost. I've tried to use independent events $\Bbb P(B\cap A) = \Bbb P(A) \Bbb P(B)$ to solve the problem; however, the answer does not match the correct answer. I've tried various different methods (different values dividing and multiplying) to reverse engineer the answer.
Without further explanation of why I cannot answer this, here's the problem:
$14$ balls are drawn from the bag with replacement.
$4$ Gold Balls
$6$ Silver Balls
$\Bbb P(Y=5) = 0.207$
$\Bbb P(Y\leq 5) = 0.486$
Given that at most five of the balls are gold, find the probability that exactly five of the balls are gold. Give the answer correct to two decimal places.
Answer: $0.43$
What should I do to approach this problem?
If I call the number of gold balls drawn $Y$, then the probability of interest is \begin{align*} P(Y = 5|Y\leq 5) &= \frac{P(Y = 5\cap Y \leq 5)}{P(Y \leq 5)} \\ &= \frac{P(Y = 5)}{P(Y\leq 5)}\\ &=\frac{\binom{14}{5}(4/10)^5(6/10)^9}{\sum_{k = 0}^5\binom{14}{k}(4/10)^k(6/10)^{14-k}}\tag 1\\ &\approx\frac{0.207}{0.486}\tag 2\\ &\approx .43 \end{align*} where in $(1)$ the denominator and numerator follow a Binomial distribution $n = 14, p = 4/10$, and in $(2)$ I used the approximations given.