I have the following probability.
probability a $0$ is sent is $0.4$
probability a $1$ is sent is $0.6$
probability that due to noise, a $0$ is changed to a $1$ during transmission is $0.2$
probability that due to noise, a $1$ is changed to a $0$ during transmission is $0.1$
Here is what I think should be the solution:
Let $A$ be $1$ was received, $B$ be $1$ was sent
$P(B) = 0.6$
$P(A|B) = 0.6 * 0.9 = 0.54$
$P(A) = 0.6 * 0.9 + 0.4 * 0.2 = 0.62$
Thus using $$P(B|A) = \frac{P(A|B)P(B)}{P(A)}$$
$P(B|A) = 52.26%$
Is my solution correct?
Not quite. $P(A | B) = 0.9$; what you wrote above is actually $P(A \cap B)$. Everything else looks fine, so you should end up with $P(B \mid A) = \frac{0.9(0.6)}{0.62} \approx 0.87$.