Find the probability that at least 2 bull's eyes are scored

A target shooter finds that on average, the target is hit nine out of every ten times and a bull's eye is scored on average once every five rounds. Four rounds are fired.

Find the probability that "at least 2 bull's eyes are scored and the target is hit on each of the four rounds".

I would really be appreciated if you can help me. The answer is 0.2158. I don't know how to approach this. :?:

If event B = bull's eye hit and T = target hit,

you can find p = P(B|T) (a probability table or tree diagram works well for that). Then p is the probability in a Binomial experiment where you're looking for at least 2 successes in 4 trials. The answer given is that resulting Binomial probability (as though the problem statement were "at least 2 bull's eyes are scored GIVEN the target is hit on each of the four rounds"). But, based on the actual wording of the problem statement, I'd think it should be that resulting Binomial probability times the probability that the target is hit on each of the four rounds.

Anyway, hope that helps.