probability and favourable conditions in dice games?

so the probability of 1 dice is 0.4

the probability of 3 dices are :

dice 1 succeed + dice 2 succeed +dice 3 succeed but we have to substract some cases that we double count : dice 1 and dice 2 succeed+ dice 1 and dice 3 succeed + dice 2 and 3 succeed , but this will make a case hasn't been counted : dice 1 and 2 and 3 succeed

so we have 0.4 + 0.4 + 0.4 - (0.4*0.4) - (0.4*0.4) - (0.4*0.4) + (0.4*0.4*0.4)=1.2 - 0.48 + 0.064 = 0.784

- 0.4*0.4 because we want both die succeed, so we need to multiply both probability, same with the 0.4*0.4*0.4

To make it clearer, you can try to make 3 circle each representing dice 1, 2, or dice 3 succeed, with some area interception between tow or three circles.

If you win with just one die showing the favourable result, then the only way you lose is if all three dice show a losing result.

This probability (all three showing a loss) is 0.6 x 0.6 x 0.6 = 0.216

So the probability of winning = 1 - 0.216 = 0.784.