1, assuming that the probability of rain in place A is p (a) = 0.3; The probability of rain in the second place is p (b) = 0.2; The probability that colleagues of Party A and Party B will rain is P(AB)=0.09.

(1), the probability of rain in at least one of A and B is p (aub) = p (a)+p (b)-p (ab) = 0.3+0.2-0.09 = 0.41.

(2) When it is known that there is at least one rain in A and B, the probability of rain in A is about 0.073 = p (a)/p (aub) = 0.3/0.41= 3/41.

(3) In the case of rain in place A, the probability of rain in place B is P=P(A)*P(B)=0.3*0.2=0.06.

2. Suppose the probability of male color blindness in this population is p (a) = 0.05; The probability of female color blindness is p (b) = 0.025; Then choose a person, the probability that this person happens to be a color-blind patient is p (color) = (p (a)+p (b))/2 = (0.05+0.025)/2 = 0.0375, then the probability that this person is a male is p (color) * (p (a)/[p (a).

3、P(AB)= P(A)-P(A-B)= 0.4-0.25 = 0. 15

P(AUB)= P(A)+P(B)-P(AB)= 0.4+0.25-0. 15 = 0.5

P(B-A)= P(B)-P(AB)= 0.25-0. 15 = 0. 1

4. Should the topic be the probability of shooting down the plane three times?

P( 1)=0.3*0.2=0.06.

P(2)=0.3*0.3*0.6=0.054 after secondary shooting.

Shot down three times P(3)=0.3*0.3*0.3=0.027.

So the probability of the plane being shot down after three launches = p (1)+p (2)+p (3) = 0.06+0.054+0.027 = 0.141.