由于$A_1,A_2,A_3,A_4$相互独立,有:
$$P(A)=P((A_1\cup A_2)A_3 A_4)=P(A_1A_3A_4\cup A_2A_3A_4)=P(A_1A_3A_4)+P(A_2A_3A_4)-P(A_1A_2A_3A_4)$$
$$\xlongequal{\text{独立}}P(A_1)P(A_3)P(A_4)+P(A_2)P(A_3)P(A_4)-P(A_1)P(A_2)P(A_3)P(A_4)$$
$$=p_1 p_3 p_4+p_2 p_3 p_4-p_1 p_2 p_3 p_4.$$