(Ⅰ)设“这位选手能过关”为事件A, 则P(A)=P(X=30)+P(X=40)= C 12 × 4 5 × 1 5 × 3 5 + 4 5 × 4 5 × 3 5 = 72 125 .(5分) (II)X可能取值为0,10,20,30,40.分布列为 P(X=0)= (1- 4 5 )×(1- 4 5 )×(1- 3 5 )= 2 125 P(X=10)= C 12 × 4 5 ×(1- 4 5 )×(1- 3 5 )= 16 125 P(X=20)= 4 5 × 4 5 ×(1- 3 5 )+(1- 4 5 )×(1- 4 5 )× 3 5 = 35 125 P(X=30)= C 12 × 4 5 × 1 5 × 3 5 = 24 125 P(X=40)= 4 5 × 4 5 × 3 5 = 48 125 所以x的分布列为 X 0 10 20 30 40 P 2 125 16 125 35 125 24 125 48 125 EX=0× 2 125 +10× 16 125 +20× 35 125 +30× 24 125 +40× 48 125 =28.(12分)
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