在10件产品中有3件次品,现任取2件,求取到到次品数X的概率分布及X的数学期

P(0)= C72/ C102=21/45
P(1)= C71 C31/ C102=21/45
P(2)= C32/ C102=3/45

§=0×P(0)+ 1×P(1)+2×P(2)
=0×21/45+1×21/45+2×21/45
=27/45=0.6

p=3c2/10c2=1/15

X 0 1 2
P [C(2,7)]/[C(2,10)] [C(1,7)×C(1,3)]/[C(2,10)] [C(2,3)]/[C(2,10)]
.7/15 7/15 1/15
E(X)=0×(7/15)＋1×(7/15)＋2×(1/15)=3/5