Question

某人拋擲一枚硬幣,出現正,反面的概率都是,構造數列{a_n},使得當第n次出現正面時,當第n次出現反面時,記S_n=a₁+a₂+…+a_n(n∈N^*).?

(1)求S₄=2的概率;

(2)記ξ=|S₆|,求ξ的概率分布及數學期望.

Answer 1

解:(1)S₄=2,需4次中有3次正面,1次反面,設其概率為P₁,?

則P₁=C³₄()³·=4()⁴=.                                                                                ?

(2)ξ∈{0,2,4,6},                                                                                         ?

P(ξ=0)=C³₆()³()³=.                                                                               ?

P(ξ=2)=C⁴₆()²()⁴+C²₆()⁴()²=.                                                               ?

P(ξ=4)=C⁵₆()⁵+C¹₆()⁵()=.                                                                    ?

P(ξ=6)=C⁶₆()⁶×2=.                                                                                         ?

∴E(ξ)=0×+2×+4×+6×=.