"四個求的標(biāo)號為奇數(shù)" = 四個球的標(biāo)號之和為奇數(shù)?
(1) N=3:2偶數(shù)3奇數(shù)
If 四個球的標(biāo)號之和為奇數(shù),then 剩下球的標(biāo)號為偶數(shù)
P(剩下球的標(biāo)號為偶數(shù))=2/5
If 四個球的標(biāo)號之和為偶數(shù),then 剩下球的標(biāo)號為奇數(shù)
P(剩下球的標(biāo)號為奇數(shù))=3/5
E(甲的得分)=2*2/5+0*3/5=0.8
(2)
乙勝:四個球的標(biāo)號之和為偶數(shù)
P(4奇0偶)= [N/(N+2)]*[(N-1)/(N+1)]*[(N-2)/N]*[(N-3)/(N-1)] = [(N-2)(N-3)]/[(N+2)(N+1)]
P(2奇2偶)= C(N,2)/C(N+2,4) = [N(N-1)/2] / [(N+2)(N+1)N(N-1)/(1*2*3*4)] = 12/[(N+2)(N+1)]
P(乙勝) = P(4奇0偶)+P(2奇2偶) = [(N-2)(N-3)]/[(N+2)(N+1)] +12/[(N+2)(N+1)]=3/7
7 [(N-2)(N-3)+12] = 3[(N+2)(N+1)] => N=5 or N=6