设随机变量X服从二项分布b(2,p),随机变量Y服从二项分布b(4,p),若P(X≥1)=5/9,则P(Y≥1)=? 设随机变量X服从二项分布b(2,p),随机变量Y服从二项分布...
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\u82e5P(X\u22651)=5/9\uff0cP(X<1)=4/9=(1-P\uff09^2\uff0cP=1/3 P(Y<2)=(1-P\uff09^4+4*1/3*(1-P\uff09^3=(16+32)/81=16/27
设随机变量X服从二项分布b(2,p),随机变量Y服从二项分布b(4,p),若P(X≥1)=5/9,则P(Y≥1)=65/81。
因为P(X≥1)=5/9,所以满足等式P(X≥1)=1-P(X=0)=1-(1-p)^2=5/9。化简可得P=1/3。所以,P(Y≥1)=1-P(Y=0)=1-(1-p)^4=1-(1-1/3)^4=1-(2/3)^4=65/81。
扩展资料:
对于固定的n以及p,当k增加时,概率P{X=k}先是随之增加直至达到最大值,随后单调减少。当(n+1)p不为整数时,二项概率P{X=k}在k=[(n+1)p]时达到最大值;当(n+1)p为整数时,二项概率P{X=k}在k=(n+1)p和k=(n+1)p-1时达到最大值。
当试验的次数趋于无穷大,而乘积np固定时,二项分布收敛于泊松分布。因此参数为λ=np的泊松分布可以作为二项分布B(n,p)的近似,近似成立的前提要求n足够大,而p足够小,np不是很小。
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