Let $X$ be a binomial random variable with $p=0.6$ and $n=10$ . What is $P(X > 8)$ ?

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Answer 1 · 2017-07-01T13:07:28

$P(X>8)=0.0463574016$

Given that, $X$ is a Binomial Random Variable, with,

parameters, $p=0.6, and, n=10,$ denoted by,

$X~~ B(10,0.6).$

$:. q=1-p=0.4$

Knowing that, for $X~~B(n,p),$

$P(X=x)=""_nC_xp^xq^(n-x), x=0,1,2,...,n.$

Now, the Reqd. Prob.= $P(X>8),$

$=P(X=9)+P(X=10),$

$=""_10C_9(0.6)^9(0.4)^1+""_10C_10(0.6)^10 (0.4)^0,$

$=10(0.6)^9(0.4)+1(0.6)^10(1),$

$=4(0.6)^9+(0.6)^10,$

$=(0.6)^9(4.6),$

$rArr P(X>8)=0.0463574016$

Let XX be a binomial random variable with p=0.6p=0.6 and n=10n=10. What is P(X > 8) P(X > 8) ?