What is $P(x)$ for $n= 6,x = 3, q=0.7$ ?

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Answer 1 · 2017-03-04T10:28:40

$P(3)=0.18522$ .

$P(x)=""_nC_xp^xq^(n-x), x=0,1,2,3,...,n; and, p+q=1.$

We have, $n=6, x=3, and, q=0.7 rArr p=1-q=0.3$

$:. P(3)=""_6C_3(0.3)^3(0.7)^(6-3)$

$=(20)(0.3)^3(0.7)^3$

$=20(0.21)^3$

$:. P(3)=20(0.009261)=0.18522$ .

What is P(x)P(x) for n= 6,x = 3, q=0.7n= 6,x = 3, q=0.7?