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

1 Answer

Mar 4, 2017

#P(3)=0.18522#.

Explanation:

#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#.

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