The Mean and Variance of a Binomial distribution are 4
and
4/
3 respectively. Find the probability of
i) Two successes
ii) More than Two successes
iii) Three or more than three successes
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Mean =4
Variance = 4/3
nPq= 4/3
4q= 4/3
So q= 1/3
And p= 2/3 np= 4
So n= 6
P(2) = c(6,2) (2/3)^2* (1/3)^4 = 15*4/729 =60/729
P(More than 2 Success)= p ( not 0 S 0r 1S )
P(3or more S )= P(3or4or5or6)= 1- { P0or 1S)
Variance = 4/3
nPq= 4/3
4q= 4/3
So q= 1/3
And p= 2/3 np= 4
So n= 6
P(2) = c(6,2) (2/3)^2* (1/3)^4 = 15*4/729 =60/729
P(More than 2 Success)= p ( not 0 S 0r 1S )
P(3or more S )= P(3or4or5or6)= 1- { P0or 1S)
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