Solution :
14C6 = 3003 combinations of 14 balls taken 6 at a time.
i)P(3 are red)= (3 red out of 4 red balls) and (3 out of total 10 balls)
14C64c3∗10C3 =3003480 =0.159
ii)
P(at least two white)=1- [P(0 white)+P(1 white)]
that is P(X≥2) =1-[P(X=0)+P(x=1)]
= 1- [14C69C6 + 14C65c1∗9c5]
=1-[300384 + 30035∗126]
=1-[3003714] = 30032289 =0.7622
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Let X have a binomial distribution with n = 4 and P = 1/3 Find the probability. (i) P (X = 1) (ii) P (X = 3/2 ) (iii) P (X = 3) (iv) P (X = 6) (v) P (X ≤ 2) *
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