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)
"\\frac{4c3*10C3}{14C6}" ="\\frac{480}{3003}" =0.159
ii)
P(at least two white)=1- [P(0 white)+P(1 white)]
that is P(X"\\geq 2)" =1-[P(X=0)+P(x=1)]
= 1- ["\\frac{9C6}{14C6}" + "\\frac{5c1*9c5}{14C6}]"
=1-["\\frac{84}{3003}" + "\\frac{5*126}{3003}]"
=1-["\\frac{714}{3003}]" = "\\frac{2289}{3003}" =0.7622
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Dear hasan, please use the panel for submitting new questions.
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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