Statistics And Probability Question 139
Question: Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals.
Options:
A) $ \frac{1}{2} $
B) $ \frac{7}{15} $
C) $ \frac{2}{15} $
D) $ \frac{1}{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] Total number of ways of arranging the balls $ =\frac{10!}{3!7!}=120 $ Favourable cases, $ xB_1yB_2zB_3t $ If $ x,y,z $ and t be the number of white balls to be Place as shown above then $ 0\le x,t\le 5 $ $ 1\le y,z\le 6 $
$ \therefore $ Number of favourable cases = coefficient of $ x^{7} $ in $ {{(1+x+x^{2}+…+x^{5})}^{2}}{{(x+x^{2}+…+x^{6})}^{2}} $ = coeff. of $ x^{7} $ in $ x^{2}{{(1+x+x^{2}+…+x^{5})}^{4}} $ = coeff. of $ x^{5} $ in $ {{( \frac{1-x^{6}}{1-x} )}^{4}}=coeff.ofx^{5}in $ $ {{(1-x)}^{-4}} $ $ {{=}^{8}}C_5=56 $
$ \therefore $ Desired probability $ =\frac{56}{120}=\frac{7}{15} $
BETA
