A bag contains 8 balls,whose colors are eithe | vedclass

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(B) Let $E$ be the event that $2$ white and $2$ black balls are drawn. Let $H_i$ be the hypothesis that the bag contains $i$ white balls and $(8-i)$ b

Vedclass · Accepted Answer

$\frac{2}{7}$

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(B) Let $E$ be the event that $2$ white and $2$ black balls are drawn. Let $H_i$ be the hypothesis that the bag contains $i$ white balls and $(8-i)$ black balls,where $i \in \{0, 1, 2, 3, 4, 5, 6, 7, 8\}$.Assuming each composition is equally likely,$P(H_i) = \frac{1}{9}$.The probability of drawing $2$ white and $2$ black balls given $H_i$ is $P(E|H_i) = \frac{{}^iC_2 	imes {}^{8-i}C_2}{{}^8C_4}$.We want to find $P(H_4|E) = \frac{P(E|H_4)P(H_4)}{\sum_{i=0}^8 P(E|H_i)P(H_i)}$.Since $P(H_i)$ is constant,$P(H_4|E) = \frac{{}^4C_2 	imes {}^4C_2}{\sum_{i=2}^6 {}^iC_2 	imes {}^{8-i}C_2}$.Calculating the numerator: ${}^4C_2 	imes {}^4C_2 = 6 	imes 6 = 36$.Calculating the denominator: $i=2: {}^2C_2 	imes {}^6C_2 = 1 	imes 15 = 15$$i=3: {}^3C_2 	imes {}^5C_2 = 3 	imes 10 = 30$$i=4: {}^4C_2 	imes {}^4C_2 = 6 	imes 6 = 36$$i=5: {}^5C_2 	imes {}^3C_2 = 10 	imes 3 = 30$$i=6: {}^6C_2 	imes {}^2C_2 = 15 	imes 1 = 15$Sum $= 15 + 30 + 36 + 30 + 15 = 126$.$P(H_4|E) = \frac{36}{126} = \frac{2}{7}$.

$A$ bag contains $8$ balls,whose colors are either white or black. $4$ balls are drawn at random without replacement and it was found that $2$ balls are white and $2$ balls are black. The probability that the bag originally contained an equal number of white and black balls is:

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