A bag contains 4 red and 4 blue balls. Four b | vedclass

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(B) Let the total number of balls be $8$ ($4$ red and $4$ blue).There are two possible sequences for alternating colors:$1$. Red,Blue,Red,Blue $(RBRB)

Vedclass · Accepted Answer

$6/35$

Vedclass · Answer

(B) Let the total number of balls be $8$ ($4$ red and $4$ blue).There are two possible sequences for alternating colors:$1$. Red,Blue,Red,Blue $(RBRB)$$2$. Blue,Red,Blue,Red $(BRBR)$The probability of sequence $RBRB$ is:$P(RBRB) = \frac{4}{8} 	imes \frac{4}{7} 	imes \frac{3}{6} 	imes \frac{3}{5} = \frac{1}{2} 	imes \frac{4}{7} 	imes \frac{1}{2} 	imes \frac{3}{5} = \frac{3}{70}$The probability of sequence $BRBR$ is:$P(BRBR) = \frac{4}{8} 	imes \frac{4}{7} 	imes \frac{3}{6} 	imes \frac{3}{5} = \frac{3}{70}$The total probability is the sum of these two mutually exclusive events:$P = \frac{3}{70} + \frac{3}{70} = \frac{6}{70} = \frac{6}{35}$

$A$ bag contains $4$ red and $4$ blue balls. Four balls are drawn one by one from the bag. Find the probability that the drawn balls are of alternating colors.

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