A bag contains 4 Red and 6 Black balls. A bal | vedclass

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(C) Let $R_1$ be the event that the first ball is red and $B_1$ be the event that the first ball is black. Let $R_2$ be the event that the second ball

Vedclass · Accepted Answer

$\frac{26}{65}$

Vedclass · Answer

(C) Let $R_1$ be the event that the first ball is red and $B_1$ be the event that the first ball is black. Let $R_2$ be the event that the second ball is red. Case $1$: First ball is black $(B_1)$. $P(B_1) = \frac{6}{10}$. After adding $3$ black balls,the bag contains $4$ red and $9$ black balls (Total $= 13$). $P(R_2 | B_1) = \frac{4}{13}$. $P(B_1 \cap R_2) = P(B_1) 	imes P(R_2 | B_1) = \frac{6}{10} 	imes \frac{4}{13} = \frac{24}{130}$. Case $2$: First ball is red $(R_1)$. $P(R_1) = \frac{4}{10}$. After adding $3$ red balls,the bag contains $7$ red and $6$ black balls (Total $= 13$). $P(R_2 | R_1) = \frac{7}{13}$. $P(R_1 \cap R_2) = P(R_1) 	imes P(R_2 | R_1) = \frac{4}{10} 	imes \frac{7}{13} = \frac{28}{130}$. Total probability $P(R_2) = P(B_1 \cap R_2) + P(R_1 \cap R_2) = \frac{24}{130} + \frac{28}{130} = \frac{52}{130} = \frac{2}{5} = \frac{26}{65}$.

$A$ bag contains $4$ Red and $6$ Black balls. $A$ ball is drawn at random from the bag,its colour is observed and this ball along with $3$ additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag,then the probability that this drawn ball is red is

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