A box contains b blue balls and r red balls. | vedclass

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(A) Let $A$ be the event that a blue ball is drawn in the first draw.Let $B$ be the event that a red ball is drawn in the first draw.Let $C$ be the ev

Vedclass · Accepted Answer

$\frac{b}{r+b}$

Vedclass · Answer

(A) Let $A$ be the event that a blue ball is drawn in the first draw.Let $B$ be the event that a red ball is drawn in the first draw.Let $C$ be the event that a blue ball is drawn in the second draw.The probability of drawing a blue ball in the first draw is $P(A) = \frac{b}{b+r}$.If a blue ball is drawn,it is returned with another blue ball,so the box now contains $b+1$ blue balls and $r$ red balls. The total number of balls is $b+r+1$. Thus,$P(C|A) = \frac{b+1}{b+r+1}$.The probability of drawing a red ball in the first draw is $P(B) = \frac{r}{b+r}$.If a red ball is drawn,it is returned with another red ball,so the box now contains $b$ blue balls and $r+1$ red balls. The total number of balls is $b+r+1$. Thus,$P(C|B) = \frac{b}{b+r+1}$.Using the law of total probability,the probability that the second ball is blue is:$P(C) = P(A) \cdot P(C|A) + P(B) \cdot P(C|B)$$P(C) = \left(\frac{b}{b+r}\right) \left(\frac{b+1}{b+r+1}\right) + \left(\frac{r}{b+r}\right) \left(\frac{b}{b+r+1}\right)$$P(C) = \frac{b(b+1) + rb}{(b+r)(b+r+1)}$$P(C) = \frac{b^2 + b + rb}{(b+r)(b+r+1)} = \frac{b(b+r+1)}{(b+r)(b+r+1)}$$P(C) = \frac{b}{b+r}$

$A$ box contains $b$ blue balls and $r$ red balls. $A$ ball is drawn randomly from the box and is returned to the box with another ball of the same colour. The probability that the second ball drawn from the box is blue,is

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