A bag contains six balls of different colours | vedclass

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Question

(B) Total balls = $6$. Since balls are drawn with replacement,the total number of outcomes for drawing $k$ balls is $6^k$.For $p$: Two balls are drawn

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(B) Total balls = $6$. Since balls are drawn with replacement,the total number of outcomes for drawing $k$ balls is $6^k$.For $p$: Two balls are drawn. Both are the same colour. There are $6$ choices for the colour,so the number of favorable outcomes is $6$. Thus,$p = \frac{6}{6^2} = \frac{6}{36} = \frac{1}{6}$.For $q$: Four balls are drawn. Exactly three are of the same colour. Step $1$: Choose the colour that appears $3$ times ($6$ ways).Step $2$: Choose the position of these $3$ balls in $4$ draws ($^4C_3 = 4$ ways).Step $3$: Choose the colour of the remaining $1$ ball ($5$ ways).Number of favorable outcomes = $6 	imes 4 	imes 5 = 120$.Thus,$q = \frac{120}{6^4} = \frac{120}{1296} = \frac{5}{54}$.Ratio $p : q = \frac{1}{6} : \frac{5}{54} = \frac{9}{54} : \frac{5}{54} = 9 : 5$.Here $m = 9$ and $n = 5$,which are coprime.Therefore,$m + n = 9 + 5 = 14$.

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