A bag contains 5 distinct Red,4 distinct Gree | vedclass

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(A) Total number of balls $= 5 + 4 + 3 = 12$. Let $R_1$ be the event that the first ball is not the particular red ball,$R_2$ be the event that the se

Vedclass · Accepted Answer

$\frac{1}{12}$

Vedclass · Answer

(A) Total number of balls $= 5 + 4 + 3 = 12$. Let $R_1$ be the event that the first ball is not the particular red ball,$R_2$ be the event that the second ball is not the particular red ball,$R_3$ be the event that the third ball is not the particular red ball,and $E$ be the event that the fourth ball is the particular red ball. The probability is given by $P(R_1 \cap R_2 \cap R_3 \cap E) = P(R_1) \cdot P(R_2|R_1) \cdot P(R_3|R_1 \cap R_2) \cdot P(E|R_1 \cap R_2 \cap R_3)$. $P(R_1) = \frac{11}{12}$ (any ball except the particular red one). $P(R_2|R_1) = \frac{10}{11}$ (any of the remaining $10$ balls except the particular red one). $P(R_3|R_1 \cap R_2) = \frac{9}{10}$ (any of the remaining $9$ balls except the particular red one). $P(E|R_1 \cap R_2 \cap R_3) = \frac{1}{9}$ (the particular red ball is now one of the $9$ remaining balls). Required probability $= \frac{11}{12} 	imes \frac{10}{11} 	imes \frac{9}{10} 	imes \frac{1}{9} = \frac{1}{12}$.

List $I$	List $II$
$(i) \ P(A \cap B)$	$(1) \ 0.4$
$(ii) \ P(A \cap \bar{B})$	$(2) \ 0.2$
$(iii) \ P(\bar{A} \cap B)$	$(3) \ 0.3$
$(iv) \ P(\bar{A} \cap \bar{B})$	$(4) \ 0.1$

$A$ bag contains $5$ distinct Red,$4$ distinct Green,and $3$ distinct Black balls. If balls are drawn one by one without replacement,what is the probability of getting a particular red ball in the fourth draw?

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