Two players play the following game: A writes | vedclass

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(C) Player $A$ has cards $\{3, 5, 6\}$. The possible products of two cards are: $(3 	imes 5) = 15$,$(3 	imes 6) = 18$,$(5 	imes 6) = 30$. Each prod

Vedclass · Accepted Answer

$\frac{4}{9}$

Vedclass · Answer

(C) Player $A$ has cards $\{3, 5, 6\}$. The possible products of two cards are: $(3 	imes 5) = 15$,$(3 	imes 6) = 18$,$(5 	imes 6) = 30$. Each product occurs with probability $\frac{1}{3}$. Player $B$ has cards $\{8, 9, 10\}$. The possible sums of two cards are: $(8 + 9) = 17$,$(8 + 10) = 18$,$(9 + 10) = 19$. Each sum occurs with probability $\frac{1}{3}$. $A$ wins if the product is greater than the sum. Let $P_A$ be the product and $S_B$ be the sum. If $P_A = 15$ (prob $\frac{1}{3}$): $A$ wins if $S_B If $P_A = 18$ (prob $\frac{1}{3}$): $A$ wins if $S_B If $P_A = 30$ (prob $\frac{1}{3}$): $A$ wins if $S_B Total probability = $\frac{1}{9} + \frac{1}{3} = \frac{1+3}{9} = \frac{4}{9}$.

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

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