Two persons A and B take part in a shooting c | vedclass

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(B) Given that,$P(A) = 0.6$ and $P(B) = 0.8$. Therefore,$P(A') = 0.4$ and $P(B') = 0.2$. $A$ wins if $A$ hits on the $1^{st}$ shot,or $A$ misses,$B$ m

Vedclass · Accepted Answer

$\frac{15}{23}$

Vedclass · Answer

(B) Given that,$P(A) = 0.6$ and $P(B) = 0.8$. Therefore,$P(A') = 0.4$ and $P(B') = 0.2$. $A$ wins if $A$ hits on the $1^{st}$ shot,or $A$ misses,$B$ misses,and $A$ hits on the $3^{rd}$ shot,or $A$ misses,$B$ misses,$A$ misses,$B$ misses,and $A$ hits on the $5^{th}$ shot,and so on. Probability that $A$ wins $= P(A) + P(A')P(B')P(A) + P(A')P(B')P(A')P(B')P(A) + \dots$ $= 0.6 + (0.4)(0.2)(0.6) + (0.4)^2(0.2)^2(0.6) + \dots$ This is an infinite geometric series with first term $a = 0.6$ and common ratio $r = (0.4)(0.2) = 0.08$. The sum of an infinite geometric series is $S = \frac{a}{1-r}$. $P(A 	ext{ wins}) = \frac{0.6}{1 - 0.08} = \frac{0.6}{0.92} = \frac{60}{92} = \frac{15}{23}$.

List-$I$	List-$II$
$(A) P(\overline{A} \cup B)$	$(I) \frac{2}{3}$
$(B) P(\frac{A}{\overline{B}})$	$(II) \frac{11}{15}$
$(C) P(A \cup B)$	$(III) \frac{3}{5}$

Two persons $A$ and $B$ take part in a shooting competition. $A$ can hit the target with a probability of $0.6$. $B$ can hit the target with a probability of $0.8$. $A$ takes the first shot,after which they shoot alternately. The probability that $A$ wins the competition is

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