A bag contains 2n+1 coins. It is known that n | vedclass

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(A) Let $H$ be the event that the toss results in a head. The bag contains $2n+1$ coins in total. There are $n$ coins with heads on both sides (biased

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$10$

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(A) Let $H$ be the event that the toss results in a head. The bag contains $2n+1$ coins in total. There are $n$ coins with heads on both sides (biased) and $n+1$ fair coins. The probability of picking a biased coin is $P(B) = \frac{n}{2n+1}$ and the probability of picking a fair coin is $P(F) = \frac{n+1}{2n+1}$. If a biased coin is picked,the probability of getting a head is $1$. If a fair coin is picked,the probability of getting a head is $\frac{1}{2}$. Using the law of total probability: $P(H) = P(H|B)P(B) + P(H|F)P(F)$ $\frac{31}{42} = (1) 	imes \frac{n}{2n+1} + \left(\frac{1}{2}ight) 	imes \frac{n+1}{2n+1}$ $\frac{31}{42} = \frac{2n + n + 1}{2(2n+1)} = \frac{3n+1}{4n+2}$ $31(4n+2) = 42(3n+1)$ $124n + 62 = 126n + 42$ $20 = 2n$ $n = 10$

Column $I$	Column $II$
$(a)$ Proliferative Phase	$(i)$ Breakdown of endometrium
$(b)$ Secretory Phase	$(ii)$ Follicular phase
$(c)$ Menstruation	$(iii)$ Luteal Phase

$A$ bag contains $2n+1$ coins. It is known that $n$ of these coins have a head on both sides,whereas the remaining $n+1$ coins are fair. $A$ coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\frac{31}{42}$,then $n$ is equal to

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Match the items given in Column $I$ with those in Column $II$ and select the correct option given below:

Column $I$ Column $II$

$(a)$ Proliferative Phase $(i)$ Breakdown of endometrium

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