એક સિક્કો m + n વખત ઉછાળવામાં આવે છે,જ્યાં m | vedclass

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(B) Let $P(H) = P(T) = \frac{1}{2}.$ The total number of outcomes is $2^{m+n}.$
Case $1$: The sequence of $m$ consecutive heads starts from the $1^{st

Vedclass · Accepted Answer

$\frac{n + 2}{2^{m + 1}}$

Vedclass · Answer

(B) Let $P(H) = P(T) = \frac{1}{2}.$ The total number of outcomes is $2^{m+n}.$
Case $1$: The sequence of $m$ consecutive heads starts from the $1^{st}$ throw.
The sequence is $(H, H, \dots, H 	ext{ (m times)}, X, X, \dots, X 	ext{ (n times)}).$
The probability is $\frac{1}{2^m}.$
Case $2$: The sequence of $m$ consecutive heads starts from the $(r+1)^{th}$ throw,where $1 \le r \le n.$
For this to happen,the $r^{th}$ throw must be a Tail $(T)$,and the next $m$ throws must be Heads $(H)$.
The sequence is $(X, X, \dots, T, H, H, \dots, H, X, X, \dots, X).$
The probability of this specific event is $\frac{1}{2} 	imes \frac{1}{2^m} = \frac{1}{2^{m+1}}.$
Since there are $n$ such possible starting positions for the sequence of $m$ consecutive heads (from $r=1$ to $r=n$),and these events are mutually exclusive:
Total Probability $= \frac{1}{2^m} + n 	imes \frac{1}{2^{m+1}}$
$= \frac{2}{2^{m+1}} + \frac{n}{2^{m+1}} = \frac{n + 2}{2^{m+1}}.$

એક સિક્કો $m + n$ વખત ઉછાળવામાં આવે છે,જ્યાં $m \ge n.$ ઓછામાં ઓછા $m$ ક્રમિક છાપ (heads) મેળવવાની સંભાવના કેટલી છે?

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