Let $X$ be a random variable such that
\[ P(X=i)=2P(X=i-1), \quad i=2,3,\ldots,n,\ n\geq7, \]
and $\displaystyle\sum_{i=1}^{n}P(X=i)=1$. Then $(2^n-1)P(X=7)$ equals ________ (answer in integer).

Question

P and Q play chess frequently against each other. Of these matches, P has won 80% of the matches, drawn 15% of the matches, and lost 5% of the matches.
If they play 3 more matches, what is the probability of P winning exactly 2 of these 3 matches?

Answer 1

Step 1: Guess a form that fits the doubling rule.
Since each term is twice the previous one, any sequence of the form $P(X=i)=C\cdot 2^{i}$ for a constant $C$ automatically satisfies $P(X=i)=2P(X=i-1)$, because $C\cdot 2^{i}=2\cdot(C\cdot 2^{i-1})$.

Step 2: Use the total probability condition to find C.
\[ \sum_{i=1}^{n}P(X=i)=C\sum_{i=1}^{n}2^{i}=1. \]

Step 3: Sum the geometric series starting from $2^1$.
\[ \sum_{i=1}^{n}2^{i}=2+4+\cdots+2^{n}=2(2^{n}-1)=2^{n+1}-2. \]
So
\[ C(2^{n+1}-2)=1 \quad \Rightarrow \quad C=\frac{1}{2(2^{n}-1)}. \]

Step 4: Write P(X=7) using this C.
\[ P(X=7)=C\cdot 2^{7}=\frac{128}{2(2^{n}-1)}=\frac{64}{2^{n}-1}. \]

Step 5: Multiply by $(2^n-1)$.
\[ (2^{n}-1)P(X=7)=(2^{n}-1)\cdot\frac{64}{2^{n}-1}=64. \]

Final Answer:
The extra constant factor cancels neatly, giving the same clean number both ways. \[ \boxed{64} \]

Let \(X\) be a random variable such that
\[ P(X=i)=2P(X=i-1), \quad i=2,3,\ldots,n,\ n\geq7, \]
and \(\displaystyle\sum_{i=1}^{n}P(X=i)=1\). Then \((2^n-1)P(X=7)\) equals ________ (answer in integer).

Show Hint

Correct Answer: 64

Solution and Explanation

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Let \(X\) be a random variable such that\[ P(X=i)=2P(X=i-1), \quad i=2,3,\ldots,n,\ n\geq7, \]and \(\displaystyle\sum_{i=1}^{n}P(X=i)=1\). Then \((2^n-1)P(X=7)\) equals ________ (answer in integer).