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Let the three games in a series be called $G_1, G_2$ and $G_3$ respectively, and the probability of winning the game $x$ be denoted as $P(x)$

**You can win the series if and only if: You win **$G_2$** and you win atleast one of **$\{G_1, G_3\}$**.**

$\large\boxed{P\left ( \substack{\text{winning}\\\text{the series}\\ \\{G_{1}G_{2}G_{3}}} \right ) = P(G_2) \times \Bigl ( P(G_1) + P(G_3) - P(G_1) \times P(G_3) \Bigr )\;}$

Let the probability of winning against player $A$ be $a$ and the probability of winning against player $B$ be $b$.

**Then, **$a < b.(A \text{ is a stronger player than} B,\text{ so probability of winning against } A$

$ \text{ is smaller compared to}\;B)$

Let $P(xyz)$ be the probability of winning the series in which the games played are against $x, y$ and $z$ in order.

- $P(AAB) = a(a+b-ab) = a^2 + ab - a^2b$
- $P(ABA) = b(a+a-aa) = 2ab - a^2b$
- $P(BAB) = a(b+b-b^2) = 2ab - ab^2$
- $P(BAA) = a(b+a-ba) = a^2 + ab - a^2b$

We can see that **not all probabilities are equal, so option E is not correct**.

We can also see that options **A and D result in the same value, so they are not correct either**.

**Comparing option B and option C.**

Since $a < b$ and $a,b\geq 0$, we have that $2ab - a^{2}b > 2ab - ab^{2}$

**Hence, option B is the correct answer.**

Pragy Agarwal Sir why Option A and D cant be the answer? Why twi different sequences can have same probability

2. How to solve if the question would have asked for Finding the sequence with highest probability amongst all possible sequence in which atleast one A must be there

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