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?

The correct option is 48/125.

To find the probability of P winning exactly 2 of the 3 matches, we can use the concept of binomial probability. Let's break down the solution step-by-step:

Step 1: Determine the probability of P winning a single match
The problem states that P wins 80% of the matches. Expressing this percentage as a fraction:

$P\left(\text{Win}\right)=80\%=\frac{80}{100}=\frac{4}{5}$

Step 2: Determine the probability of P not winning a single match
Not winning a match means either drawing the match (15% probability) or losing it (5% probability). Therefore, the probability of P not winning is:

$P\left(\text{Not Win}\right)=15\%+5\%=20\%=\frac{20}{100}=\frac{1}{5}$

Alternatively, we can calculate this as:

$P\left(\text{Not Win}\right)=1-P\left(\text{Win}\right)=1-\frac{4}{5}=\frac{1}{5}$

Step 3: Calculate the probability of winning exactly 2 out of 3 matches
We need P to win exactly 2 matches and not win exactly 1 match in a total of 3 played matches. The number of ways to choose which 2 matches P wins out of the 3 matches is given by the combination formula:

$\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$

These 3 possible combinations correspond to the following scenarios: (Win, Win, Not Win), (Win, Not Win, Win), and (Not Win, Win, Win).

The probability for any one of these specific sequences is:

${\left(\frac{4}{5}\right)}^2\times {\left(\frac{1}{5}\right)}^1=\frac{16}{25}\times \frac{1}{5}=\frac{16}{125}$

Multiplying the probability of one sequence by the 3 possible ways to arrange them gives the final probability:

$\text{Total Probability}=3\times \frac{16}{125}=\frac{48}{125}$