Solution
- In order to determine whether this method of starting the game is fair or not, we simply need to calculate the probability of choosing two cards of the same color consecutively and the probability of choosing two cards of different colors consecutively.
Probability of choosing two cards of the same color consecutively:
![\begin{gathered} P(red)=(26)/(52)\text{ \lparen Since there are 26 red cards out of 52\rparen} \\ \\ P(black)=(26)/(52)\text{ \lparen Since there are 26 black cards out of 52\rparen} \\ \\ \text{ Choosing two red cards consecutively and choosing two black cards } \\ \text{ consecutively is the same. } \\ \\ P(red\text{ twice})=P(red\text{ first time})\text{ AND }P(red\text{ second time\rparen} \\ P(red\text{ twice\rparen}=(26)/(52)*(25)/(51)=(25)/(102) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/t0ugmvi6d74x0qlgx1o2zg9bg1iporygrw.png)
Probability of choosing two cards of the different colors consecutively:
![\begin{gathered} \text{ The probability of choosing a red card first and then a black card is the same} \\ \text{ as choosing a black card first and then a red card.} \\ \\ P(red\text{ AND then black\rparen}=P(red\text{ first})* P(black\text{ second\rparen} \\ P(red\text{ AND then black})=(26)/(52)*(26)/(51)=(26)/(102) \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vf7j4mvyrno62vz1jorfkdz4jo1p0eo3s0.png)
- Now we can see that:
[tex]P(red\text{ twice\rparen}
- This implies that the method of starting the game is not fair
- Thus, there is a higher chance that Sasha starts first.
Final Answer
The answer is OPTION B