A bag contains 2 red and 3 white marbles - HSC - SSCE Mathematics Advanced - Question 9 - 2024 - Paper 1

First, let's identify the total number of marbles and the effective probabilities. There are 5 marbles total (2 red and 3 white).

The probability of selecting a red marble first is: $P(R_1) = \frac{2}{5}$

If the first marble selected is red, there will be 1 red and 3 white marbles remaining. The probability of selecting a second red marble after one red is picked is: $P(R_2|R_1) = \frac{1}{4}$

Now, calculating the joint probability of selecting red for both marbles is: $P(R_1) \times P(R_2|R_1) = \frac{2}{5} \times \frac{1}{4} = \frac{2}{20} = \frac{1}{10}$

Next, we need to calculate the probability of having selected one red marble and one white marble. The scenarios can be:

1. Red first, white second: (P(R_1) \times P(W_2|R_1) = \frac{2}{5} \times \frac{3}{4} = \frac{6}{20} = \frac{3}{10})
2. White first, red second: (P(W_1) \times P(R_2|W_1) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10})

Therefore, the total probability of selecting one red and one white marble: $P(RW) = \frac{3}{10} + \frac{3}{10} = \frac{6}{10} = \frac{3}{5}$

Now, we can utilize Bayes’ theorem to find the probability that if one marble is red, the other is also red: $P(R_2|R_1) = \frac{P(R_1 \text{ and } R_2)}{P(R_1 \text{ and } R_2) + P(R_1 \text{ and } W_2)}$

Substituting the values we calculated: $P(R_2|R_1) = \frac{\frac{1}{10}}{\frac{1}{10} + \frac{3}{10}} = \frac{\frac{1}{10}}{\frac{4}{10}} = \frac{1}{4}$

Hence, the probability that the other marble is also red given that one of them is red is (\frac{1}{4}), which corresponds to option C.