Three bags, A, B and C, each contain 1 red marble and some green marbles - Edexcel - A-Level Maths Statistics - Question 1 - 2019 - Paper 1

To find the probability that Sasha selects 3 green marbles, we need to compute the path where he selects G repeatedly from each bag:

1. From Bag A, the probability of selecting G is:

   $P(G|A) = \frac{9}{10}$

2. If G is selected from Bag A, he will go to Bag B:

   • The probability of selecting G from Bag B is:

   $P(G|B) = \frac{4}{5}$

3. If G is selected from Bag B, he then selects from Bag C:

   • The probability of selecting G from Bag C is:

   $P(G|C) = \frac{2}{3}$

The total probability of selecting 3 green marbles in sequence is:

$P(3G) = P(G|A) \times P(G|B) \times P(G|C) = \frac{9}{10} \times \frac{4}{5} \times \frac{2}{3} = \frac{72}{150} = \frac{12}{25} = 0.48$

To find the probability that Sasha selects at least 1 red marble, we can find the probability of the complementary event (selecting no red marbles) and subtract it from 1:

1. Selecting no red marbles means only selecting green marbles:

   • This is when he picks G from Bag A, Bag B, and Bag C consecutively:

   $P(no R) = P(G|A) \times P(G|B) \times P(G|C) = \frac{9}{10} \times \frac{4}{5} \times \frac{2}{3} = \frac{72}{150} = \frac{12}{25}$

2. Therefore, the probability of selecting at least 1 red marble is:

   $P(at least 1 R) = 1 - P(no R) = 1 - \frac{12}{25} = \frac{13}{25} ext{ or } 0.52$

To find the required probability, we can use Bayes' theorem:

1. Let A represent the event of selecting from Bag B, and R represent the event of selecting a red marble. We want to find:

   $P(A|R) = \frac{P(R|A) \cdot P(A)}{P(R)}$

2. Components needed:

   • $P(R|A)$: The probability of selecting a red marble from Bag B:

     $P(R|B) = \frac{1}{5}$

   • $P(A)$: The probability of reaching Bag B:

     $P(A) = P(G|A) \cdot P(R|B) = \frac{9}{10} \cdot \frac{1}{5} = \frac{9}{50}$

   • $P(R)$: Total probability of selecting a red marble:

     $P(R) = P(R|A)P(A) + P(R|B)P(B) + P(R|C)P(C) \text{ calculated using individual probabilities}$

3. Now substituting into Bayes' theorem gives:

   $P(A|R) = \frac{P(R|B) \cdot P(B)}{P(R)} = \frac{\left( \frac{1}{5} \cdot \frac{4}{10} \right)}{P(R)}$

   • Calculating this will yield the final answer which can be evaluated based on the previous parts.