High School

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Consider there are three baskets, Basket I, Basket II and Basket III with each basket containing rings of red color and green color. Basket I contains 6 red rings and 5 green rings. Basket II contains 3 green rings and 2 red rings while Basket III contains 6 rings which are all red. A person chooses a ring randomly from a basket. If the ring picked is red, find the probability that it was taken from Basket I.

Answer :

To solve the problem, we use Bayes' theorem to find the probability that the red ring was taken from Basket I.

First, we find the probability of picking a red ring from each basket individually:

  1. Basket I contains 6 red rings and 5 green rings, making a total of 11 rings. The probability of picking a red ring from Basket I is given by:
    [tex]P(R \mid I) = \frac{6}{11}[/tex]

  2. Basket II contains 3 green rings and 2 red rings, making a total of 5 rings. The probability of picking a red ring from Basket II is given by:
    [tex]P(R \mid II) = \frac{2}{5}[/tex]

  3. Basket III contains 6 rings, all of which are red. The probability of picking a red ring from Basket III is given by:
    [tex]P(R \mid III) = 1[/tex]

Next, we find the overall probability of picking a red ring ([tex]P(R)[/tex]) from any basket. We assume each basket is chosen with equal probability, i.e., [tex]\frac{1}{3}[/tex]. Thus:
[tex]P(R) = \left(\frac{1}{3}\right) \cdot P(R \mid I) + \left(\frac{1}{3}\right) \cdot P(R \mid II) + \left(\frac{1}{3}\right) \cdot P(R \mid III)[/tex]
[tex]P(R) = \frac{1}{3} \cdot \frac{6}{11} + \frac{1}{3} \cdot \frac{2}{5} + \frac{1}{3} \cdot 1[/tex]

Simplifying the expression:
[tex]P(R) = \frac{1}{3} \left( \frac{6}{11} + \frac{2}{5} + 1 \right)[/tex]

Calculating the terms:
[tex]= \frac{1}{3} \left( \frac{6}{11} + \frac{22}{22} + \frac{2}{5} \right)[/tex]

Finding a common denominator:
[tex]= \frac{1}{3} \left( \frac{30}{55} + \frac{44}{55} + \frac{110}{110} \right)[/tex]

Continuing the calculations:
[tex]= \frac{1}{3} \left( \frac{30}{55} + \frac{22}{55} + 1 \right)[/tex]

After simplifying the fractions, we calculate:
[tex]= \frac{1}{3} \left( \frac{52}{55} + 1 \right) \approx \frac{137}{165}[/tex]

Now, we use Bayes' theorem to calculate the probability that the red ring was taken from Basket I:
[tex]P(I \mid R) = \frac{P(R \mid I) \cdot P(I)}{P(R)}[/tex]

Substituting in the known values:
[tex]= \frac{\frac{6}{11} \cdot \frac{1}{3}}{\frac{137}{165}}[/tex]

Finally, simplifying the expression gives us the probability:
[tex]P(I \mid R) \approx \frac{30}{137}[/tex]

Therefore, the probability that a red ring was picked from Basket I is approximately [tex]\frac{30}{137}[/tex].

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Rewritten by : Barada