How to find the probability of a specific draw from the urns?

Question

There are 3 urns each containing 10 balls whose colors are either blue or red. In an experiment a ball is drawn from the first urn. If its color is red a second ball is drawn from the second urn. And if the color of the second ball is red, then a third ball is drawn from the third urn.

My approach:

I understood that each color of the ball is equally likely and balls are drawn without replacement so I tried to use hypergeometric distribution formula. However, I am not sure about the number of success in the population(r) and number of success in the number of trials n(x). Also we know that,

P(R) = P(B) = 1/2

P(ball) = 1/10

I have drawn a tree like diagram but could not continue further. Can anyone help me with this problem?

Answer 1

Although the original posting specified that $~p(R) = p(B) = 1/2,~$ I am unsure how to interpret this.

As indicated by the comment of callculus you need to know the relative proportion of red balls in each urn.

So, I will assume that in each of the three urns, exactly (1/2) of the balls in that urn are red. I am making this assumption because I see no way of attacking the problem, without either this assumption, or a similar assumption.

At the end of my answer, I will provide a more generic answer, that makes no assumptions on the relative proportions of red and blue balls in each urn.

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So the various outcomes are:

• Urn-1 : Blue
  Probability $~= (1/2).$

• Urn-1 : Red, Urn-2 : Blue
  Probability $~= (1/2) \times (1/2) = (1/4).$

• Urn-1 : Red, Urn-2 : Red, Urn-3: Blue
  Probability $~= (1/2) \times (1/2) \times (1/2)= (1/8).$

• Urn-1 : Red, Urn-2 : Red, Urn-3: Red
  Probability $~= (1/2) \times (1/2) \times (1/2)= (1/8).$

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The formula for a Hypergeometric distribution is not needed here, specifically because you never draw more than one ball out of any specific urn.

Let $~p_k~$ denote the probability that the ball drawn from Urn-k is red, where $~k \in \{1,2,3\}.$

That is, I am assuming that in Urn-k, $~p_k~$ is equal to the ratio $~\displaystyle \frac{\text{# red balls in urn}}{\text{total # balls in urn}}.$

let $~q_k = 1 - p_k.$

Then, the various outcomes are:

• Urn-1 : Blue
  Probability $~= q_1.$

• Urn-1 : Red, Urn-2 : Blue
  Probability $~= p_1 \times q_2.$

• Urn-1 : Red, Urn-2 : Red, Urn-3: Blue
  Probability $~= p_1 \times p_2 \times q_3.$

• Urn-1 : Red, Urn-2 : Red, Urn-3: Red
  Probability $~= p_1 \times p_2 \times p_3.$