Question

An urn B1 contains 2 white and 3 black balls and another urn B2 contains 3 white and 4 black balls. One urn is selected at random and a ball is drawn from it. If the ball drawn is found black, find the probability that the urn chosen was B1.

I tried to solve the above question as given below.
Total no of Black balls = 7
Ball from urn B1 = 3
Balls from urn B2 = 4
So probability of taking a ball from Urn1 = 3/7
But answer was given as 21/41 using Bayes theorem... How? Can anyone explain??

Answer 1

Create a Frequency Table:
$$
\begin{matrix}
& White &Black&\bf{Total} \\
B1& 2 &3&\bf{5} \\
B2& 3 &4&\bf{7}\\
\bf{Total}&\bf{5}&\bf{7} &\bf{12}
\end{matrix}
$$
Create a probability tree:

Then, since 3/5 is the probability of drawing a black from B1 and 4/7 is the probability of drawing a black from B2, the probability that the black came from B1 is:
$$
P[B1|black]=\cfrac{3/5}{3/5 +4/7}=\cfrac{3}{5}\times \cfrac{35}{41}=\cfrac{21}{41}.
$$

Answer 2

A better way to write the probability statement in Bayesian form:

$$
P[\text {black is from B1}]=\cfrac{P[black|B1]}{P[black|B1]+P[black|B2]}.
$$

Answer 3

Thank u YB. I got your explanation. But can you see any flaw in my reasoning. Any concept missing???

Answer 4

The problem in your original logic was that you looked at the problem by counting only black balls, second column in table below, while in fact, you should have used in \(addition\), the total number of balls in each urn, the totals in row 1 and 2 in table below. You might surmise that this was required because you are not ONLY selecting black balls, but you are also randomly selecting urns; these black balls come from urns each with different total populations:

$$
\begin{matrix}
& White &Black&\bf{Total} \\
B1& 2 &3&\bf{5} \\
B2& 3 &4&\bf{7}\\
\bf{Total}&\bf{5}&\bf{7} &\bf{12}
\end{matrix}
$$

When we look at your answer, 3/7, it's sort of like thinking of the problem where you have only 1 Urn with 7 balls, three of one type and 4 of another type. And 3/7 is the probability of selecting a ball of the 1st type.

But there is something else you must consider. Although the probability of getting a black ball of type one from a SINGLE urn was 3/7, the probability of getting a black ball from an urn filled with type 1 balls relative to getting a black ball from an urn filled with type 2 balls is heavily influenced by the total number of balls in each respective urn.

Say, for example that an urn with type 1 balls had 1 million balls but only 1 was black. While, another Urn with type 2 balls had 2 balls with only 1 black. Then the \(correct\) probability of selecting a black ball of type 2 (i.e. from the second urn), where we are not ONLY selecting a ball at random but we are ALSO selecting an urn at random (TWO implicit random choices) would be

$$
\cfrac{1/2}{1/2+ 1/10^6}\approx 1
$$

While if we just mixed type 1 and type 2 black balls in a single urn, the probability of a black ball of type 2 would be

$$
\cfrac{2}{2+1} = \cfrac{2}{3}
$$

So the results are quite different when the number of balls in each urn is significantly unequal.

Hope this helps.

Answer 5

Thank you :-)