Q.Three bags contain a number of red and white balls as follows: Bag 1: $3$ red balls, Bag 2: $2$ red balls and $1$ white ball, Bag 3: $3$ white balls. The probability that bag $i$ will be chosen and a ball is selected from it is $\dfrac{i}{6}$, $i = 1, 2, 3$. What is the probability that

Concept: Conditional Probability & Law of Total Probability

We are given the probability of choosing each bag:

$P(B_1) = \frac{1}{6}$, $P(B_2) = \frac{2}{6}$, $P(B_3) = \frac{3}{6}$.

Step 1 – Probability of red from each bag

Bag 1: all 3 red → $P(R|B_1) = 1$

Bag 2: 2 red out of 3 → $P(R|B_2) = \frac{2}{3}$

Bag 3: 0 red → $P(R|B_3) = 0$

Step 2 – Total probability of red

$P(R) = \sum P(B_i) P(R|B_i)$

$= \frac{1}{6}(1) + \frac{2}{6}\left(\frac{2}{3}\right) + \frac{3}{6}(0)$ …

We have three bags, each with a different composition of red and white balls. The chance of picking a particular bag is not equal — it depends on the bag number $i$, with probability $\frac{i}{6}$. This is a classic setup for conditional probability and the law of total probability.

The key idea: the overall probability of an event (like "draw a red ball") is the weighted average of the probabilities of that event under each possible condition (which bag was chosen), where the weights are the probabilities of those conditions.

Let’s define events clearly:

• $B_i$: bag $i$ is chosen, for $i = 1, 2, 3$.
• $R$: a red ball is drawn.
• $W$: a white ball is drawn.

We are given:

• $P(B_1) = \frac{1}{6}$, $P(B_2) = \frac{2}{6}$, $P(B_3) = \frac{3}{6}$.
• Bag 1: 3 red, 0 white → $P(R \mid B_1) = 1$, $P(W \mid B_1) = 0$.
• Bag 2: 2 red, 1 white → $P(R \mid B_2) = \frac{2}{3}$, $P(W \mid B_2) = \frac{1}{3}$.
• Bag 3: 0 red, 3 white → $P(R \mid B_3) = 0$, $P(W \mid B_3) = 1$.

Now we apply the law of total probability.

1. For a red ball:

Substitute:

Simplify:

2. For a white ball: Either use the same method or note that $P(W) = 1 - P(R)$ since only red and white balls exist. $P(W) = P(B_1) \cdot P(W \mid B_1) + P(B_2) \cdot P(W \mid B_2) + P(B_3) \cdot P(W \mid B_3)$ …

Method: Law of total probability across mutually exclusive sources

Use this when an outcome can arise through several exclusive "channels" (bags, machines, boxes), each with its own selection probability and its own conditional chance of the outcome.

Steps

Step 1: List priors and likelihoods

Write $P(B_i)$ (the chance of selecting each source) and $P(\text{event}\mid B_i)$ (the chance of the outcome given that source). Read the selection probabilities exactly as stated — they need not be equal.

Step 2: Combine with the total-probability formula …

Common Mistakes

Mistake 1: Treating the sources as equally likely

Why it's wrong: the bags here are chosen with probability $\tfrac{i}{6}$, not $\tfrac13$ each, so equal weighting gives the wrong total. Correct approach: use the given selection probabilities.

Mistake 2: Wrong conditional counts within a source

Why it's wrong: e.g. bag 2 (2 red, 1 white) gives $P(\text{red})=\tfrac23$, not $\tfrac12$; a miscount corrupts the whole sum. Correct approach: recount reds/whites per bag.

Mistake 3: Recomputing the complement inconsistently …