Q.Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are $0.6$ and $0.4$ respectively. Further, if the first group wins, the probability of introducing a new product is $0.7$ and the corresponding probability is $0.3$ if the second group wins. Find the probability that the new product introduced was by the second group.

Concept: Conditional Probability (Bayes' Theorem)

We want $P(\text{second group wins} \mid \text{new product introduced})$.

Let $A_1$ = first group wins, $A_2$ = second group wins, $B$ = new product introduced.

Given:

$P(A_1) = 0.6$, $P(A_2) = 0.4$,

$P(B \mid A_1) = 0.7$, $P(B \mid A_2) = 0.3$.

Step 1: Find $P(B)$ using total probability:

Step 2: Apply Bayes' theorem: …

We start with the idea of conditional probability. The problem gives us probabilities in one direction (win → product launch) but asks for the probability in the opposite direction (product launch → which group won). That’s exactly what Bayes’ theorem is built for.

Let’s define the events clearly:

• Let $G_1$ = event that the first group wins.
• Let $G_2$ = event that the second group wins.
• Let $N$ = event that a new product is introduced.

We are told:

• $P(G_1) = 0.6$, $P(G_2) = 0.4$
• $P(N \mid G_1) = 0.7$
• $P(N \mid G_2) = 0.3$

We need $P(G_2 \mid N)$ — the probability that the second group introduced the product, given that a new product was introduced.

1. Find the total probability of introducing a new product. The new product can be introduced either when the first group wins or when the second group wins. These are mutually exclusive and exhaustive events. So by the law of total probability:

Substitute:

2. Apply Bayes’ theorem.

   Bayes’ theorem says:

   $P(G_2 \mid N) = \frac{P(N \mid G_2) P(G_2)}{P(N)}$ …

Method: Bayes' Theorem via a probability tree

Use this two-stage "which branch caused the observed second-stage event" pattern — win/lose then launch/no-launch, choose then observe.

Steps

Step 1: Draw the first-stage branches with their priors.

The first stage has outcomes $H_i$ (which group wins) with probabilities $P(H_i)$.

Step 2: Attach the second-stage conditional to each branch.

On each branch write $P(E \mid H_i)$, the chance of the observed second-stage event (new product) given that branch.

Step 3: Total probability of the event = sum of the branch products.

$P(E) = \sum_i P(H_i)\,P(E \mid H_i).$ …

Common Mistakes

Mistake 1: Reporting $P(N\mid G_2)=0.3$ as the answer.

Why it's wrong: that is the forward conditional; the question asks $P(G_2\mid N)$, the reverse. Correct approach: invert with Bayes' theorem.

Mistake 2: Forgetting the first group's term in $P(N)$.

Why it's wrong: a new product can be launched under either winner, so $P(N)$ must sum both branches. Correct approach: $P(N)=(0.6)(0.7)+(0.4)(0.3)=0.54$. …