Question

A group of n professional sit at a table at a conference. Before the presentation begins, they put all their cellular phones(they each have 1) into a basket in the middle of the table. One of them, Keira, has a white phone. Assume that each of the remaining n-1 phones at the tale (not belonging to Keira) has a probability p of being white, independent of each other.

At the end of the presentation, Keira reaches for a phone in the basket before everyone else.

Given that the phone she picked is white, what is the probability that the phone in her hand is actually hers?

Answer 1

To solve this problem, we can use Bayes' theorem. Let's denote:
A: The phone in her hand is her phone (event)
B: The phone she picked is white (event)

We want to find P(A|B), the probability that the phone in her hand is actually hers given that the phone she picked is white.

Using Bayes' theorem, we have:

P(A|B) = (P(B|A) * P(A)) / P(B)

P(B|A): The probability of the phone she picked being white, given that it is her phone. Since we know her phone is white, this probability is 1.

P(B): The probability of picking a white phone. This can be calculated by considering all possible cases:
1. She picks her own white phone: Probability = 1/n
2. She picks someone else's white phone: Probability = (n-1)/n * p
(There are n-1 remaining phones that could be white, each with a probability p)

Therefore, P(B) = 1/n + (n-1)/n * p

P(A): The probability that the phone in her hand is her phone. This is simply 1/n, since there are n phones and each has an equal chance of being hers.

Substituting these values into Bayes' theorem, we get:

P(A|B) = (1 * 1/n) / (1/n + (n-1)/n * p)

Simplifying further:

P(A|B) = 1 / (1 + (n-1)p)

Therefore, the probability that the phone in her hand is actually hers, given that the phone she picked is white, is 1 / (1 + (n-1)p).