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?
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).
To find the probability that the phone in Keira's hand is actually hers given that she picked a white phone, we can use Bayes' theorem.
Let's define the events:
A: Keira's phone is in her hand (it's her own phone).
B: Keira picked a white phone.
We need to find P(A|B), the probability that event A occurred given that event B occurred.
Using Bayes' theorem, we have:
P(A|B) = (P(B|A) * P(A)) / P(B)
P(B|A) is the probability that Keira picked a white phone given that her own phone is in her hand. Since she has a white phone, P(B|A) = 1.
P(A) is the probability that Keira's phone is in her hand, and since everyone's phone is put into the basket randomly, P(A) = 1/n.
P(B) is the probability that Keira picked a white phone. To calculate this, we need to consider all the possible scenarios where Keira can pick a white phone. There are two possibilities:
1. Keira's phone is in her hand, and it's white.
2. Keira picked someone else's phone, and it's white.
For the first scenario, the probability is 1/n (since P(A) = 1/n and P(B|A) = 1).
For the second scenario, the probability is ((n-1)/n) * p (since there are (n-1) phones that are not Keira's, and the probability that one of them is white is p).
Therefore, P(B) = 1/n + ((n-1)/n) * p.
Now, substituting these values into Bayes' theorem, we get:
P(A|B) = (1 * (1/n)) / (1/n + ((n-1)/n) * p)
= 1 / (1 + (n-1)p)
So the probability that the phone in Keira's hand is actually hers given that she picked a white phone is 1 / (1 + (n-1)p).