A group of n professionals sit at a table at a conference. Before the presentation begins, they all put 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 table (not belonging to Keira) has a probability p of being white, independent of each other.

Question

At the end of the presentation, Keira reached 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 define some events:

A = Keira's phone is white
B = The phone in Keira's hand is white

We are asked to find P(A|B), the probability that the phone in Keira's hand is actually hers given that it is white.

Bayes' theorem states that:

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

We need to calculate the values for:
P(B|A) = probability that Keira picks her own phone given that it is white
P(A) = probability that Keira's phone is white
P(B) = probability that any phone picked is white

To calculate P(B|A), we know that Keira will always pick a white phone if her own phone is white, so P(B|A) = 1.

To calculate P(A), we need to consider the probability that Keira's phone is white. Since each of the remaining n-1 phones at the table has a probability p of being white, the probability that Keira's phone is white is p.

To calculate P(B), we need to consider the probability that any phone picked is white. There are n phones in total (including Keira's phone). The probability that any phone picked is white can be calculated by considering the probability that a random phone picked is white. There are n-1 phones that are not Keira's phone, so the probability that any phone picked is white is (n-1) * p.

Putting it all together:

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

Therefore, the probability that the phone in Keira's hand is actually hers given that it is white is 1 / (n-1).

Answer 2

A group of n professionals sit at a table at a conference. Before the presentation begins, they all put 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 table (not belonging to Keira) has a probability p of being white, independent of each other.

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

Given that Kiera picks a white phone out of the basket, what is the probability that the phone she picks is actually hers?

Answer 3

To solve this problem, let's define some events:

A = The phone Keira picks is hers
B = The phone Keira picks is white

We are asked to find P(A|B), the probability that the phone Keira picks is actually hers given that it is white.

By Bayes' theorem, we have:

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

Now let's calculate the values:

P(B|A) = The probability that Keira picks a white phone given that it is hers. Since Keira's phone is white, this probability is 1.

P(A) = The probability that the phone Keira picks is hers. Out of the n phones in the basket, only one belongs to Keira. So P(A) = 1/n.

P(B) = The probability that Keira picks a white phone. There are two ways this can happen: either Keira picks her own white phone, or she picks someone else's white phone. The probability of the first case is 1/n, and the probability of the second case is (n-1)/n * p (since Keira's phone is white with probability p, and there are n-1 other phones that are not hers). So P(B) = 1/n + (n-1)/n * p.

Putting it all together:

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

Therefore, the probability that the phone Keira picks is actually hers given that it is white is 1 / (1 + (n-1)p).