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 the phone she picked is white, what is the probability that the phone in her hand is actually hers?
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).
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).
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?
To solve this problem, we can use Bayes' theorem. Let's denote the events:
A: The phone Keira picked is hers.
B: The phone Keira picked is white.
We need to find the conditional probability, P(A|B), which represents the probability that the phone Keira has is hers, given that the phone she picked is white.
According to Bayes' theorem, we can express P(A|B) as:
P(A|B) = P(B|A) * P(A) / P(B)
Now let's break down each term in the equation:
P(A): The probability that the phone Keira has is hers. Since each person put their phone in the basket, the probability that Keira's phone is in the basket is 1/n.
P(B|A): The probability that Keira picked a white phone, given that the phone she has is hers. Since Keira's phone is white and she can only pick her own phone, this probability is 1.
P(B): The probability that Keira picked a white phone. This can occur in two scenarios:
1. Keira picks her own white phone with probability P(A) * 1 (since P(B|A) = 1).
2. Keira picks someone else's white phone with probability (1 - P(A)) * p (since there are (n-1) phones remaining, and each has a probability p of being white, independent of each other).
Therefore, the total probability of picking a white phone is:
P(B) = P(A) * 1 + (1 - P(A)) * p
Now we can substitute these values into Bayes' theorem:
P(A|B) = 1 * P(A) / (P(A) + (1 - P(A)) * p)
Simplifying the expression, we end up with the final answer:
P(A|B) = P(A) / (P(A) + (1 - P(A)) * p)
To find the probability that the phone Keira picked is actually hers, given that it is white, we can use Bayes' theorem.
Let's break down the problem:
P(H) = Probability that the phone in Keira's hand is her own phone
P(W|H) = Probability that the picked phone is white, given that it is Keira's own phone
P(W|not H) = Probability that the picked phone is white, given that it is not Keira's own phone
We need to calculate:
P(H|W) = Probability that the phone in Keira's hand is her own phone, given that it is white.
Now, applying Bayes' theorem:
P(H|W) = (P(W|H) * P(H)) / P(W)
To compute P(H|W), we need to evaluate P(W|H), P(H), and P(W).
P(W|H) = 1, since if the phone is Keira's, it is white for sure.
P(H) = 1/n, since there is only 1 phone that belongs to Keira out of n total phones in the basket.
To find P(W), we can use the law of total probability. There are two cases to consider:
Case 1: Keira's phone is picked (which is white).
P(W|H) * P(H) = 1/n * 1 = 1/n
Case 2: One of the other (n-1) phones is picked (which is white).
P(W|not H) * P(not H) = p * (1 - 1/n) = p * (n-1)/n
Therefore, P(W) = P(W|H) * P(H) + P(W|not H) * P(not H) = 1/n + p * (n-1)/n.
Now, we can substitute all these values back into Bayes' theorem:
P(H|W) = (1 * 1/n) / (1/n + p * (n-1)/n)
Simplifying it further, we get:
P(H|W) = 1 / (1 + p * (n-1))
Thus, the probability that the phone in Keira's hand is actually hers, given that it is white, is 1 / (1 + p * (n-1)).