A group of n professionals sit at a table at a conference. Before the presentation begins, they all put their cellular phones into a basket in the middle of the table. One of them, Keira, has a white phone. At the end of the presentation, she reaches for a white phone in the basket.
However, she is not the only one who has a white phone. The probabilty of the others at her table having a white phone is p .
Given that she has picked a white phone, what is the probability that the phone in her hand is actually hers?
Thank you very much fot the hint. So, if I am not wrong, there are (p*(n-1)+1) phones in the basket. The result is 1/(p*(n-1)+1). Now am I right?
Thanks
a little hint, the sentence emphasise on the probability of 'others', not including her.
There are p*n white phones.
so what do you think?
Thanks, so if there are p*n white phones and the phone that she picked was white. It means that it was one of the p*n phones. So the results is 1/(p*n)
Am I right?
Well, that's quite a phone-omenal question! Let's see if we can clown around with some probabilities and figure it out.
To solve this, we can use Bayes' theorem. Bayes' theorem allows us to calculate a conditional probability given some additional information.
Let's define some terms first:
- H: The event that the phone in Keira's hand is actually hers.
- W: The event that Keira picks a white phone.
According to the problem, we know that the probability of others at her table having a white phone is p. Therefore, the probability of her picking a white phone can be written as P(W) = 1 - (1 - p)^n, where n is the number of professionals at the table.
Now, we want to calculate P(H|W), the probability that the phone in Keira's hand is actually hers given that she picked a white phone.
Using Bayes' theorem, we can write:
P(H|W) = (P(W|H) * P(H)) / P(W)
Here, P(W|H) is the probability of picking a white phone given that it's hers. Since she's the only one with a white phone at the table, P(W|H) = 1.
The probability of the phone being hers, P(H), can be calculated as 1/n, as there is an equal chance for her white phone being anywhere in the basket.
And we already know P(W) from earlier.
Plugging these values in, we get:
P(H|W) = (1 * 1/n) / (1 - (1 - p)^n)
So, my mathematical comedian friend, that's how you calculate the probability that the phone in Keira's hand is actually hers given that she picked a white phone. Just remember to adjust the variables n and p to fit the specific scenario you're dealing with.
To solve this problem, we can use Bayes' theorem.
Let's define the following events:
A: Keira picks her own white phone.
B: Keira picks a white phone from the basket.
We want to find P(A|B), which is the probability that Keira has picked her own white phone given that she has picked a white phone from the basket.
According to Bayes' theorem, we have:
P(A|B) = (P(B|A) * P(A)) / P(B)
Using the Law of Total Probability, we can express P(B) as the sum of two mutually exclusive events:
P(B) = P(B|A) * P(A) + P(B|not A) * P(not A)
Given that Keira's phone is white, there are two possibilities:
1. Keira picked her own white phone (A).
2. Keira picked someone else's white phone (not A).
P(A) = 1/n (since there are n professionals and Keira has an equal chance of picking any phone).
P(not A) = (n - 1)/n (the probability of Keira not picking her own phone).
Now, let's consider the probabilities of the events B|A and B|not A:
1. If Keira picked her own white phone, the probability of picking a white phone from the basket is 1 because she knows her phone is white.
P(B|A) = 1
2. If Keira picked someone else's white phone, the probability of picking a white phone from the basket is p because p is the probability that others at her table have a white phone.
P(B|not A) = p
Now, we can substitute these values into Bayes' theorem:
P(A|B) = (1 * (1/n)) / [(1 * (1/n)) + (p * ((n - 1)/n))]
Simplifying further, we get:
P(A|B) = 1 / (1 + (p * (n - 1)))
Therefore, the probability that the white phone in Keira's hand is actually hers, given that she has picked a white phone from the basket, is 1 / (1 + (p * (n - 1))).