To calculate the probability that the phone Kiera picked is actually hers, we will use Bayes' theorem.
Let's denote the event that Kiera picks her own phone as A, and the event that the phone she picks is white as B. We want to find P(A|B), the probability that Kiera's phone is picked given that it is white.
According to Bayes' theorem, P(A|B) = (P(B|A) * P(A)) / P(B)
First, let's calculate P(B|A), the probability that Kiera picks a white phone given that it is her own phone. Since Kiera has a white phone, P(B|A) = 1.
Next, let's calculate P(A), the probability that Kiera's phone is picked. Since there are n phones in total, the probability of Kiera's phone being picked is 1/n.
Now, let's calculate P(B), the probability of picking a white phone. This is equal to P(B|A) * P(A) + P(B|not A) * P(not A). The probability of picking a white phone given that it is not Kiera's phone (not A) is (n-1)/n, since there are (n-1) remaining phones that are not hers. The probability of not picking Kiera's phone, P(not A), is (n-1)/n.
Therefore, P(B) = 1 * (1/n) + ((n-1)/n) * ((n-1)/n)
Now we can substitute these values into the equation for Bayes' theorem:
P(A|B) = (P(B|A) * P(A)) / P(B)
= (1 * (1/n)) / (1 * (1/n) + ((n-1)/n) * ((n-1)/n))
Simplifying this expression, we get:
P(A|B) = 1 / (1 + (n-1)^2/n^2)
n people put their phones into a box
Kiera has a white phone.
Assume each of the remaining n-1 phones has probability p of being white.
Kiera takes a phone back out of the basket.
Given that the phone she picks is white, what is the probability that the phone she picked is actually hers?
1 answer