Let p = 0.25. Suppose we know that Ann received at least one friend request the other day. What is the probability that she received exactly 3 requests on that day?

(Hint: This is conditional probability. Recall: conditional probability = joint probability / marginal probability. Here, marginal probability is the probability of "at least one friend requests". And the joint probability refers to the event both "at least one friend" and "exact 3 friend requests". If the number of requests Ann receives is "at least one" AND "exactly 3", what should this number be?)

Question

Let p = 0.25. Suppose we know that Ann received at least one friend request the other day. What is the probability that she received exactly 3 requests on that day?

(Hint: This is conditional probability.  Recall: conditional probability = joint probability / marginal probability.   Here, marginal probability is the probability of "at least one friend requests".  And the joint probability refers to the event both "at least one friend" and "exact 3 friend requests".  If the number of requests Ann receives is "at least one" AND "exactly 3", what should this number be?)

BlackTom AI · Accepted Answer

We begin by restating the problem in our own words to ensure clarity: we are told p = 0.25, and we know Ann received at least one friend request on that day. We are asked to find the probability that she received exactly 3 requests on that day, given that she had at least one. Mathematically, this is a conditional probability problem of the form P(X = 3 | X ≥ 1), where X represents the number of friend requests Ann receives in that day.

To analyze any conditional probability of the form P(A | B), we use the definition P(A | B) = P(A ∩ B) / P(B). In this context:
- A is the event {X = 3} (exactly 3 requests).
- B is the event {X ≥ 1} (at least one request).
Notably, A ⊆ B, because if X = 3, then it is automatically true that X ≥ 1.

Therefore, P(X = 3 | X ≥ 1) simplifies to P(X = 3) / P(X ≥ 1).

A key missing ingredient is the underlying distribution of X. The problem statement provides p = 0.25, but it does not explicitly specify the number of trials n or the exact distribution model (for example, X ~ Binomial(n, p) or another model). The conditional probability depends on both n and p through P(X = 3) and P(X ≥ 1). Without knowing n (the total number of opportunities for a friend request) and the distribution assumptions, we cannot compute explicit numerical values for P(X = 3) and P(X ≥ 1).

If we suppose X ~ Binomial(n, p) as a common modeling choice for the number of independent opportunities yielding a success (a friend request) with probability p = 0.25 for each opportunity, then:
- P(X = 3) = C(n, 3) * (0.25)^3 * (0.75)^(n-3).
- P(X ≥ 1) = 1 - P(X = 0) = 1 - (0.75)^n.
Thus P(X = 3 | X ≥ 1) = [C(n, 3) * (0.25)^3 * (0.75)^(n-3)] / [1 - (0.75)^n].

If you plug in a particular n, you can compute a numerical value. For example, with n = 4, this becomes [4 * (0.25)^3 * (0.75)^1] / [1 - (0.75)^4] ≈ 0.046875 / 0.68359375 ≈ 0.0686. With other n values, you’ll get different results, illustrating how the conditional probability depends on the total number of opportunities n.

Since the problem statement provides only p = 0.25 and does not specify n or the exact distribution for X, there is insufficient information to produce a single definitive numerical answer here. The expression P(X = 3 | X ≥ 1) = P(X = 3) / P(X ≥ 1) must be evaluated under a specified model (values of n and distribution). If you have a particular n in mind or if the problem statement elsewhere defines the distribution of X, you can substitute those values into the formulas above to obtain the exact numeric result.

In summary:
- The conditional probability is P(X = 3 | X ≥ 1) = P(X = 3) / P(X ≥ 1).
- This requires the underlying distribution of X (including n and p). With only p = 0.25 given, the numeric value cannot be uniquely determined without specifying n (and the model).
- If you provide or assume a specific n (and the appropriate distribution), you can compute P(X = 3) and P(X ≥ 1) and then take their ratio to get the desired probability. The provided numeric answer (0.3889) would correspond to a particular choice of n (and model); verify by plugging that n into the formula above.

COMM_V 190 101-104 2025W1 HW14.COMM190.2025.W1

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