Conditional probability: change the reference group

A condition changes which cases remain possible

The probability of an event can change when you learn that another event occurred. P(A given B) means the chance of A among cases satisfying B. In a finite equally weighted group, remove cases outside B and count within the remaining group. The denominator becomes the size of B, not the size of the original population. The formula P(A given B) = P(A and B)/P(B) expresses this same restriction, provided P(B) is positive. Read the direction carefully: the group being conditioned on comes after 'given.'

Keep three questions separate

The probability of A, the probability of both A and B, and the probability of A given B answer different questions. They may share a numerator while using different denominators. Reversing a condition usually changes the denominator too: P(A given B) need not equal P(B given A). A practical technique is to translate the probability into a sentence beginning 'Among the cases that...' Then identify which part of that group meets the requested condition. This can prevent a memorized formula from hiding the actual meaning of the calculation.

Worked example 1: restrict a two-way record

A fictional workshop has 80 attendees. Of 50 who registered online, 30 attended a morning session and 20 attended an afternoon session. Of 30 who registered at the desk, 10 attended the morning and 20 the afternoon. Every attendee belongs to exactly one registration group and one session. Find the probability that an attendee registered online, given that the attendee attended in the afternoon.

• The condition is afternoon attendance. The relevant group contains 20 online registrants plus 20 desk registrants, or 40 attendees.
• Within that group, 20 registered online. Therefore P(online given afternoon) = 20/40 = 1/2.
• The joint probability P(online and afternoon) uses all 80 attendees: 20/80 = 1/4.
• The reversed condition uses the 50 online registrants: P(afternoon given online) = 20/50 = 2/5. The same 20-person intersection supports three different probabilities.

Worked example 2: information updates an unseen choice

One of two boxes is selected with equal probability. Box A contains one red token and three blue tokens; Box B contains three red tokens and one blue token. A token is drawn at random from the selected box and is red. What is the probability that Box B was selected? The red observation favors the box more likely to produce red, but the reasoning must include both possible routes to that observation.

• The chance of selecting A and then drawing red is (1/2)(1/4) = 1/8.
• The chance of selecting B and then drawing red is (1/2)(3/4) = 3/8.
• A red token can arise through either route, so P(red) = 1/8 + 3/8 = 1/2.
• Among red outcomes, the fraction coming from B is (3/8)/(1/2) = 3/4. This is an updated probability, not proof that B was selected.

Independence is a claim to check

Two events are independent when learning one does not change the other's probability. For example, if P(A given B) equals P(A), with the condition defined, B has not changed A's chance. Do not assume independence merely because two events have different names or occur at different times. Drawing without replacement changes what remains available, so successive draws often depend on earlier outcomes. Multiplying conditional probabilities along a route remains valid when independence does not hold. Independence only allows you to replace a conditional probability with the corresponding unconditional one.

Misconception: a likely explanation is certain

In the box example, Box B becomes more likely after a red token appears, yet Box A can also produce red. A probability of 3/4 leaves a 1/4 chance for the alternative under the model. Also, the initial equal-selection assumption matters: changing the chance of selecting each box changes the update. When a problem provides a base rate, keep it in the calculation instead of considering only how strongly an outcome fits one explanation.

Sources & further reading