A coin is tossed twice and the four possible outcomes are assumed to be equally likely. If A is the event, both head and tail have appeared, and B is the event, at most one tail is observed, find P of event A, P of event B, and P of event A given not B.

To solve this problem, we first need to identify the possible outcomes when a coin is tossed twice. The outcomes are: HH (both heads), HT (head then tail), TH (tail then head), and TT (both tails). This gives us a total of four equally likely outcomes.

Defining Events

Let's define the events:

• Event A: Both head and tail have appeared. This includes the outcomes HT and TH.
• Event B: At most one tail is observed. This includes the outcomes HH, HT, and TH.

Calculating Probabilities

Now, we can calculate the probabilities of events A and B.

Probability of Event A (P(A))

Event A has 2 favorable outcomes (HT and TH) out of 4 total outcomes:

P(A) = Number of favorable outcomes for A / Total outcomes = 2 / 4 = 1/2.

Probability of Event B (P(B))

Event B has 3 favorable outcomes (HH, HT, and TH) out of 4 total outcomes:

P(B) = Number of favorable outcomes for B / Total outcomes = 3 / 4.

Probability of Event A Given Not B (P(A | Not B))

First, we need to determine Not B, which includes the outcome TT (both tails). The outcomes that are not in B are:

• TT

Now, we find P(A | Not B). Since Not B only includes TT, there are no outcomes from A in this case. Thus:

P(A | Not B) = 0 / 1 = 0.

Summary of Probabilities

• P(A) = 1/2
• P(B) = 3/4
• P(A | Not B) = 0