Which of the following represents the probability that a die toss will be less than four given that it is an odd number?Event A: an odd number.Event B: less than 4. P(B) P(B|A) P(A|B)

The problem is asking for the conditional probability of Event B (getting a number less than 4) given that Event A (getting an odd number) has occurred.

Step 1: Identify the sample space The sample space for a die toss is {1, 2, 3, 4, 5, 6}.

Step 2: Identify the outcomes for Event A Event A is getting an odd number. The outcomes for this event are {1, 3, 5}.

Step 3: Identify the outcomes for Event B Event B is getting a number less than 4. The outcomes for this event are {1, 2, 3}.

Step 4: Identify the outcomes for both Event A and Event B The outcomes that satisfy both Event A and Event B are {1, 3}.

Step 5: Calculate the probabilities P(A) = Number of outcomes for Event A / Total outcomes = 3/6 = 0.5 P(B) = Number of outcomes for Event B / Total outcomes = 3/6 = 0.5 P(A ∩ B) = Number of outcomes for both Event A and Event B / Total outcomes = 2/6 = 0.33

Step 6: Use the formula for conditional probability The formula for conditional probability is P(B|A) = P(A ∩ B) / P(A). Plugging in the values we calculated:

P(B|A) = P(A ∩ B) / P(A) = 0.33 / 0.5 = 0.66

So, the probability that a die toss will be less than four given that it is an odd number is 0.66 or 66%.