Question

For a biased die the probabilities for the different faces to turn up are given below:

Faces : 1 2 3 4 5 6
Probabilities : 0.10 0.32 0.21 0.15 0.05 0.17

The die is tossed & you are told that either face one or face two has turned up. Then the prothat it is face one is :

Answer 1

The conditional probability that face one has turned up, given that either face one or face two has turned up, is approximately 0.2381 or 23.81% for a biased die.

Step-by-step explanation:

Understanding Conditional Probability with a Biased Die

The question revolves around the concept of conditional probability in a scenario with a biased die. The probabilities for each face of the die are: 1 (0.10), 2 (0.32), 3 (0.21), 4 (0.15), 5 (0.05), 6 (0.17). We are asked to find the probability that face one has turned up given that either face one or face two has turned up.

To solve this, we can apply the formula for conditional probability, which is P(A|B) = P(A and B) / P(B), where A is the event that face one has turned up and B is the event that either face one or face two has turned up.

First, we find P(B), the probability that either face one or face two has turned up, which is the sum of the individual probabilities of the two events: P(B) = P(face 1) + P(face 2) = 0.10 + 0.32 = 0.42.

Next, since we are given that face one or face two has turned up, P(A and B) is simply the probability of face one, which is 0.10.

Now we can calculate P(A|B):

P(A|B) = P(A and B) / P(B) = 0.10 / 0.42 ≈ 0.2381

So, the conditional probability that it is face one given that face one or face two has turned up is approximately 0.2381 or 23.81%.