A biased die is tossed and the respective probabilities for various faces to turn up are
`{:("Face",:,1,2,3,4,5,6),("Probability",:,0.1,0.24,0.19,0.18,0.15,0.14):}`
If an even face has turned up, then the probability that it is face 2 or face 4, is

Options

A. `0.25`
B. `0.42`
C. `0.75`
D. `0.9`

Answer

To solve the problem step by step, we will use the concept of conditional probability.

### Step 1: Define the Events
Let:
- Event A: An even face has turned up.
- Event B: The face is either 2 or 4.

### Step 2: Identify the Sample Space and Probabilities
The sample space for a die is {1, 2, 3, 4, 5, 6} with the following probabilities:
- P(1) = 0.1
- P(2) = 0.24
- P(3) = 0.19
- P(4) = 0.18
- P(5) = 0.15
- P(6) = 0.14

### Step 3: Determine Event A
The even faces in the sample space are {2, 4, 6}. Therefore, event A consists of these outcomes.

### Step 4: Calculate P(A)
To find P(A), we sum the probabilities of the even faces:
\[ P(A) = P(2) + P(4) + P(6) = 0.24 + 0.18 + 0.14 = 0.56 \]

### Step 5: Determine Event B
Event B consists of the outcomes {2, 4}.

### Step 6: Calculate P(B ∩ A)
The intersection of events B and A, denoted as B ∩ A, includes the outcomes that are in both events. Thus:
\[ B ∩ A = \{2, 4\} \]
Now, we calculate P(B ∩ A):
\[ P(B ∩ A) = P(2) + P(4) = 0.24 + 0.18 = 0.42 \]

### Step 7: Calculate P(B | A)
Now we can find the conditional probability P(B | A) using the formula:
\[ P(B | A) = \frac{P(B ∩ A)}{P(A)} = \frac{0.42}{0.56} \]
To simplify:
\[ P(B | A) = \frac{42}{56} = \frac{3}{4} = 0.75 \]

### Final Answer
Thus, the probability that if an even face has turned up, it is face 2 or face 4 is **0.75**.

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