Question

For a biased die, the probabilities for different faces are given by P(1)=0.1, P(2)=0.32, P(3)=0.21, P(4)=0.15, P(5)=0.05, P(6)=0.17. The die is tossed and it is known that either face 1 or 2 turned up. The probability that it is face 1 is:

• A. \(\frac{10}{33} \)
• B. \(\frac{5}{21} \)
• C. \(\frac{8}{21} \)
• D. \(\frac{1}{42} \)

Hint:

When the sample space is restricted by prior knowledge, the new probability is simply the specific case divided by the sum of all cases in the restricted set.

Answer 1

The Correct Option is B

Step 1: Recognise it as conditional probability.
We are told the result is face 1 or 2, and we want the chance it is face 1. The tool is $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$.
Step 2: Name the events.
Let $E_1$ be face 1 with $P(E_1)=0.1$, and $E_2$ be face 2 with $P(E_2)=0.32$. The condition $B$ is the outcome being in $\{1,2\}$.
Step 3: Find $P(B)$.
Faces are mutually exclusive, so $P(B)=P(E_1)+P(E_2)=0.1+0.32=0.42$.
Step 4: Note that $E_1$ sits inside $B$.
Since face 1 is one of the allowed outcomes, $E_1\cap B=E_1$, so the numerator is just $P(E_1)=0.1$.
Step 5: Form the ratio.
\[ P(E_1\mid B)=\frac{0.1}{0.42}=\frac{10}{42}. \]
Step 6: Simplify.
Dividing top and bottom by $2$ gives $\tfrac{5}{21}$, which is option 2.
\[ \boxed{\tfrac{5}{21}} \]