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:
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:
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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.
Updated On: Jun 9, 2026
- \(\frac{10}{33} \)
- \(\frac{5}{21} \)
- \(\frac{8}{21} \)
- \(\frac{1}{42} \)
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The Correct Option is B
Solution and Explanation
Concept:
The problem asks for conditional probability. When we are given information that an event $B$ has occurred, and we wish to find the probability of event $A$ given that $B$ has occurred, we use the formula:
\[
P(A|B) = \frac{P(A \cap B)}{P(B)}
\]
Step 1: Define the specific events for this problem.
Let $E_1$ be the event that the outcome is face 1. Given $P(E_1) = 0.1$.
Let $E_2$ be the event that the outcome is face 2. Given $P(E_2) = 0.32$.
Let $B$ be the condition provided, which is that "either face 1 or 2 turned up".
Step 2: Calculate the probability of the condition \(B\).
Since the outcomes of a single die roll are mutually exclusive, the probability of the union of these events is the sum of their individual probabilities:
\[ P(B) = P(E_1 \cup E_2) = P(E_1) + P(E_2) \] \[ P(B) = 0.1 + 0.32 = 0.42 \]
Step 3: Calculate the conditional probability \(P(E_1|B)\).
We want to find the probability that the face is 1, given that the face is in the set $\{1, 2\}$.
\[ P(E_1|B) = \frac{P(E_1 \cap B)}{P(B)} \] Since $E_1$ is entirely contained within the event $B$, $E_1 \cap B = E_1$.
\[ P(E_1|B) = \frac{0.1}{0.42} \] To simplify, multiply the numerator and denominator by 100:
\[ \frac{10}{42} = \frac{5}{21} \] center minipage0.3
Probability = 5/21 minipage center
Step 1: Define the specific events for this problem.
Let $E_1$ be the event that the outcome is face 1. Given $P(E_1) = 0.1$.
Let $E_2$ be the event that the outcome is face 2. Given $P(E_2) = 0.32$.
Let $B$ be the condition provided, which is that "either face 1 or 2 turned up".
Step 2: Calculate the probability of the condition \(B\).
Since the outcomes of a single die roll are mutually exclusive, the probability of the union of these events is the sum of their individual probabilities:
\[ P(B) = P(E_1 \cup E_2) = P(E_1) + P(E_2) \] \[ P(B) = 0.1 + 0.32 = 0.42 \]
Step 3: Calculate the conditional probability \(P(E_1|B)\).
We want to find the probability that the face is 1, given that the face is in the set $\{1, 2\}$.
\[ P(E_1|B) = \frac{P(E_1 \cap B)}{P(B)} \] Since $E_1$ is entirely contained within the event $B$, $E_1 \cap B = E_1$.
\[ P(E_1|B) = \frac{0.1}{0.42} \] To simplify, multiply the numerator and denominator by 100:
\[ \frac{10}{42} = \frac{5}{21} \] center minipage0.3
Probability = 5/21 minipage center
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