Question:
Two fair dice are tossed simultaneously. Let \(A\) be the event that the sum of the numbers is prime, and \(B\) be the event that at least one die shows a number 5. The odds against the occurrence of event \(B\) given that event \(A\) has already occurred is:
Two fair dice are tossed simultaneously. Let \(A\) be the event that the sum of the numbers is prime, and \(B\) be the event that at least one die shows a number 5. The odds against the occurrence of event \(B\) given that event \(A\) has already occurred is:
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Count outcomes with a prime sum and those that also contain a 5, find P(B given A), and convert to odds against.
Updated On: Nov 12, 2026
- \(11:4\)
- \(11:15\)
- \(11:7\)
- \(4:11\)
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The Correct Option is A
Solution and Explanation
Step 1: Count event A
Prime sums are 2, 3, 5, 7, 11 with \(1+2+4+6+2=15\) outcomes.
Step 2: Count A and B
Outcomes with a prime sum and at least one 5: \((2,5),(5,2)\) with sum 7 and \((5,6),(6,5)\) with sum 11. So 4 outcomes.
Step 3: Conditional probability
\(P(B|A)=\frac{4}{15}\), so \(P(\text{not }B|A)=\frac{11}{15}\).
Step 4: Odds against
Odds against \(B\) given \(A\) are \(\frac{11/15}{4/15}\), that is \(11:4\).
Final Answer:
The odds against are \(11:4\). \[ \boxed{\text{(A) }11:4} \]
Prime sums are 2, 3, 5, 7, 11 with \(1+2+4+6+2=15\) outcomes.
Step 2: Count A and B
Outcomes with a prime sum and at least one 5: \((2,5),(5,2)\) with sum 7 and \((5,6),(6,5)\) with sum 11. So 4 outcomes.
Step 3: Conditional probability
\(P(B|A)=\frac{4}{15}\), so \(P(\text{not }B|A)=\frac{11}{15}\).
Step 4: Odds against
Odds against \(B\) given \(A\) are \(\frac{11/15}{4/15}\), that is \(11:4\).
Final Answer:
The odds against are \(11:4\). \[ \boxed{\text{(A) }11:4} \]
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