Question:
A die is thrown twice and found that the sum of the numbers is 6. Find the conditional probability of getting number 4 at least once.
A die is thrown twice and found that the sum of the numbers is 6. Find the conditional probability of getting number 4 at least once.
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Restrict the sample space to the 5 outcomes summing to 6, then count how many contain a 4.
Updated On: Sep 23, 2026
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Solution and Explanation
Step 1: Understanding the Concept:
Let \(E\)=\"at least one 4 appears\" and \(F\)=\"the sum of the two throws is 6\". We need \(P(E/F)\), which restricts attention only to outcomes where the sum is 6.
Step 2: Listing outcomes with sum 6:
The ordered pairs \((\text{die1},\text{die2})\) summing to 6 are: \((1,5),(2,4),(3,3),(4,2),(5,1)\) — 5 outcomes in total, forming the event \(F\).
Step 3: Picking out those with at least one 4:
Among these, the pairs containing a 4 are \((2,4)\) and \((4,2)\) — 2 outcomes.
Step 4: Computing the conditional probability:
Since all outcomes of a fair die are equally likely, \(P(E/F)=\dfrac{n(E\cap F)}{n(F)}=\dfrac{2}{5}\).
Final Answer:
\(P(\text{at least one }4\mid\text{sum}=6)=\boxed{\dfrac{2}{5}}\).
Let \(E\)=\"at least one 4 appears\" and \(F\)=\"the sum of the two throws is 6\". We need \(P(E/F)\), which restricts attention only to outcomes where the sum is 6.
Step 2: Listing outcomes with sum 6:
The ordered pairs \((\text{die1},\text{die2})\) summing to 6 are: \((1,5),(2,4),(3,3),(4,2),(5,1)\) — 5 outcomes in total, forming the event \(F\).
Step 3: Picking out those with at least one 4:
Among these, the pairs containing a 4 are \((2,4)\) and \((4,2)\) — 2 outcomes.
Step 4: Computing the conditional probability:
Since all outcomes of a fair die are equally likely, \(P(E/F)=\dfrac{n(E\cap F)}{n(F)}=\dfrac{2}{5}\).
Final Answer:
\(P(\text{at least one }4\mid\text{sum}=6)=\boxed{\dfrac{2}{5}}\).
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