Question

It is given that the numbers obtained on throwing two dice are different. Find the probability that the sum of both numbers is 4.

Hint:

Restrict to the 30 outcomes with unequal dice, then count how many sum to 4.

Answer 1

Step 1: Setting up the conditioning event:
Let \(A\)= event that the two dice show different numbers, \(B\)= event that the sum is 4. We need \(P(B\mid A)\).

Step 2: Counting outcomes in A:
Out of 36 equally likely outcomes, 6 have equal numbers (1,1),(2,2),...,(6,6); so \(n(A)=36-6=30\).

Step 3: Counting outcomes in A∩B:
Pairs summing to 4: \((1,3),(2,2),(3,1)\). Of these, \((2,2)\) has equal numbers, so it's excluded from \(A\). Remaining: \((1,3),(3,1)\), giving \(n(A\cap B)=2\).

Step 4: Applying the conditional probability formula:
\(P(B\mid A)=\dfrac{n(A\cap B)}{n(A)}=\dfrac{2}{30}\).

Final Answer:
\[ \boxed{P(B\mid A)=\dfrac{1}{15}} \]