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: Working with probabilities directly instead of counts:
\(P(A)=\dfrac{30}{36}=\dfrac{5}{6}\) (probability both dice differ).

Step 2: Finding P(A∩B):
Sum \(=4\) with different numbers: only \((1,3)\) and \((3,1)\) out of 36 total outcomes, so \(P(A\cap B)=\dfrac{2}{36}=\dfrac1{18}\).

Step 3: Applying Bayes'/conditional formula:
\(P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{1/18}{5/6}=\dfrac{1}{18}\times\dfrac{6}{5}=\dfrac{6}{90}=\dfrac{1}{15}\).

Final Answer:
\[ \boxed{\dfrac{1}{15}} \]