+116 Two number cubes are rolled for two separate events:
Event A is the event that the sum of numbers on both cubes is less than 10.
Event B is the event that the sum of numbers on both cubes is a multiple of 3.
Complete the conditional-probability formula for event B given that event A occurs first by writing A and B in the blanks:
P( a0 | a1) =
P( a2 U a3)
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P( a4)
+6248 \(\text{well.. the main formula they are after is}\\ P[A|B] = \dfrac{P[B|A]P[A]}{P[B]}\)
\(\text{here}\\ P[B|A] = P[3\cup 6 \cup 9]\\ P[A] = P[3\cup6\cup 9\cup 12]\\ P[B] = P[2-9]\)
\(\text{with some simple counting }\\ P[3 \cup 6 \cup 9] = \dfrac{2+5+2}{36} = \dfrac{11}{36}\\ P[3\cup 6 \cup 9 \cup 12] = \dfrac{12}{36}=\dfrac 1 3\\ P[2-9]=\dfrac{1+2+3+4+5+6+5+4}{36} = \dfrac{30}{36}=\dfrac{5}{6}\)
\(\dfrac{P[B|A]P[A]}{P[B]} = \dfrac{\dfrac{11}{36}\dfrac{1}{3}}{\dfrac{5}{6}}=\dfrac{11}{90}\)
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