A dice is rolled three times. Let $E_1$ denote the event of getting a number larger than the previous number each time and $E_2$ denote the event that the numbers (in order) form an increasing $AP$ then:
$P(E_1) \geq P(E_2)$
$P(E_2 \cap E_1) = \frac{1}{36}$
$P(E_2 / E_1) = \frac{3}{10}$
$P(E_1) = \frac{10}{3} P(E_2)$
Step-by-Step Solution
Key Concept: For strictly increasing sequences from three dice rolls, count ordered triples (a,b,c) where a<b<c for E₁, and triples in arithmetic progression with common difference d>0 for E₂. Use conditional probability P(E₂|E₁) = P(E₂∩E₁)/P(E₁) where E₂∩E₁ consists of AP sequences that are strictly increasing.
For $P(E_1)$, selecting 5 specific cards from 6 total: $P(E_1) = \frac{\binom{6}{5}}{\binom{216}{5}} = \frac{6 \cdot C_5}{216 \cdot 54}$. For $P(E_2)$, selecting cards such that the sum is $4+2=6$: $P(E_2) = \frac{4+2}{216} = \frac{1}{36}$. The sample space consists of all ordered pairs from the given set: $\{(1,1,2,3), (2,3,4), (3,4,5), (4,5,6), (1,3,5), (2,4,6)\}$.
Correct Answer: 1,2,3,4