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: Count strictly increasing sequences (E₁) as C(6,3)=20 outcomes from choosing 3 distinct numbers from {1,2,3,4,5,6}, then count increasing AP sequences (E₂) by identifying triples with common difference d where first term a and a+(a+d)+(a+2d)≤6. Use conditional probability P(E₂|E₁) = P(E₂∩E₁)/P(E₁) where E₂∩E₁ requires the increasing sequence to also be an AP.
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