Let $\mathrm{S}=\left\{\mathrm{E}_1, \mathrm{E}_{2} \ldots \mathrm{E}_{8}\right\}$ be a sample space of random experiment such that $\mathrm{P}\left(\mathrm{E}_{\mathrm{n}}\right)=\frac{\mathrm{n}}{36}$ for every $\mathrm{n}=1,2 \ldots 8$. Then the number of elements in the set $\left\{\mathrm{A} \subset \mathrm{S}: \mathrm{P}(\mathrm{A}) \geq \frac{4}{5}\right\}$ is

Probability XII — JEE Main · 2022 · 27 Jun 2022 · Shift 2 Question with Solution

Let

S = {E_{1}, E_{2} \dots E_{8}}

be a sample space of random experiment such that

P (E_{n}) = \frac{n}{36}

for every

n = 1, 2 \dots 8

. Then the number of elements in the set

{A \subset S : P (A) \geq \frac{4}{5}}

Answer: 19

We need

P (A) \geq \frac{4}{5}

, which means

P (A^{c}) \leq \frac{1}{5} = \frac{36}{180}

.

Since

\sum_{i = 1}^{8} P (E_{i}) = \frac{1 + 2 + \dots + 8}{36} = \frac{36}{36} = 1

, we need the sum of probabilities of elements NOT in A to be at most

\frac{36}{180}

, i.e., the sum of indices not in A should be at most

7.2

.

Counting subsets by number of missing elements:

0 missing: ${E_{1}, E_{2}, \dots, E_{8}}$ — 1 way
1 missing: Can miss $E_{1}$ through $E_{7}$ (sum $\leq 7$ ) — 7 ways
2 missing: Pairs summing to at most 7: $(1, 2), (1, 3), \dots, (1, 6), (2, 3), (2, 4), (2, 5), (3, 4)$ — 9 ways
3 missing: Triples summing to at most 7: $(1, 2, 3), (1, 2, 4)$ — 2 ways

Total:

1 + 7 + 9 + 2 = 19

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Integer answer