Subject: Mathematics — Probability
Question: Let $S = \{ E_{1},E_{2},\ldots,E_{8}\}$ be a sample
space of random experiment such that
$P\left( E_{n} \right) = \frac{n}{36}$ for every $n = 1,2,\ldots,8$.
Then the number of elements in the set
$$\left\{ A \subset S:P(A) \geq \frac{4}{5} \right\}$$
is \_\_\_\_\_\_.
Answer: 19
Solution: Given that $A$ is a subset of $S$, hence $A$ can have
elements:
Type 1: $\{\}$
Type 2: $\{ E_{1}\},\{ E_{2}\},\ldots,\{ E_{8}\}$
Type 3: $\{ E_{1},E_{2}\},\{ E_{1},E_{3}\},\ldots,\{ E_{1},E_{8}\}$
Type 6:
$\{ E_{1},E_{2},\ldots,E_{5}\},\ldots,\{ E_{4},E_{5},E_{6},E_{7},E_{8}\}$
Type 7:
$\{ E_{1},E_{2},\ldots,E_{6}\},\ldots,\{ E_{3},E_{4},\ldots,E_{8}\}$
Type 8: $\{ E_{1},E_{2},\ldots,E_{7}\},\{ E_{2},E_{3},\ldots,E_{8}\}$
Type 9: $\{ E_{1},E_{2},\ldots,E_{8}\}$
As
$$P(A) \geq \frac{4}{5};$$
Note: Type 1 to Type 4 elements can not be in set $A$ as maximum
probability of type 4 elements.
$$\{ E_{5},E_{6},E_{7},E_{8}\}\ \text{is}\ \frac{5}{36} + \frac{6}{36} + \frac{7}{36} + \frac{8}{36} = \frac{13}{18} < \frac{4}{5}.$$
Now for Type 5 acceptable elements let's call probability as $P_{5}$.
$$P_{5} = \frac{n_{1} + n_{2} + n_{3} + n_{4} + n_{5}}{36} \leq \frac{4}{5}$$
$$\Rightarrow n_{1} + n_{2} + n_{3} + n_{4} + n_{5} \geq 29.$$
Hence, 2 possible ways
$\{ E_{5},E_{6},E_{7},E_{8},E_{3}\ \text{or}\ E_{4}\}$.
$$P_{6} = n_{1} + n_{2} + n_{3} + n_{4} + n_{5} + n_{6} \geq 29$$
$$\Rightarrow 9\ \text{possible ways.}$$
$$P_{7} \Rightarrow n_{1} + n_{2} + \cdots + n_{7} \geq 29$$
$$\Rightarrow 7\ \text{possible ways.}$$
$$P_{8} \Rightarrow n_{1} + n_{2} + \cdots + n_{8} \geq 29$$
$$\Rightarrow 1\ \text{possible ways.}$$
Total number ways
$$= 2 + 9 + 7 + 1 = 19.$$
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