Statistical Methods II

• Expectation of Geometric Distribution:
• Let $X$ denote the number of independent tosses required until the first head appears in tossing a fair coin.
• The probability of getting a head on any single toss is $p = 0.5$, and tails is $q = 1 - p = 0.5$.
• The random variable $X$ follows a Geometric distribution:

$$P(X = k) = q^{k-1} p, \quad k = 1, 2, 3, \dots$$

• The expected value (mean number of tosses) for a geometric distribution is:

$$E[X] = \frac{1}{p} = \frac{1}{0.5} = 2$$

• Conclusion: The expectation of the number of tosses required is 2 (Option 3).

Correct option: (3) 2

• Probability of Non-occurrence of Independent Events:
• Given $n$ mutually independent events $A_1, A_2, \dots, A_n$ with probabilities:

$$P(A_i) = \frac{1}{i + 1}, \quad \text{for } 1 \le i \le n$$

• The complementary probability that event $A_i$ does not occur is:

$$P(A_i^c) = 1 - P(A_i) = 1 - \frac{1}{i + 1} = \frac{i}{i + 1}$$

• By independence of the events, the complements $A_1^c, A_2^c, \dots, A_n^c$ are also independent:

$$P\left(\bigcap_{i=1}^n A_i^c\right) = \prod_{i=1}^n P(A_i^c) = \prod_{i=1}^n \frac{i}{i + 1}$$

• Expanding the telescoping product:

$$P(\text{none occurs}) = \left(\frac{1}{2}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{3}{4}\right) \times \dots \times \left(\frac{n-1}{n}\right) \times \left(\frac{n}{n + 1}\right)$$

• All intermediate numerator and denominator terms cancel out, leaving:

$$P(\text{none occurs}) = \frac{1}{n + 1}$$

• Conclusion: The probability that none of the $n$ events occurs is $\frac{1}{n+1}$ (Option 1).

Correct option: (1) Option 1

• Probability with Combinations:
• Composition of eligible officers:
• Production department: 3
• Purchase department: 4
• Sales department: 2
• Chartered Accountant (CA): 1
• Total pool size: $N = 3 + 4 + 2 + 1 = 10$ individuals
• Committee size: $k = 4$ individuals
• Direct Symmetry Method:
• Since each of the 10 individuals has an equal chance of being chosen in a committee of 4, the probability that any specific individual (the CA) is included is simply:

$$P(\text{CA included}) = \frac{k}{N} = \frac{4}{10} = \frac{2}{5}$$

• Combinatorial Verification:
• Total number of possible committees: $\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$
• Favorable committees (including the 1 CA, remaining 3 chosen from the other 9): $\binom{1}{1} \times \binom{9}{3} = 1 \times \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$
• Probability: $P = \frac{84}{210} = \frac{2}{5}$
• Conclusion: The probability is 2/5 (Option 4).

Correct option: (4) 215