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Calculation:

Combinations:

${}^{10}C_3 = 120$

${}^{10}C_4 = 210$

${}^{10}C_5 = 252$

${}^{10}C_6 = 210$

${}^{10}C_7 = 120$

Sum of reciprocals:

$\frac{1}{120} + \frac{1}{210} + \frac{1}{252} + \frac{1}{210} + \frac{1}{120} = 2\left(\frac{1}{120} + \frac{1}{210}\right) + \frac{1}{252}$

$= 2\left(\frac{7 + 4}{840}\right) + \frac{1}{252} = \frac{11}{420} + \frac{1}{252} = \frac{33 + 5}{1260} = \frac{38}{1260} = \frac{19}{630}$

Harmonic Mean:

$\text{H.M.} = \frac{5}{\frac{19}{630}} = \frac{3150}{19}$

Hence, the correct answer is Option 1.

Calculation:

The probability that one farmer is in debt is given as P = 0.60

Since the events are independent, the probability that three randomly selected farmers are all in debt is:

P³ = (0.60)³ = 0.216

Hence, the correct answer is Option 3.

Calculation:

Given:

$P(L) = 0.68$

$P(D) = 0.56$

$P(L \cap D) = 0.48$

Using the formula for union of two events:

⇒ $P(L \cup D) = P(L) + P(D) - P(L \cap D)$

Substituting values:

⇒ $P(L \cup D) = 0.68 + 0.56 - 0.48 = 0.76$

Hence, the correct answer is Option 2.

Calculation:

Given:

White balls = 10,Red balls= 5 ,Total balls= 15 

Total ways to draw 2 balls from 15:

$\binom{15}{2}=\frac{15\times 14}{2}=105$

Favourable ways to draw 2 red balls from 5 red balls:

$\binom{5}{2}=\frac{5\times 4}{2}=10$

Therefore, required probability:

$P=\frac{\binom{5}{2}}{\binom{15}{2}}=\frac{10}{105}=\frac{2}{21}$

Hence, the correct answert is Option 1.

Calculation:

Given:

White balls=5, Red balls = 6, Blue balls = 4

$\text{Total balls}=5+6+4=15$

Total ways to draw 3 balls from 15:

$\binom{15}{3}=\frac{15\times14\times13}{6}=455$

Favourable ways to draw one white, one red and one blue ball:

$\binom{5}{1}\times\binom{6}{1}\times\binom{4}{1} =5\times6\times4=120$

Therefore, required probability:

$P=\frac{120}{455}=\frac{24}{91}$

Hence, the correct answer is Option 3.

Calculation:

For a binomial distribution, the required conditions are:

• The number of trials must be finite and fixed
• The trials must be independent
• Each trial must have exactly two possible outcomes (success/failure)
• Probability of success remains constant

Hence, the correct answer is Option 4.

Calculation:

We are asked to compute:

$P(X \cap Y') + P(X' \cap Y)$

Using the formula:

$= P(X)[1 - P(Y)] + [1 - P(X)]P(Y)$

Substituting values:

$= \frac{4}{5} \left(1 - \frac{5}{6}\right) + \left(1 - \frac{4}{5}\right) \cdot \frac{5}{6}$

$= \frac{4}{5} \cdot \frac{1}{6} + \frac{1}{5} \cdot \frac{5}{6} = \frac{4}{30} + \frac{5}{30} = \frac{9}{30} = \frac{3}{10}$

Hence, the correct answer is Option 1.

Calculation:

Let:

$P(P)=\frac{2}{3},\quad P(P\cap E)=\frac{11}{15},\quad P(P\cup E)=\frac{4}{5}$

Using the formula:

⇒ $P(P\cup E)=P(P)+P(E)-P(P\cap E)$

Substitute the given values:

⇒ $\frac{4}{5}=\frac{2}{3}+P(E)-\frac{11}{15}$

Convert fractions to a common denominator:

⇒ $\frac{4}{5}=\frac{12}{15},\quad \frac{2}{3}=\frac{10}{15}$

Simplifying:

⇒ $\frac{12}{15}=\frac{10}{15}+P(E)-\frac{11}{15}$

⇒ $\frac{12}{15}=P(E)-\frac{1}{15}$

Therefore,

 $P(E)=\frac{12}{15}+\frac{1}{15}=\frac{13}{15}$

Hence, the correct answer is Option 2.

Calculation:

Let:

$P(X) = p$

$P(Y) = q$

X and Y are independent ⇒ $P(X \cap Y) = pq$

Probability of exactly one event occurring:

$P(\text{exactly one}) = P(X) + P(Y) - 2P(X \cap Y) = p + q - 2pq$

So, statement I is incorrect.

Probability of at least one event occurring:

$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = p + q - pq$

So, statement II is also incorrect.

Hence, the Correct answer is Option 4.

Calculation:

Given:

Number of black faces = 3

Number of white faces = 2

Number of red faces = 1

Probability that black, white, and red appear in first, second, and third positions respectively:

$P = \frac{3}{6} \cdot \frac{2}{6} \cdot \frac{1}{6} = \frac{6}{216} = \frac{1}{36}$

Hence, the correct answer is Option 1.