Probability

The first step is to define the given events clearly:

$ \begin{array}{ll} E_1 \rightarrow \text{Ball from Box I} & F_1 \rightarrow \text{Red Ball} \\\\ E_2 \rightarrow \text{Ball from Box II} & F_2 \rightarrow \text{Green Ball} \end{array} $

Find total and individual probabilities.

Total number of balls in both boxes = $15 + 20 = 35$.

So, we have:

$\begin{array}{ll} P(E_1) = \frac{15}{35}, & P(E_2) = \frac{20}{35} \\\\ P(F_1) = \frac{14}{35}, & P(F_2) = \frac{21}{35} \\\\ P(E_1 \cap F_1) = \frac{6}{35}, & P(E_1 \cap F_2) = \frac{9}{35} \\\\ P(E_2 \cap F_1) = \frac{8}{35}, & P(E_2 \cap F_2) = \frac{12}{35} \end{array}$

Check the given options.

Option (A):

$P(E_1) \cdot P(F_1) = \frac{15}{35} \times \frac{14}{35} = \frac{6}{35} = P(E_1 \cap F_1)$

This statement is true.

Option (B):

$P(E_2) \cdot P(F_2) = \frac{20}{35} \times \frac{21}{35} = \frac{12}{35} = P(E_2 \cap F_2)$

This statement also appears correct numerically, but let’s recheck later with the next steps.

Option (C):

Find the conditional probabilities:

$\begin{array}{r} P(F_1 | E_1) = \frac{\frac{6}{35}}{\frac{15}{35}} = \frac{6}{15} = \frac{2}{5}, \\\\ P(F_1 | E_2) = \frac{\frac{8}{35}}{\frac{20}{35}} = \frac{4}{20} = \frac{2}{5} \end{array}$

Since both are equal, $P(F_1 | E_1) = P(F_1 | E_2)$, this statement is true.

Option (D):

$P(F_2 | E_2) = \frac{\frac{12}{35}}{\frac{20}{35}} = \frac{3}{5} > P(F_1 | E_1)$

This means more green balls are likely from Box II than red balls from Box I.

Conclusion

The correct statements are Options (A) and (C).

For option (a)

$P(E \cap F \cap {G^C}) = P(E \cap F) - P(E \cap F \cap G) \le P(E) - P(E \cap F \cap G)$

$ = {1 \over 8} - {1 \over {10}} = {1 \over {40}}$

For option (b)

$P({E^C} \cap F \cap G) = P(F \cap G) - P(E \cap F \cap G) \le P(F) - P(E \cap F \cap G)$

$ = {1 \over 6} - {1 \over {10}} = {1 \over {15}}$

For option (c)

$P(E \cup F \cup G) \le P(E) + P(F) + P(G)$

$ = {1 \over 8} + {1 \over 6} + {1 \over 4} = {{3 + 4 + 6} \over {24}} = {{13} \over {24}}$

For option (d)

$P({E^C} \cap {F^C} \cap {G^C}) = 1 - P(E \cap F \cap G) \ge 1 - {{13} \over {24}}$

$P({E^C} \cap {F^C} \cap {G^C}) \ge {{11} \over {24}} > {5 \over {12}}$

It is given that there are three bags B₁, B₂ and B₃ and probabilities of being chosen B₁, B₂ and B₃ are respectively.

$ \therefore $ $P({B_1}) = {3 \over {10}},\,P({B_2}) = {3 \over {10}}$ and $P({B_3}) = {4 \over {10}}$.



Now, probability that the chosen ball is green, given that selected bag is

${B_3} = P\left( {{G \over {{B_3}}}} \right) = {3 \over 8}$

Now, probability that the selected bag is B₃, given that the chosen ball is green = $P\left( {{{{B_3}} \over G}} \right)$

$ = {{P\left( {{G \over {{B_3}}}} \right)P({B_3})} \over {P\left( {{G \over {{B_1}}}} \right)P({B_1}) + P\left( {{G \over {{B_2}}}} \right)P({B_2}) + P\left( {{G \over {{B_3}}}} \right)P({B_3})}}$

[by Baye's theorem]

= ${{\left( {{3 \over 8} \times {4 \over {10}}} \right)} \over {\left( {{5 \over {10}} \times {3 \over {10}}} \right) + \left( {{5 \over 8} \times {3 \over {10}}} \right) + \left( {{3 \over 8} \times {4 \over {10}}} \right)}}$

$ = {{{1 \over 2}} \over {{1 \over 2} + {5 \over 8} + {1 \over 2}}} = {4 \over {13}}$

Now, probability that the chosen ball is green = P(G)

= $P({B_1})P\left( {{G \over {{B_1}}}} \right) + P({B_2})P\left( {{G \over {{B_2}}}} \right) + P({B_3})P\left( {{G \over {{B_3}}}} \right)$

[By using theorem of total probability]

= $\left( {{3 \over {10}} \times {5 \over {10}}} \right) + \left( {{3 \over {10}} \times {5 \over 8}} \right) + \left( {{4 \over {10}} \times {3 \over 8}} \right)$

= ${3 \over {20}} + {3 \over {16}} + {3 \over {20}} = {{12 + 15 + 12} \over {80}} = {{39} \over {80}}$

Now, probability that the selected bag is B₃ and the chosen ball is green

= $P({B_3}) \times P\left( {{G \over {{B_3}}}} \right) = {4 \over {10}} \times {3 \over 8} = {3 \over {20}}$

Hence, options (a) and (c) are correct.

$P(X) = {1 \over 3}$

$P\left( {{X \over Y}} \right) = {{P(X \cap Y)} \over {P(Y)}} = {1 \over 2}$

$P\left( {{Y \over X}} \right) = {{P(X \cap Y)} \over {P(X)}} = {2 \over 5}$

$P(X \cap Y) = {2 \over {15}}$

$P(Y) = {4 \over {15}}$

$P\left( {{{X'} \over Y}} \right) = {{P(Y) - P(X \cap Y)} \over {P(Y)}}$

$ = {{{4 \over {15}} - {2 \over {15}}} \over {{4 \over {15}}}} = {1 \over 2}$

$P(X \cup Y) = $${1 \over 3} + {4 \over {15}} - {2 \over {15}}$

$ = {7 \over {15}} = {7 \over {15}}$

$\therefore$ P (drawing red ball from B₁) = ${1 \over 3}$

$ \Rightarrow \left( {{{{n_1} - 1} \over {{n_1} + {n_2} - 1}}} \right)\left( {{{{n_1}} \over {{n_1} + {n_2}}}} \right) + \left( {{{{n_2}} \over {{n_1} + {n_2}}}} \right)\left( {{{{n_1}} \over {{n_1} + {n_2} - 1}}} \right) = {1 \over 3}$

$ \Rightarrow {{n_1^2 + {n_1}{n_2} - {n_1}} \over {({n_1} + {n_2})({n_1} + {n_2} - 1)}} = {1 \over 3}$

Clearly, options (c) and (d) satisfy.

Box I : red balls $\to$ n₁

black balls $\to$ n₂

Box II : red balls $\to$ n₃

black balls $\to$ n₄

$P(R) = {1 \over 2}.{{{n_1}} \over {{n_1} + {n_2}}} + {1 \over 2}.{{{n_3}} \over {{n_3} + {n_4}}}$

P(Box II / R) $ = {{{1 \over 2}.{{{n_3}} \over {{n_3} + {n_4}}}} \over {{1 \over 2}.{{{n_1}} \over {{n_1} + {n_2}}} + {1 \over 2}.{{{n_3}} \over {{n_3} + {n_4}}}}}$

$ = {1 \over {1 + \left( {{{{{{n_1}} \over {{n_1} + {n_2}}}} \over {{{{n_3}} \over {{n_3} + {n_4}}}}}} \right)}} = {1 \over 3}$

Thus, ${{{n_1}} \over {{n_1} + {n_2}}} = 2{{{n_3}} \over {{n_3} + {n_4}}}$

i.e. $2\left( {1 + {{{n_2}} \over {{n_1}}}} \right) = 1 + {{{n_4}} \over {{n_3}}}$ i.e. ${{{n_4}} \over {{n_3}}} - 2{{{n_2}} \over {{n_1}}} = 1$

Let's analyze the given conditions and evaluate which options are correct.

Given:

1. $ P(X|Y) = \frac{1}{2} $


2. $ P(Y|X) = \frac{1}{3} $


3. $ P(X \cap Y) = \frac{1}{6} $

Now, let's proceed step by step through each option.

Option A: $ P(X \cup Y) = \frac{2}{3} $

We can use the formula for the union of two events:

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

To find $ P(X) $ and $ P(Y) $, we use the definitions of conditional probability:

$ P(X|Y) = \frac{P(X \cap Y)}{P(Y)} $

$ \frac{1}{2} = \frac{\frac{1}{6}}{P(Y)} $

$ P(Y) = \frac{1}{6} \times 2 = \frac{1}{3} $

Similarly,

$ P(Y|X) = \frac{P(X \cap Y)}{P(X)} $

$ \frac{1}{3} = \frac{\frac{1}{6}}{P(X)} $

$ P(X) = \frac{1}{6} \times 3 = \frac{1}{2} $

Now, substituting these values into the union formula:

$ P(X \cup Y) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} $

To add these fractions, we need a common denominator. The common denominator is 6:

$ P(X \cup Y) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3} $

Hence, Option A is correct.

Option B: $ X $ and $ Y $ are independent

For two events $ X $ and $ Y $ to be independent, the following condition should hold:

$ P(X \cap Y) = P(X) \times P(Y) $

We already found that:

$ P(X) = \frac{1}{2} $

$ P(Y) = \frac{1}{3} $

$ P(X \cap Y) = \frac{1}{6} $

Checking the independence condition:

$ P(X) \times P(Y) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} $

Since this is equal to $ P(X \cap Y) $, $ X $ and $ Y $ are indeed independent.

Hence, Option B is correct.

Option C: $ X $ and $ Y $ are not independent

From the previous analysis, we have shown that $ X $ and $ Y $ are independent.

Hence, Option C is not correct.

Option D: $ P((X^c) \cap Y) = \frac{1}{3} $

To find $ P((X^c) \cap Y) $, we use the fact that:

$ P(Y) = P((X \cap Y) \cup (X^c \cap Y)) $

Since $ (X \cap Y) $ and $ (X^c \cap Y) $ are disjoint events, we can write:

$ P(Y) = P(X \cap Y) + P(X^c \cap Y) $

Given:

$ P(Y) = \frac{1}{3} $

$ P(X \cap Y) = \frac{1}{6} $

Substituting these,

$ \frac{1}{3} = \frac{1}{6} + P(X^c \cap Y) $

Solving for $ P(X^c \cap Y) $:

$ P(X^c \cap Y) = \frac{1}{3} - \frac{1}{6} = \frac{2}{6} - \frac{1}{6} = \frac{1}{6} $

This contradicts Option D, as it says $ \frac{1}{3} $.

Hence, Option D is not correct.

In conclusion:

Option A and Option B are correct.

Option C and Option D are not correct.

$\begin{array}{r} \text { Given, } \mathrm{P}\left(\mathrm{X}_1\right)=\frac{1}{2} \Rightarrow \mathrm{P}\left(\overline{\mathrm{X}}_1\right)=1-\frac{1}{2}=\frac{1}{2} \\ \mathrm{P}\left(\mathrm{X}_2\right)=\frac{1}{4} \Rightarrow \mathrm{P}\left(\bar{X}_2\right)=1-\frac{1}{4}=\frac{3}{4} \\ \mathrm{P}\left(\mathrm{X}_3\right)=\frac{1}{4} \Rightarrow \mathrm{P}\left(\bar{X}_3\right)=1-\frac{1}{4}=\frac{3}{4} \end{array}$

$\begin{aligned} & \text { Now, } P(X)=P\left(E_1 E_2 E_3\right)+P\left(\bar{E}_1 E_2 E_3\right)+P\left(E_1\right. \\ & \left.\overline{\mathrm{E}}_2 \mathrm{E}_3\right)+\mathrm{P}\left(\mathrm{E}_1 \mathrm{E}_2 \overline{\mathrm{E}}_3\right) \\ & =\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4} \\ \end{aligned}$

$\begin{aligned} & P(X)=\frac{1}{32}+\frac{1}{32}+\frac{3}{32}+\frac{3}{32} \\ & P(X)=\frac{1}{4} \end{aligned}$

(i) $\mathrm{P}\left(\frac{\mathrm{X}_1^c}{\mathrm{X}}\right)=\frac{\mathrm{P}\left(\mathrm{X}_1^c \cap \mathrm{X}\right)}{\mathrm{P}(\mathrm{X})}$

Now, $X_1^c \cap X$ indicates the event that the ship is operational but $\mathrm{E}_1$ is not functioning.

$\begin{aligned} & \therefore \quad \mathrm{P}\left(\mathrm{X}_1^c \cap \mathrm{X}\right)=\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}=\frac{1}{32} \\ & \therefore \mathrm{P}\left(\frac{\mathrm{X}_1^c}{\mathrm{X}}\right)=\frac{\frac{1}{32}}{\frac{1}{4}}=\frac{1}{8} \end{aligned}$

(ii) P (Exactly two engines of the ship are functioning $\mid X$)

$\begin{aligned} & =\frac{\mathrm{P}\left(\overline{\mathrm{E}}_1 \mathrm{E}_2 \mathrm{E}_3\right)+\mathrm{P}\left(\mathrm{E}_1 \overline{\mathrm{E}}_2 \mathrm{E}_3\right)+1\left(\mathrm{E}_1 \mathrm{E}_2 \overline{\mathrm{E}}_3\right)}{\mathrm{P}(\mathrm{X})} \\ & =\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{4}} \\ & =\frac{7}{8} \end{aligned}$

$\begin{aligned} \text { (iii) } & P\left(X \mid X_2\right)=\frac{P\left(X \cap X_2\right)}{P\left(X_2\right)} \\ & =\frac{P\left(E_1 E_2 E_3\right)+P\left(\bar{E}_1 E_2 E_3\right)+P\left(E_1 E_2 \bar{E}_3\right)}{P\left(X_2\right)} \\ & =\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{4}} \\ & =\frac{5}{8} \end{aligned}$

$\begin{aligned} \text { (iv) } & P\left(X \mid X_1\right)=\frac{P\left(X \cap X_1\right)}{P\left(X_1\right)} \\ = & \frac{P\left(E_1 E_2 E_3\right)+P\left(E_1 \bar{E}_2 E_3\right)+P\left(E_1 E_2 \bar{E}_3\right)}{P\left(X_1\right)} \\ = & \frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{2}} \\ = & \frac{7}{16} \end{aligned}$

Let $P(E) = e$ and $P(F) = f$.

$P(E \cup F) - P(E \cap F) = {{11} \over {25}}$

$ \Rightarrow e + f - 2ef = {{11} \over {25}}$ ...... (1)

$P(\overline E \cap \overline F ) = {2 \over {25}}$

$ \Rightarrow (1 - e)(1 - f) = {2 \over {25}}$

$ \Rightarrow 1 - e - f + ef = {2 \over {25}}$ ...... (2)

From Eqs. (1) and (2), we get

$ef = {{12} \over {25}}$ and $e + f = {7 \over 5}$

Solving, we get

$e = {4 \over 5},\,f = {3 \over 5}$ or $e = {3 \over 5},\,f = {4 \over 5}$