Probability

We start with the given probabilities:

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

$ P(V) = \frac{1}{10} $

$ P(W) = \frac{1}{12} $

The event $ U $ indicates that at least one of the students can solve the problem. Thus, its complement $ U' $ (none of the students can solve the problem) is expressed as:

$ \mathrm{P}(U') = \mathrm{P}(S_1' \cap S_2' \cap S_3') = 1 - P(U) = \frac{1}{2} $

The expression for the probability of the complement is:

$ \mathrm{P}(S_1') \cdot \mathrm{P}(S_2') \cdot \mathrm{P}(S_3') = \frac{1}{2} $

From this, we understand:

$ (1 - \mathrm{P}(S_1))(1 - \mathrm{P}(S_2))(1 - \mathrm{P}(S_3)) = \frac{1}{2} \quad \text{(Equation 1)} $

Given $ P(V) $, which is the probability that $ S_1 $ solves the problem given that neither $ S_2 $ nor $ S_3 $ can, we have:

$ \mathrm{P}(V) = \frac{\mathrm{P}(S_1 \cap S_2' \cap S_3')}{\mathrm{P}(S_2' \cap S_3')} = \frac{1}{10} $

$ \Rightarrow \mathrm{P}(S_1) \cdot \mathrm{P}(S_2') \cdot \mathrm{P}(S_3') = \frac{1}{10} \cdot \mathrm{P}(S_2' \cap S_3') $

Now, calculate $ \mathrm{P}(S_2' \cap S_3') $:

From $ P(W) $:

$ \mathrm{P}(W) = \mathrm{P}(S_2 \cap S_3') = \frac{1}{12} $

Combining Equations from conditions of complementary events:

$ \mathbf{\frac{P(S_1)}{1 - \mathrm{P}(S_1)} = \frac{1}{10}} $

Next, let's solve:

$ \begin{aligned} & (1 - \mathrm{P}(S_2)) (1 - \mathrm{P}(S_3)) = \frac{5}{9} \\ & \text{Given, using } \mathrm{P}(W), \; \mathrm{P}(S_2) \cdot \mathrm{P}(S_3') = \frac{1}{12} \\ & \text{Therefore, from equivalence} \left(\frac{\mathrm{P}(S_2)}{1 - \mathrm{P}(S_2)} = \frac{1}{12} \times \frac{9}{5}\right) \end{aligned} $

From simplification:

$ \begin{aligned} & \mathrm{P}(S_2) = \frac{3}{23} \\ & \frac{3}{23} \cdot (1 - \mathrm{P}(S_3)) = \frac{1}{12} \\ & 1 - \mathrm{P}(S_3) = \frac{23}{36} \\ & \mathrm{P}(S_3) = \frac{13}{36} \end{aligned} $

Thus, the probability that student $ S_3 $ can solve the problem, denoted by $ P(T) $, is:

$ P(T) = 1 - \mathrm{P}(S_3') = \frac{13}{36} $

We can solve this problem using Bayes' theorem. Let's define the following events :

$A$: The student knows the answer of a randomly chosen question.

$B$: The student guesses the answer of a randomly chosen question.

$C$: The student gives the correct answer to a question.

We are asked to find the probability that the student knows the answer of a randomly chosen question, i.e., $P(A | C)$. According to Bayes' theorem,

$ P(A | C) = \frac{P(C | A) \cdot P(A)}{P(C)} $

We need to find the values of $P(C | A)$, $P(A)$, and $P(C)$. Let's proceed step by step :

1. $P(C | A)$ is the probability that the student gives the correct answer given that he knows the answer. Since he knows the answer, he will definitely give the correct answer. Therefore,

$ P(C | A) = 1 $

2. $P(A)$ is the probability that the student knows the answer of a randomly chosen question. We don't have this value directly, and we will denote it as $P(A)$. Likewise, the probability that the student guesses the answer is $P(B) = 1 - P(A)$.

3. $P(C)$ is the total probability that the student gives the correct answer. This can be calculated using the Law of Total Probability :

$ P(C) = P(C | A) \cdot P(A) + P(C | B) \cdot P(B) $

Where:

$ P(C | B) = \frac{1}{2} $

(The probability that the student gives the correct answer given that he guessed it), and

$ P(B | C) = \frac{1}{6} $

(The probability that the answer was guessed given the student's answer is correct).

We also have the relation :

$ P(B | C) = \frac{P(C | B) \cdot P(B)}{P(C)} $

Substitute the known values and simplify :

$ \frac{1}{6} = \frac{\frac{1}{2} \cdot P(B)}{P(C)} $

Solving for $P(C)$ gives :

$ P(C) = 3 \cdot P(B) $

From $P(B) = 1 - P(A)$, we get :

$ P(C) = 3 \cdot (1 - P(A)) $

We also know from our Law of Total Probability calculation :

$ P(C) = P(A) + \frac{1}{2} \cdot (1 - P(A)) $

Equate the two expressions for $P(C)$ :

$ P(A) + \frac{1}{2} \cdot (1 - P(A)) = 3 \cdot (1 - P(A)) $

Simplify this equation:

$ P(A) + \frac{1}{2} - \frac{1}{2} \cdot P(A) = 3 - 3 \cdot P(A) $

$ P(A) - \frac{1}{2} \cdot P(A) + \frac{1}{2} = 3 - 3 \cdot P(A) $

$ \frac{1}{2} \cdot P(A) + \frac{1}{2} = 3 - 3 \cdot P(A) $

$ \frac{7}{2} \cdot P(A) = \frac{5}{2} $

$ P(A) = \frac{5}{7} $

Thus, the probability that the student knows the answer of a randomly chosen question is :

$ \boxed{\frac{5}{7}} $

We are given a biased coin with the probability of getting heads (H) as $\frac{1}{3}$ and thus, the probability of getting tails (T) is $\frac{2}{3}$ (since the total probability of getting either heads or tails should sum to 1).

The experiment involves tossing this coin repeatedly until we obtain two identical outcomes in a row. We are interested in the case where the experiment ends with two consecutive heads.

Let's consider two different scenarios for the sequences of coin tosses :

1. Sequences that begin with a head and end with HH (like HH, HTHH, HTHTHH, and so on).
2. Sequences that begin with a tail and end with HH (like THH, THTHH, THTHTHH, and so on).

Both types of sequences will result in the experiment ending, so we need to consider both in our final probability calculation.

Scenario 1 : Sequences beginning with H

The simplest sequence is getting HH right away. The probability of this event is $(\frac{1}{3})^2 = \frac{1}{9}$.

The next sequence is getting a head, then a tail, and then HH (i.e., HTHH). The probability of this sequence is $\frac{1}{3} \cdot \frac{2}{3} \cdot (\frac{1}{3})^2 = \frac{2}{81}$.

We can see a pattern in these sequences. Each sequence in this scenario can be described as getting one head, followed by some number of "tail, head" pairs, and ending with HH.

Hence, the sequences in this scenario form a geometric series, where the ratio between consecutive terms is $r = P(HT) = \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9}$.

The sum of an infinite geometric series is given by the formula $\frac{a}{1 - r}$, where $a$ is the first term of the series and $r$ is the common ratio.

In this scenario, the first term is the probability of getting HH immediately, which is $\frac{1}{9}$. So, the sum of this series is :

$S_1 = \frac{1/9}{1 - 2/9} = \frac{1/9}{7/9} = \frac{1}{7}.$

Scenario 2 : Sequences beginning with T

In this scenario, we first get a tail, and then we follow the same pattern as in the first scenario.

The simplest sequence is THH, with a probability of $\frac{2}{3} \cdot (\frac{1}{3})^2 = \frac{2}{27}$.

The next sequence is getting a tail, a head, another tail, and then HH (i.e., THTHH). The probability of this sequence is $\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} \cdot (\frac{1}{3})^2 = \frac{4}{243}$.

As before, these sequences form a geometric series, with the ratio between consecutive terms is $r = P(TH) = \frac{2}{9}$.

In this scenario, the first term is the probability of getting THH, which is $\frac{2}{27}$. So, the sum of this series is :

$S_2 = \frac{2/27}{1 - 2/9} = \frac{2/27}{7/9} = \frac{2}{21}.$

Finally, to calculate the total probability of the experiment ending with two consecutive heads, we need to sum up the probabilities calculated in both scenarios:

$P(\text{HH}) = S_1 + S_2 = \frac{1}{7} + \frac{2}{21} = \frac{5}{21}.$

So, the answer to the problem is $\frac{5}{21}$ (Option B).

To solve this problem, we need to find the total number of possible solutions to the equation $ x + y + z = 10 $ where $ x, y, $ and $ z $ are nonnegative integers.

First, we employ the "stars and bars" method to determine the number of nonnegative integer solutions to this equation. This method states that the number of ways to partition $ n $ identical items (our sum) into $ k $ distinct groups (our variables) is given by:

$ \binom{n + k - 1}{k - 1} $

For $ x + y + z = 10 $, we have $ n = 10 $ and $ k = 3 $. Plugging into the formula, we get:

$ \binom{10 + 3 - 1}{3 - 1} = \binom{12}{2} $

We calculate the binomial coefficient:

$ \binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66 $

So, there are 66 possible solutions to the equation $ x + y + z = 10 $.

Next, we need to find the number of solutions for which $ z $ is even. Let $ z = 2k $ where $ k $ is a nonnegative integer. Then the equation becomes:

$ x + y + 2k = 10 $

Rearranging it, we get:

$ x + y = 10 - 2k $

Here, $ 10 - 2k $ must be nonnegative, so $ k $ can take values $ 0, 1, 2, 3, 4, 5 $ making total 6 possible values of k.

For each value of $ k $, $ x + y $ must equal the corresponding $ 10 - 2k $. The number of nonnegative integer solutions to $ x + y = m $ for any nonnegative integer $ m $ is given by:

$ \binom{m + 1}{1} = m + 1 $

We will sum the solutions for each valid $ k $:

When $ k = 0 $: $ x + y = 10 $, the number of solutions is $ 11 $.

When $ k = 1 $: $ x + y = 8 $, the number of solutions is $ 9 $.

When $ k = 2 $: $ x + y = 6 $, the number of solutions is $ 7 $.

When $ k = 3 $: $ x + y = 4 $, the number of solutions is $ 5 $.

When $ k = 4 $: $ x + y = 2 $, the number of solutions is $ 3 $.

When $ k = 5 $: $ x + y = 0 $, the number of solutions is $ 1 $.

Summing these, we get:

$ 11 + 9 + 7 + 5 + 3 + 1 = 36 $

Thus, there are 36 solutions where $ z $ is even out of a total of 66 solutions. Hence, the probability that $ z $ is even is given by:

$ \frac{36}{66} = \frac{6}{11} $

Thus, the correct answer is:

Option C: $ \frac{6}{11} $

$\bullet$ Probability of wining of T₁ against T₂ is = 1/2.

$\bullet$ Probability of drawing of T₁ against T₂ is = 1/6.

$\bullet$ Probability of losing of T₁ against T₂ is = 1/3.

$\bullet$3 points for win.

$\bullet$1 point for draw.

$\bullet$ 0 point for loss.

Here, P(X > Y) = P(T₁ win) P(T₁ win) + P(T₁ win) P(draw) + P(draw) P(T₁ win)

$ = \left( {{1 \over 2} \times {1 \over 2}} \right) + \left( {{1 \over 2} \times {1 \over 6}} \right) + \left( {{1 \over 6} \times {1 \over 2}} \right) = {5 \over {12}}$

$\bullet$ Probability of wining of T₁ against T₂ = 1/2.

$\bullet$ Probability of drawing of T₁ against T₂ is = 1/6.

$\bullet$ Probability of losing of T₁ against T₂ is = 1/3.

$\bullet$ 3 points for win.

$\bullet$ 1 point for draw.

$\bullet$ 0 point for loss.

P[X = Y] = P(draw) . P(draw) + P(T₁ win) P(T₂ win) + P(T₂ win) . P(T₁ win)

= (1/6 $\times$ 1/6) + (1/2 $\times$ 1/3) + (1/3 $\times$ 1/2) = 13/36

Given three Boxes:

Given, that a card is drow from each Box and $x_i$ denote the number on the card drawn from $i^{\text {tht }}$ box, $i=1,2,3$.

Sample space $={ }^3 \mathrm{C}_1 .{ }^5 \mathrm{C}_1 .{ }^7 \mathrm{C}_1=3 \cdot 5 \cdot 7=105$.

Let A be the event when $x_1+x_2+x_3$ is odd.

Case I: When all $x_1, x_2, x_3$ are odd

No. of ways $={ }^2 \mathrm{C}_1 .{ }^3 \mathrm{C}_1 .{ }^4 \mathrm{C}_1=2.3 \cdot 4=24$

Case II: When $x_1$ is odd and $x_2, x_3$ are even

No. of ways $={ }^2 \mathrm{C}_1 .{ }^2 \mathrm{C}_1 .{ }^3 \mathrm{C}_1=12$

Case III: When $x_2$ is odd and $x_1, x_3$ are even

No. of ways $={ }^1 \mathrm{C}_1 .{ }^3 \mathrm{C}_1 .{ }^3 \mathrm{C}_1=9$

Case IV: When $x_3$ is odd and $x_1, x_2$ are even

No. of ways $={ }^1 \mathrm{C}_1 .{ }^2 \mathrm{C}_1 . \mathrm{C}_1 \mathrm{C}_1=8$

No. of favorable cases for A is

$\begin{aligned} 24+12+9+8 & =53 \\ \text { Required Probability } & =\mathrm{P}(\mathrm{A})=\frac{53}{105} \end{aligned}$

Hint:

Recall that sum of three numbers is odd only when

(i) All three numbers are odd.

(ii) Two numbers are even and one is odd.

Given three Boxes :

Given, that a card is drow from each box and $x_i$ denote the number on the card drawn from $i^{\text {th }}$ box, $i=1,2,3$.

Sample Space $={ }^3 \mathrm{C}_1 .{ }^5 \mathrm{C}_1 .{ }^7 \mathrm{C}_1=3 \cdot 5 \cdot 7=105$.

Let B be the event when $x_1, x_2, x_3$ are in A.P.

$\Rightarrow 2 x_2=x_1+x_3$

Here, $x_1+x_3$ is even number and for every even value of $x_1+x_3$ there is only one possible value of $x_2$

$x_1+x_2$ is even in two cases

Case I: When both $x_1$ and $x_2$ are even

No. of ways $={ }^1 \mathrm{C}_1$. ${ }^3 \mathrm{C}_{1=}=3$

Case II: When both $x_1$ and $x_2$ are odd

No. of ways $={ }^2 \mathrm{C}_1 .{ }^4 \mathrm{C}_1=2 \times 4=8$

Number of favorable cases for $\mathrm{B}=3+8=11$

Required probability $=\frac{11}{105}$

Hint:

$2 x_2=x_1+x_3$ indicate, $x_1+x_3$ is even and for every even value of $x_1+x_3$ there is only one possible value of $x_2$. Now, $x_1+x_3$ is even in two cases:

(i) When both $x_1$ and $x_2$ are even.

(ii) When both $x_1$ and $x_2$ are odd.

Given, 3 boys and 2 girls stand in a queue.

Sample space $n(s)=\left| \!{\underline {\, {5} \,}} \right.=120$

According to the given condition, following cases may arise.

BGGBB $\quad \left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.$

GGBBB $\quad \left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.$

GBGBB $\quad \left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.$

GBBGB $\quad \left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.$

BGBGB $\quad \left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.$

So, number of favorable ways $=5\left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.=60$

Required probability $=\frac{60}{120}=\frac{1}{2}$

Let $E$ be the event of one white and one red balls $A_1, A_2, A_3$ be the events of both balls are from box $\mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3$ respectively

Now, the probability of one white and one red both balls are drawn from box $B_2$ is $P\left(\frac{A_2}{E}\right)$.

$\Rightarrow P\left(\frac{A_2}{E}\right)=\frac{P\left(A_2 \cap E\right)}{P(E)}$

$\begin{aligned} & \Rightarrow P\left(\frac{A_2}{E}\right)=\frac{P\left(A_2\right) \cdot P\left(\frac{E}{A_2}\right)}{P\left(A_1\right) \cdot P\left(\frac{E}{A_1}\right)+P\left(A_2\right).\mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}_2}\right)+\mathrm{P}\left(\mathrm{A}_3\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}_3}\right)} \\ & \Rightarrow \mathrm{P}\left(\frac{\mathrm{A}_2}{\mathrm{E}}\right)=\frac{\frac{1}{3} \times \frac{{ }^2 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_1}{{ }^9 \mathrm{C}_2}}{\frac{1}{3} \times \frac{{ }^1 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_1}{{ }^6 \mathrm{C}_2}+\frac{1}{3} \times \frac{{ }^2 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_1}{{ }^9 \mathrm{C}_2}+\frac{1}{3} \times \frac{{ }^3 \mathrm{C}_1 \cdot{ }^4 \mathrm{C}_1}{{ }^{12} \mathrm{C}_2} } \\ & \Rightarrow \mathrm{P}\left(\frac{\mathrm{A}_2}{\mathrm{E}}\right)=\frac{\frac{1}{6}}{\frac{1}{5}+\frac{1}{6}+\frac{2}{11}} \\ & \Rightarrow \mathrm{P}\left(\frac{\mathrm{A}_2}{\mathrm{E}}\right)=\frac{55}{181} \\ \end{aligned}$

Hints:

$\Rightarrow$ Recall the Bay's theorem

$ \begin{gathered} P\left(\frac{A_2}{E}\right)=\frac{P\left(A_2\right) \cdot P\left(\frac{E}{A_2}\right)}{P\left(A_1\right) \cdot P\left(\frac{E}{A}\right)+P\left(A_2\right) \cdot P\left(\frac{E}{A_2}\right)+P\left(A_3\right) \cdot P\left(\frac{E}{A_3}\right)}\\ \end{gathered}$

$\begin{aligned} & \text { Let } \mathrm{P} \text { be the probability of same colour balls } \\ & \text { drawn from each boxes } \mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3 \\ & \therefore \quad \mathrm{P}=\mathrm{P} \text { (all are white) }+\mathrm{P} \text { (all are red) }+\mathrm{P} \text { (all } \\ & \quad \text { are Black) } \\ & \Rightarrow P=\frac{1}{6} \times \frac{2}{9} \times \frac{3}{12}+\frac{3}{6} \times \frac{3}{9} \times \frac{4}{12}+\frac{2}{6} \times \frac{4}{9} \times \frac{5}{12} \\ & \Rightarrow P=\frac{6}{648}+\frac{36}{648}+\frac{40}{648} \\ & \Rightarrow P=\frac{82}{648} \end{aligned}$

Hints:

The probability of one ball is drawn from each boxes $B_1, B_2, B_3$ is equal to the sum of probabilities of all are white balls, all are red balls and all are Black balls.

Let A, B, C, D are the events of solving a problem.

Given: $\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{3}{4}, \mathrm{P}(\mathrm{C})=\frac{1}{4}, \mathrm{P}(\mathrm{D})=\frac{1}{8}$

Also given A, B, C, D are independent events.

Let P is the probability of solved by at least one person.

$\Rightarrow \mathrm{P}=1-$ (Probability that none of the person can solve the problem)

$\begin{aligned} & \Rightarrow \mathrm{P}=1-\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}} \cap \overline{\mathrm{C}} \cap \overline{\mathrm{D}}) \\ & \Rightarrow \mathrm{P}=1-\mathrm{P}(\overline{\mathrm{A}}) \cdot \mathrm{P}(\overline{\mathrm{B}}) \cdot \mathrm{P}(\overline{\mathrm{C}}) \cdot \mathrm{P}(\overline{\mathrm{D}}) \\ & \Rightarrow \mathrm{P}=1-(1-\mathrm{P}(\mathrm{A}))(1-\mathrm{P}(\mathrm{B}))(1-\mathrm{P}(\mathrm{C}))(1-\mathrm{P}(\mathrm{D})) \\ & \Rightarrow \mathrm{P}=1-\left(1-\frac{1}{2}\right)\left(1-\frac{3}{4}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right) \\ & \Rightarrow \mathrm{P}=1-\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4} \times \frac{7}{8} \\ & \Rightarrow \mathrm{P}=1-\frac{21}{256}\\ & \Rightarrow \mathrm{P}=\frac{235}{256} \end{aligned}$

Hints:

$\Rightarrow$ (i) If $A$ and $B$ are independent events, then

$\begin{aligned} & \rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \\ & \rightarrow \mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A}}) \cdot \mathrm{P}(\overline{\mathrm{B}})=(1-\mathrm{P}(\mathrm{A}))(1-\mathrm{P}(\mathrm{B})) \\ & \rightarrow \mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B})=\mathrm{P}(\overline{\mathrm{A}}) \cdot \mathrm{P}(\mathrm{B})=(1-\mathrm{P}(\mathrm{A})) \mathrm{P}(\mathrm{B}) \\ & \rightarrow \mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\overline{\mathrm{B}})=\mathrm{P}(\mathrm{A}) \cdot(1-\mathrm{P}(\mathrm{B})) \end{aligned}$

(ii) Probability of at least one person can solve the problem $= 1-$ Probability that none of the persons can solve that problem.

For the given condition, the sample space $=6^4$

For favourable condition

Case I : For $D_4$ there are ${ }^6 C_1$ way. Now, it appears on any one of $D_1, D_2, D_3$ i.e. ${ }^3 C_1 \cdot 1$, for other two there are $5 \times 5$ ways.

$\Rightarrow{ }^6 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_1 \cdot 1 \cdot 5^2=450 \text { Ways }$

Case II : For $\mathrm{D}_4$ there are ${ }^6 \mathrm{C}_1$ ways. Now, it appears on any two of $D_1, D_2, D_3$ i.e. ${ }^3 \mathrm{C}_2 .1 .1$, for other one there are 5 ways.

$\Rightarrow{ }^6 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_2 \cdot 1^2 \cdot 5=90 \text { Ways }$

Case III : For $\mathrm{D}_4$ there are ${ }^6 \mathrm{C}_1$ ways. Now, it appears on all i.e. $1^3$.

$\Rightarrow{ }^6 \mathrm{C}_1 \cdot 1^3=6$

Total number of favourable cases

$\begin{aligned} & =450+90+6 \\ & =546 \text { ways } \end{aligned}$

Thus, Probability $=\frac{546}{6^4}=\frac{91}{216}$

$P\left( {{H \over W}} \right) = {{P(W/H) \times P(H)} \over {P(W/T)\,.\,P(T) + (W/H)\,.\,P(H)}}$

$ = {{{1 \over 2}\left( {{3 \over 5} \times 1 + {2 \over 5} \times {1 \over 2}} \right)} \over {23/30}} = {{12} \over {23}}$

H $\to$ One ball from U₁ to U₂.

T $\to$ Two balls from U₁ to U₂.

E : One ball drawn from U₂.

P/W from

${U_2} = {1 \over 2} \times \left( {{3 \over 5} \times 1} \right) + {1 \over 2} \times \left( {{2 \over 5} \times {1 \over 2}} \right) + {1 \over 2} \times \left( {{{{}^3{C_2}} \over {{}^5{C_2}}} \times 1} \right) + {1 \over 2} \times \left( {{{{}^2{C_2}} \over {{}^5{C_2}}} \times {1 \over 3}} \right) + {1 \over 2} \times \left( {{{{}^3{C_1}\,.\,{}^2{C_1}} \over {{}^5{C_2}}} \times {2 \over 3}} \right) = {{23} \over {30}}$

Sample space A dice is thrown thrice, $n(s) = 6 \times 6 \times 6$.

Favorable events ${\omega ^{{r_1}}} + {\omega ^{{r_2}}} + {\omega ^{{r_3}}} = 0$

i.e. $({r_1},{r_2},{r_3})$ are ordered 3-triples which can take values,

$\left. {\matrix{ {(1,2,3),} & {(1,5,3),} & {(4,2,3),} & {(4,5,3)} \cr {(1,2,6),} & {(1,5,6),} & {(4,2,6),} & {(4,5,6)} \cr } } \right\}$

i.e. 8 ordered pairs and each can be arranged in 3! ways = 6

$\therefore$ $n(E) = 8 \times 6$

$ \Rightarrow P(E) = {{8 \times 6} \over {6 \times 6 \times 6}} = {2 \over 9}$

From the tree-diagram it follows that

$P({B_G}) = {{46} \over {80}}$

$P({B_G}|G) = {{10} \over {16}} = {5 \over 8}$

$\therefore$ $P({B_G} \cap G) = {5 \over 8} \times {4 \over 5} = {1 \over 2}$

$P(G|{B_G}) = {{{1 \over 2}} \over {P({B_G})}} = {1 \over 2} \times {{80} \over {46}} = {{20} \over {23}}$

The required probability is $P(X = 3)$

$\left( {{5 \over 6}} \right)\left( {{5 \over 6}} \right){1 \over 6} = {{25} \over {216}}$

The probability $P(X \ge 3)$ is nothing but the probability of $P(X \le 2)$:

${1 \over 6} + {5 \over 6} \times {1 \over 6} = {{11} \over {36}}$

The required probability is

$1 - {{11} \over {36}} = {{25} \over {36}}$

$X \ge 6$: The probability for

${{{5^5}} \over {{6^6}}} + {{{5^6}} \over {{6^7}}}\, + \,...\, + \,\infty = {{{5^5}} \over {{6^6}}}\left( {{1 \over {1 - 5/6}}} \right) = {\left( {{5 \over 6}} \right)^5}$

For $X>3$, we have

${{{5^3}} \over {{6^4}}} + {{{5^4}} \over {{6^5}}}\, + {{{5^5}} \over {{6^6}}}\, + ...\, + \,\infty = {\left( {{5 \over 6}} \right)^3}$

Hence, the conditional probability is

${{{{(5/6)}^6}} \over {{{(5/6)}^3}}} = {{25} \over {36}}$

Given,

An experiment has 10 equally likely outcomes,

Let B have n no of outcomes,

So, $P(B) = {n \over {10}}$

$P(A) = {4 \over {10}}$

$P(A \cap B) = {4 \over {10}} \times {n \over {10}} = {{{{2n} \over 5}} \over {10}}$

Since, A and B are independent

n is an integer

n = 5 or 10

We have,

$ax + by = 0$

$cx + dy = 0$

since, the system of homogeneous equation is always consistent and has a solution.

Therefore, statement 2 is true.

Now, $\Delta = \left| {\matrix{ a & b \cr c & d \cr } } \right|$ and $a,b,c,d \in \{ 0,1\} $

$ = ad - bc$

No. of ways of selecting $a,b,c,d$ from the set {0, 1} is 2 $\times$ 2 $\times$ 2 $\times$ 2 = 16

If the system has unique solution,

Then $\Delta\ne0$

$\Rightarrow$ either $ad = 1,bc = 0$ or $ad = 0,bc = 1$ favourable case = 6

Therefore, probability that system of equation has unique solution is ${6 \over {16}} = {3 \over 8}$. Statement 1 is True.

Hence, the Statement 2 is True but is not a correct explanation of statement 1.

$\begin{aligned} & \mathrm{E}_{1} f_{1} G \text { are pairwise independent events. } \\\\ & \therefore \quad \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{F}) \\\\ & \mathrm{P}(\mathrm{F} \cap \mathrm{G})=\mathrm{P}(\mathrm{F}) \cdot \mathrm{P}(\mathrm{G}) \\\\ & \mathrm{P}(\mathrm{G} \cap \mathrm{E})=\mathrm{P}(\mathrm{G}) \cdot \mathrm{P}(\mathrm{E}) \\\\ & \mathrm{P}\left(\frac{\mathrm{E}^{\mathrm{c}} \cap \mathrm{F}^{\mathrm{c}}}{\mathrm{G}}\right)=\mathrm{P}\left(\frac{\left(E^{\mathrm{c}} \cap \mathrm{F}^{\mathrm{c}}\right) \cap G}{\mathrm{P}(\mathrm{G})}\right) \\\\ &=\frac{\mathrm{P}(\mathrm{G})-\mathrm{P}(\mathrm{G} \cap \mathrm{E})-\mathrm{P}(\mathrm{G} \cap \mathrm{F})}{\mathrm{P}(\mathrm{G})} \\\\ &=1-\mathrm{P}(\mathrm{E})-\mathrm{P}(\mathrm{F}) \\\\ &=\mathrm{P}\left(\mathrm{E}^{\mathrm{c}}\right)-\mathrm{P}(\mathrm{F}) . \end{aligned}$

Let E = event when each American man is seated adjacent to his wife and

A = event when Indian man is seated adjacent to his wife.

Now,

$n(A\cap E)=(4!)\times(2!)^5$

Event when each American man is seated adjacent to his wife.

Again $n(E)=(5!)\times(2!)^4$

$p\left( {{A \over E}} \right) = {{n(A \cap E)} \over {n(E)}}$

$ = {{(4!) \times {{(2!)}^5}} \over {(5!) \times {{(2!)}^4}}} = {2 \over 5}$

Statement 1 : If P(H$_i$ $\cap$ E) = O for some, then

$P\left( {{{{H_i}} \over E}} \right) = P\left( {{E \over {{H_i}}}} \right) = 0$

If P(H$_i$ $\cap$ E) $\ne o~\forall i=1,2,3.......,n$ then

$P\left( {{{{H_i}} \over E}} \right) = {{P({H_i} \cap E)} \over {P({H_i})}} \times {{P({H_i})} \over {P(E)}}$

$ = {{P\left( {{E \over {{H_i}}}} \right) \times P({H_i})} \over {P(E)}}$

$ > P\left( {{E \over {{H_i}}}} \right) \times P({H_i})$

Hence, statement 1 may not always be true.

Statement 2 : Clearly, we can write as

H$_1$ $\cup$ H$_2$ $\cup$ H$_1$ $\cup$ ------------------ $\cup$ H$_n$ = S

$\Rightarrow$ P(H$_1$) + P(H$_2$) + ---------------- + P(H$_n$) = 1

Hence, statement 2 is true.

$\because \mathrm{P}\left(u_{i}\right) \propto i$

$\mathrm{P}\left(u_{i}\right)=\mathrm{K} i$

We know that $\sum \mathrm{P}\left(\mathrm{u}_{i}\right)=1$

$\begin{aligned} \mathrm{K} \frac{n(n+1)}{2} & =1 \\ \mathrm{~K} & =\frac{2}{n(n+1)} \\ \mathrm{P}\left(u_{i}\right) & =\frac{2 i}{n(n+1)} \\ \mathrm{P}\left(\frac{w}{u_{2 i}}\right) & =\frac{2 i}{n+1} \end{aligned}$

Now, $\quad \mathrm{P}(w)=\sum_\limits{1^{i}=1}^{n} \mathrm{P}\left(u_{i}\right) \cdot \mathrm{P}\left(\frac{w}{u_{i}}\right)$

$ = \sum\limits_{{1^i} = 1}^n {{{2i} \over {n(n + 1)}}{i \over {(n + 1)}}} $

$ = \sum\limits_{{1^i} = 1}^n {{{2{i^2}} \over {n{{(n + 1)}^2}}}} $

$ = {2 \over {n{{(n + 1)}^2}}}\sum\limits_{{1^i} = 1}^n {{i^2}} $

$ = {2 \over {n{{(n + 1)}^2}}}{{n(n + 1)(2n + 1)} \over 6}$

$ = {{2n + 1} \over {3(n + 1)}}$

Now, $\mathop {\lim }\limits_{n \to \infty } {{2n + 1} \over {3(n + 1)}} = \mathop {\lim }\limits_{n \to \infty } {{\left( {2 + {1 \over n}} \right)} \over {3\left( {{1 \over n} + 1} \right)}} = {2 \over 3}$

Since, $\mathrm{P}\left(u_{i}\right)=\mathrm{C}$ means

All events are equally likely

$\mathrm{P}\left(u_{i}\right)=\mathrm{C}=\frac{1}{n} \quad$ for $i=1,2,3, \ldots . n$

$\mathrm{P}(u_n)=\frac{1}{n}$

$\mathrm{P}\left(\frac{w}{u_{n}}\right)=\frac{n}{n+1}$

$\begin{aligned} \mathrm{P}(w) & =\sum \mathrm{P}\left(u_{i}\right) \cdot \mathrm{P}\left(\frac{w}{u_{n}}\right) \\ & =\sum\left(\frac{1}{n} \frac{2^{i}}{(n+1)}\right)=\frac{1}{n(n+1)} \cdot \frac{n(n+1)}{2}=\frac{1}{2} \\ \mathrm{P}\left(\frac{u_{\mathrm{n}}}{w}\right) & =\frac{\mathrm{P}\left(u_{\mathrm{n}}\right) \times \mathrm{P}\left(\frac{w}{u_{\mathrm{n}}}\right)}{\mathrm{P}(w)} \\ & =\frac{\frac{1}{n} \times \frac{n}{(n+1)}}{\frac{1}{2}}=\frac{2}{n+1} \end{aligned}$

Total number of ball

Even numbered urns $=\frac{n}{2}(n+1)$

Total no. of white balls = $2+4+6+....+n$.

$=2(1+2+3+...+m)$

where $m=\frac{n}{2}$

$=2 \times \frac{m(m+1)}{2}$

$=\frac{n}{2} \cdot\left(\frac{n}{2}+1\right)=\frac{n(n+2)}{4}$

$\therefore \quad \mathrm{P}\left(\frac{w}{\mathrm{E}}\right)=\frac{\frac{n}{4}(n+2)}{\frac{n}{2}(n+1)}$

$=\frac{(n+2)}{2(n+1)}$

$\begin{aligned} P\left(E_{\text {car }}\right)=\frac{1}{7} \quad & P\left(E_{S_{\text {cooter }}}\right)=\frac{3}{2} \\ P\left(E_{\text {bus}}\right)=\frac{2}{7} \quad & P\left(E_{\text {train }}\right)=\frac{1}{7} \\ P\left(\frac{E_{\text {late }}}{E_{\text {car }}}\right) & =\frac{2}{9} \\ P\left(\frac{E_{\text {late }}}{E_{\text {scooter }}}\right) & =\frac{1}{9} \\ E\left(\frac{E_{\text {late }}}{E_{\text {bus }}}\right) & =\frac{4}{9} \\ E\left(\frac{E_{\text {late }}}{E_{\text {train }}}\right) & =\frac{1}{9}\end{aligned}$

Using Total probability theorem.

$\begin{aligned} & P\left(E_{\text {late }}\right)=P\left(E_{\text {car }}\right) \times P\left(\frac{E_{\text {late }}}{E_{\text {car }}}\right)+P\left(E_{\text {scooter }}\right) \\ & \times P\left(\frac{E_{\text {late }}}{E_{\text {scooter }}}\right)+P\left(E_{\text {Bus }}\right) \times P\left(\frac{E_{\text {late }}}{E_{\text {Bus }}}\right) \\ & +P\left(E_{\text {train }}\right) \times P\left(\frac{E_{\text {late }}}{E_{\text {train }}}\right) \\ & =\frac{1}{7} \times \frac{2}{9}+\frac{3}{2} \times \frac{1}{9}+\frac{2}{7} \times \frac{4}{9}+\frac{1}{7} \times \frac{1}{9} \\ & =\frac{2}{63}+\frac{3}{18}+\frac{8}{63}+\frac{1}{63}=\frac{11}{63}+\frac{3}{18}=\frac{43}{126} \\ & \mathrm{P}\left(\mathrm{E}_{\text {ontime }}\right)=1-\frac{43}{126}=\frac{83}{126} \\ & P\left(\frac{E_{\text {car }}}{E_{\text {ontime }}}\right)=\frac{P\left(E_{\text {car }} \cap E_{\text {ontime }}\right)}{P\left(E_{\text {ontime }}\right)} \\ & =\frac{P\left(E_{\text {car }}\right) \times P\left(E_{\text {ontime }}\right)}{P\left(E_{\text {ontime }}\right)}=P\left(E_{\text {car }}\right)=\frac{1}{7} \end{aligned}$