Probability

The expected value of a discrete random variable is given by the formula :

$ E(X)=\sum X_i \cdot P\left(X_i\right) $

Using the data from the table :

$ E(X)=\left(4 k \cdot \frac{2}{15}\right)+\left(\frac{30 k}{7} \cdot \frac{1}{15}\right)+\left(\frac{32 k}{7} \cdot \frac{2}{15}\right)+\left(\frac{34 k}{7} \cdot \frac{1}{5}\right)+\left(\frac{36 k}{7} \cdot \frac{1}{15}\right)+\left(\frac{40 k}{7} \cdot \frac{1}{5}\right)+\left(6 k \cdot \frac{1}{15}\right) $

Take $\frac{k}{15}$ factor common

$ \begin{gathered} E(X)=\frac{k}{15}\left[8+\frac{30}{7}+\frac{64}{7}+\frac{102}{7}+\frac{36}{7}+\frac{76}{7}+\frac{120}{7}+6\right] \\ E(X)=\frac{k}{15}\left[14+\frac{30+64+102+36+76+120}{7}\right] \\ E(X)=\frac{k}{15}\left[14+\frac{428}{7}\right] \end{gathered} $

$ E(X)=\frac{k}{15}\left[\frac{98+428}{7}\right]=\frac{k}{15} \cdot \frac{526}{7}=\frac{526 k}{105} $

Given that $E(X)=\frac{263}{15}$ :

$ \begin{gathered} \frac{263}{15}=\frac{526 k}{105} \\ 263=\frac{526 k}{7} \\ k=\frac{263 \times 7}{526}=\frac{1 \times 7}{2}=\frac{7}{2}=3.5 \end{gathered} $

Identify $X$ values where $X<20$

Substitute $k=\frac{7}{2}$ into each $X$ :

$ \begin{array}{ll} \text { } X_1=4 \times \frac{7}{2}=14(<20) & X_4=\frac{34}{7} \times \frac{7}{2}=17(<20) \\ \text { } X_2=\frac{30}{7} \times \frac{7}{2}=15(<20) & X_5=\frac{36}{7} \times \frac{7}{2}=18(<20) \\ \text { } X_3=\frac{32}{7} \times \frac{7}{2}=16(<20) & X_6=\frac{38}{7} \times \frac{7}{2}=19(<20) \\ \text { } & X_7=\frac{40}{7} \times \frac{7}{2}=20(\text { Not }<20) \\ \text { }& X_8=6 \times \frac{7}{2}=21(\text { Not }<20) \end{array} $

Calculate $P(X<20)$ :

Sum the probabilities for $X=14,15,16,17,18$, and 19:

$ P(X<20)=P(X=14)+P(X=15)+P(X=16)+P(X=17)+P(X=18)+P(X=19) $

$\Rightarrow $ $ P(X<20)=\frac{2}{15}+\frac{1}{15}+\frac{2}{15}+\frac{3}{15}+\frac{1}{15}+\frac{2}{15} $

$\Rightarrow $ $ P(X<20)=\frac{11}{15} $

Given Terms

• Total Balls ( $n$ ) : 10

• Red Balls ( $k$ ) : Variable from 0 to 10 .

• Black Balls : $(10-k)$.

• Sample Drawn ( $r$ ) : 3 balls, without replacement.

• Observed Event (A) : All 3 balls drawn are black.

• Target Event : Find the probability that the bag contained exactly $\mathbf{1}$ red and $\mathbf{9}$ black balls (which means $k=1$ ).

Let $E_k$ be the event that the bag contains $k$ red balls. Since $k$ can be any integer from 0 to 10 , there are 11 possible compositions. Assuming they are equally likely:

$ P\left(E_k\right)=\frac{1}{11} $

The event $A$ is drawing 3 black balls. The probability of this happening depends on how many black balls $(10-k)$ were in the bag :

$ P\left(A \mid E_k\right)=\frac{{ }^{10-k} C_3}{{ }^{10} C_3} $

Note : For this to be possible, there must be at least 3 black balls, so $10-k \geq 3$, meaning $k \leq 7$.

Using Bayes' Theorem

We need $P\left(E_1 \mid A\right)$, which is the probability of having 1 red ball given that 3 black balls were drawn :

$ P\left(E_1 \mid A\right)=\frac{P\left(E_1\right) \cdot P\left(A \mid E_1\right)}{\sum\limits_{k=0}^7 P\left(E_k\right) \cdot P\left(A \mid E_k\right)} $

Since $P\left(E_k\right)$ is constant $\left(\frac{1}{11}\right)$ for all $k$, it cancels out :

$ P\left(E_1 \mid A\right)=\frac{{ }^{10-1} C_3}{{ }^{10} C_3+{ }^9 C_3+\cdots+{ }^3 C_3} $

denominator ${ }^{10} C_3+{ }^9 C_3+\cdots+{ }^3 C_3$ can be written as ${ }^3 C_3+{ }^4 C_3+{ }^5 C_3 \cdots{ }^{10} C_3$

using identity ${ }^k C_k+{ }^{k+1} C_k+{ }^{k+2} C_k+\ldots{ }^n C_k={ }^{n+1} C_{k+1}$

so, denominator become $={ }^{11} C_4=\frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1}=330$

Numerator : ${ }^{10-1} C_3={ }^9 C_3=\frac{9 \times 8 \times 7}{3 \times 2 \times 1}=84$

$P\left(E_1 \mid A\right)=\frac{84}{330}=\frac{14}{55}$

total number of defective bulbs $n(D)=10$

total number of non defective bulbs $n(\bar{D})=90$

total bulbs $=n(D)+n(\bar{D})=100$

probability of picking a defective bulbs: $p=\frac{10}{100}=\frac{1}{10}$

probability of picking a non-defective bulbs $q=\frac{90}{100}=\frac{9}{10}$

number of trials $n=8$ bulbs are selected

let $X$ be the event of getting atleast 7 defective bulbs

$ \begin{aligned} & P(X)=P(X=7)+P(X=8) \\ & ={ }^8 C_7 p^7 \cdot q^1+{ }^8 C_8 p^8 q^0 \\ & =8 \cdot\left(\frac{1}{10}\right)^7 \cdot \frac{9}{10}+1 \cdot\left(\frac{1}{10}\right)^8 \end{aligned} $

$P(X)=\left(\frac{1}{10}\right)^8[72+1]=\frac{73}{10^8}$

Bag A : 9 white, 8 black (Total $=17$ )

Bag B : 6 white, 4 black (Total $=10$ )

One ball is moved from Bag B to Bag A. There are two mutually exclusive events for this transfer:

Case 1 : The ball moved is white

The probability of choosing a white ball from Bag B is:

$ P\left(W_B\right)=\frac{6}{10} $

After adding this white ball to Bag A, Bag A now contains 10 white and 8 black balls ( 18 total). The probability of then drawing a white ball from Bag A is :

$ P\left(W_A \mid W_B\right)=\frac{10}{18} $

Case 2: The ball moved is black

The probability of choosing a black ball from Bag B is :

$ P\left(B_B\right)=\frac{4}{10} $

After adding this black ball to Bag A, Bag A now contains 9 white and 9 black balls ( 18 total). The probability of then drawing a white ball from Bag A is :

$ P\left(W_A \mid B_B\right)=\frac{9}{18} $

Using total probability theorem

$ P\left(W_A\right)=P\left(W_B\right) \cdot P\left(W_A \mid W_B\right)+P\left(B_B\right) \cdot P\left(W_A \mid B_B\right) $

$\Rightarrow $$P\left(W_A\right)=\left(\frac{6}{10} \times \frac{10}{18}\right)+\left(\frac{4}{10} \times \frac{9}{18}\right)$

$ P\left(W_A\right)=\frac{60}{180}+\frac{36}{180}=\frac{96}{180}=\frac{8}{15} $

It is given that $P\left(W_A\right)=\frac{p}{q}$.

$ \begin{aligned} & \operatorname{gcd}(8,15)=1 \\\\ & \text { So } p=8 \& q=15 \end{aligned} $

$p+q=8+15=23$

Total ways to choose two distinct numbers from $1,2,\dots,50$ is

$ \binom{50}{2}=1225. $

The product $ab$ is not divisible by $3$ only if neither $a$ nor $b$ is divisible by $3$.

Multiples of $3$ in $1$ to $50$: $\left\lfloor \frac{50}{3}\right\rfloor=16$.

So numbers not divisible by $3$: $50-16=34$.

Unfavorable pairs:

$ \binom{34}{2}=\frac{34\cdot 33}{2}=561. $

Therefore favorable pairs:

$ 1225-561=664, $

and the required probability is

$ \frac{664}{1225}. $

Answer: $\boxed{\frac{664}{1225}}$ (Option C).

$ \begin{aligned} & 2 K+K+3 K+2 K^2+2 K+K^2+K+7 K^2=1 \\ & (K+1)(10 K-1)=0 \\ & K=-1 \text { or } K=\frac{1}{10} \\ & P(3 < x \leq 6)=\frac{2}{100}+\frac{2}{10}+\frac{1}{100}+\frac{1}{10}=0.33 \end{aligned} $

$ \begin{aligned} & \text { Mean }=8 \\ & \frac{2+4+10+x+y+12+14}{7}=8 \\ & \Rightarrow x+y=14 \\ & \text { variance }=16 \\ & 16=\frac{2^2+4^2+10^2+x^2+y^2+12^2+14^2+y^2}{7}-8^2 \\ & \Rightarrow x^2+y^2=100 \\ & \therefore x=8, y=6 \\ & \text { New set }:\{1,2,3,4,5,6\} \\ & \text { Required probability }=1-P(\text { both number are } \geq 4) \\ & =1-\frac{{ }^3 C_2}{{ }^6 C_2} \\ & =1-\frac{3}{15} \\ & =\frac{4}{5} \end{aligned} $

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).

$ \begin{array}{|l|l|l|l|l|l|l|l|} \hline \text { Conveyance } & \text { Bus } & \text { Scooter } & \text { Car } & \text { Arrive late } & \text { Bus } & \text { Scooter } & \text { Car } \\ \hline \text { Probability } & \frac{2}{5} & \frac{1}{5} & \frac{2}{5} & \text { Probability } & \frac{1}{5} & \frac{1}{3} & \frac{1}{4} \\ \hline \end{array} $

$ \begin{aligned} &\text { We label the events in the chart below : }\\ &\begin{array}{|l|l|l|l|l|l|l|l|} \hline \text { Conveyance } & \text { Bus } & \text { Scooter } & \text { Car } & \text { Arrive late by } & \text { Bus } & \text { Scooter } & \text { Car } \\ \hline \text { Event } & B & S & C & \text { Event } & \boldsymbol{L}_b & L_s & \boldsymbol{L}_c \\ \hline \end{array} \end{aligned} $

We label the event that the candidate arrives late be $\boldsymbol{L}$

$ \begin{array}{|l|c|c|c|c|c|c|} \hline \text { Event } & \boldsymbol{B} & \boldsymbol{S} & \boldsymbol{C} & \boldsymbol{L}_b & \boldsymbol{L}_s & \boldsymbol{L}_c \\ \hline \text { Probability } & \frac{2}{5} & \frac{1}{5} & \frac{2}{5} & \frac{1}{5} & \frac{1}{3} & \frac{1}{4} \\ \hline \end{array} $

$\Rightarrow P(L \cap B)=P(L) \cdot P(B / L)=P(B \cap L)$

$\Rightarrow $ $P(B / L)=\frac{P(B \cap L)}{P(L)}$

$=\frac{P(B) \cdot P(L / B)}{P(B \cap L)+P(S \cap L)+P(C \cap L)}$

$=\frac{P(B) \cdot P(L / B)}{P(B) \cdot P(L / B)+P(S) \cdot P(L / S)+P(C) \cdot P(C / B)}$

$P(B / L)=\frac{P(B) \cdot P\left(L_b\right)}{P(B) \cdot P\left(L_b\right)+P(S) \cdot P\left(L_s\right)+P(C) \cdot P\left(L_c\right)}$

$=\frac{\left(\frac{2}{5}\right) \cdot\left(\frac{1}{5}\right)}{\left(\frac{2}{5}\right) \cdot\left(\frac{1}{5}\right)+\left(\frac{1}{5}\right) \cdot\left(\frac{1}{3}\right)+\left(\frac{2}{5}\right) \cdot\left(\frac{1}{4}\right)}$

$=\left(\frac{2}{25}\right) \cdot\left(\frac{300}{24+20+30}\right)=\frac{12}{37}$

We want every pair drawn to have one blue and one green. Since there are 6 blue and 6 green balls, we will draw a total of 6 pairs.

For each pair to be mixed (one of each colour), we must match the 6 blue balls with the 6 green balls in pairs. The number of such “good” pairings is counted by choosing and matching one ball of one colour with one ball of the other colour.

The total number of ways to split 12 balls into 6 pairs (without caring about the order of drawing inside a pair) is counted separately. The required probability is:

Probability $=\frac{\lfloor 6 \times\lfloor 6}{\left(\frac{12}{\left(\lfloor 2)^6\lfloor 6\right.}\right) \times\lfloor 6}=\frac{16}{231}$

So, the probability that each drawn pair consists of one blue and one green ball is $\frac{16}{231}$.

Let $E_1$ be the event of selecting a fair coin.

Let $E_2$ be the event of selecting the two-headed coin.

Let $A$ be the event of getting a 'Head' when the selected coin is tossed.

We are given that there are $N$ fair coins and $1$ two-headed coin.

Hence, the total number of coins in the bag is $N + 1$.

Since a coin is selected at random, the probability of choosing a fair coin is:

$P(E_1) = \frac{N}{N + 1}$

The probability of choosing the two-headed coin is:

$P(E_2) = \frac{1}{N + 1}$

Now, let us find the conditional probabilities of getting a 'Head' for each type of coin:

If a fair coin is selected, the probability of getting a 'Head' is:

$P(A|E_1) = \frac{1}{2}$

If the two-headed coin is selected, the probability of getting a 'Head' is a certainty (since both sides are 'Head'):

$P(A|E_2) = 1$

According to the Law of Total Probability, the total probability of getting a 'Head', denoted as $P(A)$, is given by:

$P(A) = P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)$

We are given that the probability of getting a 'Head' is $\frac{9}{16}$. Let us substitute the known values into the equation:

$\frac{9}{16} = \left(\frac{N}{N + 1} \cdot \frac{1}{2}\right) + \left(\frac{1}{N + 1} \cdot 1\right)$

Let us simplify the equation:

$\frac{9}{16} = \frac{N}{2(N + 1)} + \frac{1}{N + 1}$

To add the terms on the right side, let us take the common denominator as $2(N + 1)$:

$\frac{9}{16} = \frac{N + 2(1)}{2(N + 1)}$

$\frac{9}{16} = \frac{N + 2}{2N + 2}$

Now, we can solve for $N$ by cross-multiplying:

$9 \cdot (2N + 2) = 16 \cdot (N + 2)$

$18N + 18 = 16N + 32$

Transpose the terms containing $N$ to one side and the constant terms to the other side:

$18N - 16N = 32 - 18$

$2N = 14$

Divide both sides by $2$:

$N = 7$

Therefore, the value of $N$ is equal to $7$.

Option B is the correct answer.

Let

• $A$ = event that player A is selected as captain

• $B$ = event that player B is selected as captain

• $W$ = event that the team wins the tournament

Given:

$ P(A)=0.6,\qquad P(B)=0.4 $

Also,

$ P(W|A)=0.8,\qquad P(W|B)=0.7 $

We have to find $P(W)$.

Using the formula of total probability:

$ P(W)=P(A)\,P(W|A)+P(B)\,P(W|B) $

Substituting the values:

$ P(W)=0.6\times 0.8+0.4\times 0.7 $

$ P(W)=0.48+0.28 $

$ P(W)=0.76 $

So, the required probability is

$ \boxed{0.76} $

Hence, the correct option is Option B.

Let

• $A$ = event that the letter came from ANANTPUR

• $K$ = event that the letter came from KANPUR

• $B$ = event that on the envelope, the two visible consecutive letters are AN

We have to find:

$ P(A \mid B) $

Using Bayes’ theorem,

$ P(A \mid B)=\frac{P(A)\,P(B\mid A)}{P(A)\,P(B\mid A)+P(K)\,P(B\mid K)} $

Since the letter is known to have come either from ANANTPUR or KANPUR, and no other information is given, we take

$ P(A)=P(K)=\frac{1}{2} $

So,

$ P(A \mid B)=\frac{P(B\mid A)}{P(B\mid A)+P(B\mid K)} $

Now we find these probabilities.

1. Probability of seeing AN in ANANTPUR

Word: ANANTPUR

Its consecutive pairs are:

$ AN,\ NA,\ AN,\ NT,\ TP,\ PU,\ UR $

There are $7$ consecutive pairs in all, and AN occurs $2$ times.

Therefore,

$ P(B\mid A)=\frac{2}{7} $

2. Probability of seeing AN in KANPUR

Word: KANPUR

Its consecutive pairs are:

$ KA,\ AN,\ NP,\ PU,\ UR $

There are $5$ consecutive pairs in all, and AN occurs $1$ time.

Therefore,

$ P(B\mid K)=\frac{1}{5} $

3. Apply Bayes’ theorem

$ P(A \mid B)=\frac{\frac{2}{7}}{\frac{2}{7}+\frac{1}{5}} $

Take LCM:

$ \frac{2}{7}+\frac{1}{5}=\frac{10+7}{35}=\frac{17}{35} $

So,

$ P(A \mid B)=\frac{\frac{2}{7}}{\frac{17}{35}} =\frac{2}{7}\cdot\frac{35}{17} =\frac{10}{17} $

Therefore, the required probability is

$ \boxed{\frac{10}{17}} $

So, the correct option is B.

$\begin{aligned} & \mathrm{P}=(6 \text { tail })+(4 \text { tail }+1 \text { head })+(2 \text { tail }+2 \text { head })+(3 \text { head }) \\ & \mathrm{P}=\left(\frac{1}{2}\right)^6+\frac{5!}{1!4!}\left(\frac{1}{2}\right)^5+\frac{4!}{2!2!}\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^3 \\ & \mathrm{P}=\frac{1}{64}+\frac{5}{32}+\frac{3}{8}+\frac{1}{8} \\ & \mathrm{P}=\frac{1+10+24+8}{64}=\frac{43}{64}=\frac{\mathrm{m}}{\mathrm{n}} \\ & \mathrm{m}+\mathrm{n}=107\end{aligned}$

$\begin{aligned} & \mathrm{P}(\mathrm{~A})=\frac{7}{10}, \mathrm{P}(\mathrm{~B})=\frac{4}{10} \\ & \mathrm{P}(\mathrm{~A} \cup \overline{\mathrm{~B}})=\frac{5}{10} \\ & \mathrm{P}\left(\frac{\mathrm{~B}}{\mathrm{~A} \cup \overline{\mathrm{~B}}}\right)=\frac{\mathrm{P}(\mathrm{~B} \cap(\mathrm{~A} \cup \overline{\mathrm{~B}}))}{\mathrm{P}(\mathrm{~A} \cup \overline{\mathrm{~B}})} \\ & =\frac{\mathrm{P}((\mathrm{~B} \cap \overline{\mathrm{~B}}) \cup(\mathrm{B} \cap \mathrm{~A}))}{\mathrm{P}(\mathrm{~A} \cup \overline{\mathrm{~B}})}=\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~A} \cup \overline{\mathrm{~B}})} \end{aligned}$

$\begin{aligned} & =\frac{\mathrm{P}(\mathrm{~A})-\mathrm{P}(\mathrm{~A} \cap \overline{\mathrm{~B}})}{\mathrm{P}(\mathrm{~A})+\mathrm{P}(\overline{\mathrm{~B}})-\mathrm{P}(\mathrm{~A} \cap \overline{\mathrm{~B}})}=\frac{\frac{7}{10}-\frac{5}{10}}{\frac{7}{10}+\left(1-\frac{4}{10}\right)-\frac{5}{10}} \\ & =\frac{2}{8}=\frac{1}{4} \end{aligned}$

Define the Events:

Let $A$ be the event that an unbiased coin is drawn.

Let $B$ be the event that the coin with heads on both sides is drawn.

Let $H$ be the event that a head turns up when the coin is tossed.

State the Given Probabilities:

$P(A) = \frac{19}{20}$ (since there are 19 unbiased coins out of 20 total coins)

$P(B) = \frac{1}{20}$ (since there is 1 coin with two heads out of 20 total coins)

$P(H|A) = \frac{1}{2}$ (probability of getting a head given an unbiased coin is drawn)

$P(H|B) = 1$ (probability of getting a head given the coin with two heads is drawn)

Use Bayes' Theorem:

We want to find $P(A|H)$, which is the probability that the drawn coin was unbiased given that a head turned up. Bayes' Theorem states:

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

Calculate $P(H)$:

We can find $P(H)$ using the law of total probability:

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

$P(H) = \left(\frac{1}{2}\right) \cdot \left(\frac{19}{20}\right) + (1) \cdot \left(\frac{1}{20}\right) = \frac{19}{40} + \frac{1}{20} = \frac{19}{40} + \frac{2}{40} = \frac{21}{40}$

Apply Bayes' Theorem to find $P(A|H)$:

$P(A|H) = \frac{P(H|A) \cdot P(A)}{P(H)} = \frac{\left(\frac{1}{2}\right) \cdot \left(\frac{19}{20}\right)}{\frac{21}{40}} = \frac{\frac{19}{40}}{\frac{21}{40}} = \frac{19}{21}$

Determine $m$ and $n$:

Since $P(A|H) = \frac{m}{n}$ and $\gcd(m, n) = 1$, we have $m = 19$ and $n = 21$.

Calculate $n^2 - m^2$:

$n^2 - m^2 = 21^2 - 19^2 = (21 + 19)(21 - 19) = (40)(2) = 80$

Therefore, $n^2 - m^2 = 80$.

So, the correct answer is:

Option B: 80

$\begin{aligned} & 2 p+2 q=\frac{1}{2} \\ & p+q \\ & E\left(x^2\right)=\sum_{i=0}^3 x_i^2 p\left(x_i\right)=0 \cdot p+1 . p+4 \cdot q+9 q \\ & =p+13 q \\ & E(x)=\sum_{i=0}^3 x_i^2 p\left(x_i\right)=0 . p+1 . p+2 q+3 q=p+5 q \\ & p+13 q=2(p+5 q) \\ & p=3 q \\ & \text { So, } q=\frac{1}{8} \& p=\frac{3}{8} \\ & \text { So, } 8 p-1=2 \end{aligned}$

$\begin{array}{lll} 3 E, & 1 D, & 8 P \\ 3 E, & 2 D, & 7 P \\ 4 E, & 1 D, & 7 P \\ 4 E, & 2 D, & 6 P \end{array}$

$\begin{aligned} & P=\frac{{ }^4 C_3 \cdot{ }^2 C_1 \cdot{ }^{10} C_8+{ }^4 C_3 \cdot{ }^2 C_2 \cdot{ }^{10} C_7+{ }^4 C_4 \cdot{ }^2 C_1 \cdot{ }^{10} C_7+{ }^4 C_4 \cdot{ }^2 C_2 \cdot{ }^{10} C_6}{{ }^{10} C_{12}} \\ & =\frac{129}{182} \end{aligned}$

$\begin{aligned} & \mu=\sum X_i P\left(X_i\right)=0+\frac{7}{15}+\frac{2}{15}=\frac{3}{5} \\ & \text { Variance }(X)= \\ & \sum P_i\left(X_i-\mu\right)^2=\frac{7}{15}\left(\frac{9}{25}\right)+\frac{7}{15}\left(\frac{4}{25}\right)+\frac{2}{15}\left(\frac{49}{25}\right)=\frac{28}{75} \end{aligned}$

To find $ P(X \geq 3) $, we first determine the constant $ k $ using the total probability for $ X $.

The probability $ P(X = x) $ is given by:

$ P(X = x) = k(x+1) \cdot 3^{-x}, \quad x = 0, 1, 2, 3, \ldots $

The total probability must equal 1:

$ s = \sum_{x = 0}^{\infty} k(x+1) \cdot 3^{-x} $

Calculating that series:

$ s = \frac{k}{3^0} + \frac{2k}{3} + \frac{3k}{3^2} + \ldots $

Therefore, dividing the series by 3:

$ \frac{s}{3} = \frac{k}{3} + \frac{2k}{3^2} + \ldots $

Subtracting these:

$ s - \frac{s}{3} = k + \frac{k}{3} + \frac{k}{3^2} + \ldots $

The resulting series is a geometric series:

$ \frac{2s}{3} = k \left( 1 + \frac{1}{3} + \frac{1}{3^2} + \ldots \right) $

The sum of the infinite geometric series is:

$ \frac{2s}{3} = k \cdot \frac{1}{1-\frac{1}{3}} = \frac{3k}{2} $

Equating:

$ s = \frac{9k}{4} = 1 $

Thus, solving for $ k $:

$ k = \frac{4}{9} $

Next, compute $ P(X \geq 3) $:

$ P(X \geq 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2)) $

Calculating these:

$ P(X = 0) = k = \frac{4}{9} $

$ P(X = 1) = \frac{2k}{3} = \frac{8}{27} $

$ P(X = 2) = \frac{3k}{9} = \frac{4}{27} $

Adding these probabilities:

$ P(X = 0) + P(X = 1) + P(X = 2) = \frac{4}{9} + \frac{8}{27} + \frac{4}{27} = \frac{8}{9} $

Finally, calculate $ P(X \geq 3) $:

$ P(X \geq 3) = 1 - \frac{8}{9} = \frac{1}{9} $

Probability that a Red ball comes from Bag I ($p$):

$ p(B_1 / R) = \frac{p(B_1) \cdot p(R / B_1)}{p(R)} $

Substituting the values,

$ p(B_1 / R) = \frac{\frac{1}{3} \times \frac{3}{10}}{\frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}} $

Simplify to find $ p $:

$ = \frac{\frac{1}{10}}{\frac{1}{3}(1.2)} = \frac{1}{4} $

So, $ p = \frac{1}{4} $.

Probability that a Green ball comes from Bag III ($q$):

$ p(B_3 / G) = \frac{p(B_3) \cdot p(G / B_3)}{p(G)} $

Substitute the values,

$ p(B_3 / G) = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{5}{10} + \frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10}} $

Simplify to find $ q $:

$ = \frac{\frac{2}{15}}{\frac{1}{3} \times 1.2} = \frac{1}{3} $

So, $ q = \frac{1}{3} $.

Calculation of $\left(\frac{1}{p} + \frac{1}{q}\right)$:

$ \frac{1}{p} = 4, \quad \frac{1}{q} = 3 $

Therefore,

$ \left(\frac{1}{p} + \frac{1}{q}\right) = 4 + 3 = \boxed{7} $

$\begin{aligned} & \text { Bag } 1=\{4 \mathrm{~W}, 5 \mathrm{~B}\} \\ & \text { Bag } \mathbf{2}=\{\mathbf{n W}, \mathbf{3 B}\} \\ & \mathrm{P}\left(\frac{\mathrm{~W}}{\mathrm{Bag} 2}\right)=\frac{29}{45} \\ & \Rightarrow \mathrm{P}\left(\frac{\mathrm{~W}}{\mathrm{~B}_1}\right) \times \mathrm{P}\left(\frac{\mathrm{~W}}{\mathrm{~B}_2}\right)+\mathrm{P}\left(\frac{\mathrm{~B}}{\mathrm{~B}_1}\right) \times \mathrm{P}\left(\frac{\mathrm{~W}}{\mathrm{~B}_2}\right)=\frac{29}{45} \\ & \frac{4}{9} \times \frac{\mathrm{n}+1}{\mathrm{n}+4}+\frac{5}{9} \times \frac{\mathrm{n}}{\mathrm{n}+4}=\frac{29}{45} \end{aligned}$

$\mathrm{n = 6}$

$E_1: B a g B_1$ is selected

$\begin{array}{lll} B_1 & B_2 & B_3 \\ 6 \mathrm{~W} 4 \mathrm{~B} & 4 \mathrm{~W} 6 \mathrm{~B} & 5 \mathrm{~W} 5 \mathrm{~B} \end{array}$

$E_2:$ bag $B_2$ is selected

$E_3: B a g B_3$ is selected

A : Drawn ball is white

We have to find $P\left(\frac{E_2}{A}\right)$

$\begin{aligned} & P\left(\frac{E_2}{A}\right)=\frac{P\left(E_2\right) P\left(\frac{A}{E_2}\right)}{P\left(E_1\right) P\left(\frac{A}{E_1}\right)+P\left(E_2\right) P\left(\frac{A}{E_2}\right)+P\left(E_3\right) P\left(\frac{A}{E_3}\right)} \\ & =\frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{6}{10}+\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}} \\ & =\frac{4}{15} \end{aligned}$

A, E,G R D N

$\text { Probabllity }(\mathrm{P})=\frac{\text { favourable case }}{\text { Total case }}$

(when A & E are in order)

Total case $=6$ !

Favourable case $={ }^6 \mathrm{C}_2 .4$ !

$P=\frac{(15) 4!}{(30) 4!}$

Probablity when not in order $=1-\frac{1}{2}=\frac{1}{2}$

$i^{k_1}+i^{k_2} \neq 0 \quad i^{k_1} \rightarrow 4$ option for $\mathrm{i},-1,-\mathrm{i}, 1$

Total cases $\Rightarrow 4 \times 4=16$

Unfovourble cases $\Rightarrow \mathrm{i}^{\mathrm{k}_1}+\mathrm{i}^{\mathrm{k}_2}=0$

$\left\{\begin{array}{c} 1,-1 \\ -1,1 \\ i,-i \\ -i, i \end{array}\right\}$

4 Cases $\Rightarrow$ Probability $=\frac{16-4}{16}=\frac{3}{4}$

$\begin{aligned} &\text { Probability distribution }\\ &\begin{array}{c|c} \mathrm{x}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} \\ \hline \mathrm{x}=0 & \frac{{ }^7 \mathrm{C}_2}{{ }^{10} \mathrm{C}_2}=\frac{42}{90} \\ \hline \mathrm{x}=1 & \frac{{ }^7 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1}{{ }^{10} \mathrm{C}_2}=\frac{42}{90} \\ \hline \mathrm{x}=2 & \frac{{ }^3 \mathrm{C}_2}{{ }^{10} \mathrm{C}_2}=\frac{6}{90} \end{array} \end{aligned}$

$\begin{aligned} &\text { Now, }\\ &\begin{aligned} & \mu=\sum \mathrm{x}_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=\frac{42}{90}+\frac{12}{90}=\frac{54}{90} \\ & \sigma^2=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_1^2-\mu^2=\frac{42}{90}+\frac{24}{90}-\left(\frac{54}{90}\right)^2 \\ & \Rightarrow \frac{66}{90}-\left(\frac{54}{90}\right)^2 \\ & \sigma^2 \Rightarrow \frac{28}{75} \therefore(1) \end{aligned} \end{aligned}$

C-I $\left|\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right| \rightarrow 4$ ways

C-II $\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| \&\left|\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right| \rightarrow 2$ ways

$\mathrm{P}=\frac{\text { favourable }}{\text { total }}=\frac{6}{16}=\frac{3}{8}$

$\begin{aligned} & \mathrm{p}\left(\mathrm{~S}_5\right)=\frac{1}{9} \\ & \mathrm{p}\left(\mathrm{~S}_5\right)=\frac{5}{36} \\ & \text { required prob }=\frac{1}{9}+\frac{8}{9} \cdot \frac{31}{36} \cdot \frac{1}{9}+\left(\frac{8}{9} \cdot \frac{31}{36}\right)^2 \cdot \frac{1}{9}+\ldots \infty \\ & =\frac{\frac{1}{9}}{1-\frac{62}{81}}=\frac{9}{19} \end{aligned}$

Total ways for selecting any two squares $={ }^{16} \mathrm{C}_2=120$

Total ways for selecting common side squares

$\begin{aligned} &=24\\ &\text { so required probability }\\ &\begin{aligned} & =1-\frac{24}{120} \\ & =\frac{4}{5} \end{aligned} \end{aligned}$

$\mathrm{a}=$ number or dice 1

$\mathrm{b}=$ number on dice 2

$(a, b)=(1,3),(3,1),(2,2),(2,3),(3,2),(1,4),(4,1)$

Required probability

$\begin{aligned} & =\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{1}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{2}{6} \\ & =\frac{18}{36}=\frac{1}{2} \end{aligned}$

To solve this problem, start by considering the equation given for the probabilities:

$ 12x^2 - 7x + 1 = 0 $

The roots of this equation are:

$ x = \frac{1}{3}, \frac{1}{4} $

Assume $ P(A \mid B) = \frac{1}{3} $ and $ P(B \mid A) = \frac{1}{4} $.

From the definitions of conditional probability, we have:

$ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{1}{3} $

$ P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{1}{4} $

Given that $ P(A \cap B) = 0.1 $, we can use these equations to find $ P(B) $ and $ P(A) $:

From $ \frac{0.1}{P(B)} = \frac{1}{3} $, we find $ P(B) = 0.3 $.

From $ \frac{0.1}{P(A)} = \frac{1}{4} $, we find $ P(A) = 0.4 $.

Now, calculate $ P(A \cup B) $ using the formula for the union of two events:

$ P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0.1 = 0.6 $

The task is to find the value of:

$ \frac{P(\overline{A} \cup \overline{B})}{P(\overline{A} \cap \overline{B})} $

Using De Morgan's laws and the complements:

$ P(\overline{A} \cup \overline{B}) = P(\overline{A \cap B}) = 1 - P(A \cap B) = 1 - 0.1 = 0.9 $

$ P(\overline{A} \cap \overline{B}) = 1 - P(A \cup B) = 1 - 0.6 = 0.4 $

Finally, compute the ratio:

$ \frac{P(\overline{A} \cup \overline{B})}{P(\overline{A} \cap \overline{B})} = \frac{0.9}{0.4} = \frac{9}{4} $

In a bag containing 4 white and 6 black balls, two balls are selected one by one without replacement. We need to find the probability that the first ball is black, given that the second ball is black. Let’s define the events:

A: The first ball selected is black.

B: The second ball selected is black.

The probability $ P(A|B) $ is given by the formula:

$ P(A|B) = \frac{P(A \cap B)}{P(B)} $

Let's compute each part separately:

Probability of both balls being black $ P(A \cap B) $:

The chance that the first ball is black is $\frac{6}{10}$. Given that one black ball is removed, the probability that the second is also black is $\frac{5}{9}$.

$ P(A \cap B) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3} $

Probability that the second ball is black $ P(B) $:

This can occur if either the first ball was white or black:

First ball white, second black: $\frac{4}{10} \times \frac{6}{9}$

Both balls black: $\frac{6}{10} \times \frac{5}{9}$

$ P(B) = \frac{4}{10} \times \frac{6}{9} + \frac{6}{10} \times \frac{5}{9} = \frac{24}{90} + \frac{30}{90} = \frac{54}{90} = \frac{3}{5} $

Now, using these probabilities:

$ P(A|B) = \frac{\frac{1}{3}}{\frac{3}{5}} = \frac{1}{3} \times \frac{5}{3} = \frac{5}{9} $

Here, $ \frac{5}{9} $ is in its simplest form ($\operatorname{gcd}(5, 9) = 1$), so $ m = 5$ and $ n = 9 $.

Therefore, $ m + n = 5 + 9 = 14 $.

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} $

Total number of function

$ =5 \times 5 \times 5=125 $

Numbers of one-one function

$ ={ }^5 C_3 \times 3! $

∴ Probability of one-one function

$ =\frac{10 \times 6}{125}=\frac{12}{25} $

$ \begin{aligned} & \text { } P(B)=0.4 \\ & \Rightarrow P(A \cap \bar{B})=P(A)-P(A \cap B)=0.5 \\ & \Rightarrow P(A \cup B)+\frac{P(B \cap A \cup \bar{B})}{P(A \cup \bar{B})}=1.15 \end{aligned} $

$ \begin{aligned} & P(A \cup B)+\frac{P(A \cap B)}{P(A)+1-P(A \cup B)}=1.15 \\ & \text { Let } P(A)=x \\ & P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ & =0.4+0.5=0.9 \\ & \Rightarrow 0.9+\frac{x-0.5}{x+1-0.9}=1.15 \\ & \Rightarrow \frac{x-0.5}{x+0.1}=0.25 \Rightarrow \frac{10 x-5}{10 x+1}=\frac{1}{4} \\ & \Rightarrow 40 x-20=10 x+1 \\ & \Rightarrow 30 x=21 \Rightarrow x=0.7 \end{aligned} $

Probability of ball to be black

$ \begin{aligned} & =\frac{1}{2} \times \frac{{ }^x C_1}{{ }^{10} C_1}+\frac{1}{2} \times \frac{{ }^y C_1}{{ }^{10} C_1} \\ & =\frac{x+y}{20} \end{aligned} $

Probability of ball to be black and drawn from box 2

$ \begin{aligned} & =\frac{\frac{1}{2} \times \frac{{ }^y C_1}{{ }^{10} C_1}}{\frac{x+y}{20}}=\frac{1}{5} \\ \Rightarrow \quad \frac{\frac{y}{20}}{\frac{x+y}{20}} & =\frac{1}{5} \Rightarrow 5 y=x+y \\ \Rightarrow \quad x & =4 y \end{aligned} $

Number of black balls in box 1 will be 4 or 8 .

Here, $n=3$

$ p=\frac{3}{5} $

It is a binomial distribution

$ \begin{aligned} \therefore \text { Mean } & =n p \\ & =3 \times \frac{3}{5}=\frac{9}{5} \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} P(X=k) & =\frac{e^{-\lambda} \cdot \lambda^k}{k!} \\ P(X=5) & =\frac{e^{-\lambda} \cdot \lambda^5}{5!} \\ P(X=2) & =\frac{e^{-\lambda} \cdot \lambda^2}{2!} \\ \frac{P(X=5)}{P(X=2)} & =\frac{\lambda^3}{60}=\frac{1}{7500} \\ \lambda^3 & =\frac{1}{125} \\ \lambda & =\frac{1}{5} \end{aligned} \end{aligned} $

$ \begin{aligned} n(S) & ={ }^{64} C_2=\frac{64 \times 63}{2} \\ n(E) & =\frac{64 \times 63}{2}-2 \times{ }^8 C_1 \times{ }^7 C_1 \\ & =2016-112=1904 \end{aligned} $

$ \begin{aligned} P(E) & =\frac{1904}{2016} \\ & =\frac{17}{18} \end{aligned} $

$ \begin{aligned} P(A \cap \bar{B}) & =P(A)-P(A \cap B) \\ P(\bar{A} \cap B) & =P(B)-P(A \cap B) \\ P(\bar{A} \cap \bar{B}) & =1-P(A \cup B) \\ P(A \cap B) & =P(B)-P(\bar{A} \cap B) \\ & =0.5-0.2=0.3 \\ P(A) & =P(A \cap \bar{B})+P(A \cap B) \\ & =0.1+0.3=0.4 \\ P(A \cup B) & =0.4+0.5-0.3 \\ & =0.6 \\ \Rightarrow \quad P(\bar{A} \cap \bar{B}) & =0.4 \end{aligned} $