Let’s consider E to be the event of getting even number on tossing a die.

Let’s consider X to be the random variable such that X = \[0,1,2\] Now, let E = the event of drawing an ace And, F = the event of drawing non – ace So,

Given, n coins are two headed coins and the remaining \[(n+1)\] coins are fair. Let  \[{{E}_{1}}\] : the event that unfair coin is selected \[{{E}_{2}}\] : the event that the fair coin is selected...

The probability distribution of random variable X is given by: We know that \[\sum\limits_{i=1}^{n}{P({{X}_{i}})=1}\] So, \[k\text{ }+\text{ }4k\text{ }+\text{ }9k\text{ }+\text{ }8k\text{ }+\text{...

(i) We know that: (ii) Now, the distribution becomes \[E({{X}^{2}})\text{ }=\text{ }1\text{ }\times \text{ 1/2 }+\text{ }4\text{ }\times \text{ }1/5\text{ }+\text{ }16\text{ }\times 3/25\text{...

(i) Given, \[P\left( X\text{ }=\text{ }x \right)\text{ }=\text{ }k\left( x\text{ }+\text{ }1 \right)\]for \[x\text{ }=\text{ }1,\text{ }2,\text{ }3,\text{ }4\] So, \[P\left( X\text{ }=\text{ }1...

Let’s consider: \[{{E}_{1}}\] = The event that the item is manufactured on machine A \[{{E}_{2}}\] = The event that the item is manufactured on machine B \[{{E}_{3}}\] = The event that the item is...

Let \[{{E}_{1}}\] = Event that a person has TB \[{{E}_{2}}\] = Event that a person does not have TB And H = Event that the person is diagnosed to have TB. So, \[P({{E}_{1}})\text{ }=\text{...

Given, we have \[3\] urns: Urn \[1\] = \[2\] white and \[3\] black balls Urn \[2\] = \[3\] white and 2 black balls Urn \[3\] = \[4\] white and \[1\] black balls Now, the probabilities of choosing...

Let \[{{E}_{1}}\] be the event of selecting Bag \[1\] and \[{{E}_{2}}\] be the event of selecting Bag \[2\]. Also, let \[{{E}_{3}}\] be the event that black ball is selected Now,...

Let E1 be the event that the letter comes from TATA NAGAR, E2 be the event that the letter comes from CALCUTTA And, E3 be the event that on the letter, two consecutive letters TA are visible Now,...

Given that: \[{{A}_{1}}:\text{ }{{A}_{2}}:\text{ }{{A}_{3}}~=\text{ }4:\text{ }4:\text{ }2\] So, the probabilities will be \[P({{A}_{1}})\text{ }=\text{ }4/10,\text{ }P({{A}_{2}})\text{ }=\text{...

Referring the Question 41, here we will use Baye’s Theorem Therefore, the required probabilities are \[2/11\] and \[9/11\].

Given: Bag \[1:3\] red balls, Bag \[2:2\] red balls and \[1\] white ball Bag \[3:3\]  white balls Now, let E1, E2 and E3 be the events of choosing Bag \[1\], Bag \[2\] and Bag \[3\] respectively and...

Let’s consider A to be the event of having m white and n black balls \[{{E}_{1}}\] = First ball drawn of white colour \[{{E}_{2}}\] = First ball drawn of black colour \[{{E}_{3}}\]  = Second ball...

Given, two events A and B are independent such that \[\mathbf{A}\text{ }=\text{ }\left\{ \left( x,~y \right)\text{ }:~x~+~y~=\text{ }\mathbf{11} \right\}\text{ }\mathbf{B}\text{ }=\text{ }\left\{...

Solution: Let’s take A1 to be the event of getting a total of \[6\] \[{{A}_{1}}~=\text{ }\left\{ \left( 2,\text{ }4 \right),\text{ }\left( 4,\text{ }2 \right),\text{ }\left( 1,\text{ }5...

We know that, Variance(X) = \[E({{X}^{2}})\text{ }\text{ }{{\left[ E\left( X \right) \right]}^{2}}\] Therefore, the required variance is \[665/324\].

Given, \[X\text{ }=\text{ }0,\text{ }1,\text{ }2\]and \[P\left( X\text{ }=\text{ }0 \right)\text{ }=\text{ }P\text{ }\left( X\text{ }=\text{ }1 \right)\text{ }=~p\] Let P(X) at \[X\text{ }=\text{...

Let X be the random variable scores when a die is thrown twice. \[X\text{ }=\text{ }1,\text{ }2,\text{ }3,\text{ }4,\text{ }5,\text{ }6\] And \[S\text{ }=\text{ }\left\{ \left( 1,\text{ }1...

Given, \[\mathbf{S}=\left\{ \mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{3},\text{ }\ldots .,~n \right\}\] So, \[P\left( r\text{ }\le \text{ }p/s\text{ }\le \text{ }p \right)\text{ }=\text{...

Let’s assume \[{{E}_{1}}\] = The event that a person selected is of blood group O \[{{E}_{2}}\] = The event that the people selected is of other group And H = The event that selected person is left...

Let’s consider \[{{E}_{1}}\] = Event that the coin is fair \[{{E}_{2}}\] = Event that the coin is \[2\] headed And H = Event that the tossed coin gets head. Now, \[P({{E}_{1}})\text{ }=\text{...

Let’s assume X to be the random variable denoting a bulb to be defective. Here, \[n\text{ }=\text{ }10,\text{ }p\text{ }=\text{ }1/50,\text{ }q\text{ }=\text{ }1\text{ }\text{ }1/50\text{ }=\text{...

The probability distribution of random variable X is We know that, \[E(X)=\sum\limits_{i=1}^{n}{{{P}_{i}}{{X}_{i}}}\] \[E\left( X \right)\text{ }0.\text{ }1/5\text{ }+\text{ }1.\text{ }2/5\text{...

Therefore, the required probability distribution is

Here, we have \[X\text{ }=\text{ }0,\text{ }1,\text{ }2,\text{ }3\] [Since, die is thrown \[3\] times] And \[p\text{ }=\text{ }1/6,\text{ }q\text{ }=\text{ }5/6\] Therefore, the required expression...

Here, random variable \[X\text{ }=\text{ }0,\text{ }1,\text{ }2\] \[P\left( X\text{ }=\text{ }2 \right)\text{ }=\text{ }P\left( 4 \right).P\left( 4 \right)\text{ }=\text{ }1/10\text{ }x\text{...

We know that: Standard deviation (S.D.) = \[\sqrt{Variance}\] So, \[Var\left( X \right)\text{ }=\text{ }E({{X}^{2}})\text{ }\text{ }{{\left[ E\left( X \right) \right]}^{2}}\] \[E\left( X...

Given We know that: \[\mathbf{Var}\left( \mathbf{X} \right)\text{ }=\text{ }\mathbf{E}({{\mathbf{X}}^{\mathbf{2}}})\text{ }\text{ }{{\left[ \mathbf{E}\left( \mathbf{X} \right)...

Given: Total number of watches = \[100\] and number of defective watches = \[10\] So, the probability of selecting a detective watch = \[10/100\text{ }=\text{ }1/10\] Now, n = \[8\], \[p\text{...

Here, we have n = \[7\], \[p\text{ }=\text{ }0.25\text{ }=\text{ }25/100\text{ }=\text{ 1/4}\] and \[q=1-1/4=3/4\] \[P\left( X\text{ }\ge \text{ }2 \right)\text{ }=\text{ }1\text{ }\text{ }\left[...

Here we have, \[n=10\], \[p=1/2\] and \[q=1-1/2=1/2\] \[P\left( X\text{ }\ge \text{ }8 \right)\text{ }=\text{ }P\left( x\text{ }=\text{ }8 \right)\text{ }+\text{ }P\left( x\text{ }=\text{ }9...

Here, \[p\text{ }=\text{ }1/6\text{ }+\text{ }1/6\text{ }+\text{ }1/6\text{ }=\text{ 1/2}\Rightarrow q\text{ }=\text{ }1\text{ }\text{ 1/2 }=\text{ 1/2}\] and \[n=5\] Now, \[P\left( x\text{ }=\text{...

Let \[{{E}_{1}},{{E}_{2}},{{E}_{3}}\] and \[{{E}_{4}}\] be the events that first, second, third and fourth card is King respectively. Therefore, the required probability is \[1/27075\].

Given that the box has \[5\] blue and \[4\] red balls. Let us consider \[{{E}_{1}}\] be the event that first ball drawn is blue and \[{{E}_{2}}\] be the event that first ball drawn is red. And, E is...

According to the question: Bag \[1\] has \[3\]B, \[2\]W balls and Bag \[2\] has \[2\]B, \[4\]W balls. Let \[{{E}_{1}}\] = The event that bag \[1\] is selected \[{{E}_{2}}\] = The event that bag...

Let us consider  \[{{W}_{1}}\] and \[{{W}_{2}}\] to be two bags containing \[\left( 4W,\text{ }5B \right)\]and \[\left( 9W,\text{ }7B \right)\]balls respectively. Let us take \[{{E}_{1}}\] be the...

  Let’s take X to be the random variable where X = \[0,500,2000\]and \[3000\]

Given that the dice is thrown three times So, the sample space n(S) = \[{{6}^{3}}~=\text{ }216\] Let E1 be the event when the sum of number on the dice was \[6\] and \[{{E}_{2}}\]be the event when...

Let’s take X to be the random variable of profit per throw. As, she loses Rs \[1\] for giving any od \[2,4,5\]. So, \[P\left( X\text{ }=\text{ }-1 \right)\text{ }=\text{ }1/6\text{ }+\text{...

For a probability distribution, we know that if \[{{P}_{i}}~\ge \text{ }0\]  

Here, \[p({{\mathbf{E}}_{\mathbf{1}}})\text{ }=~{{p}_{\mathbf{1}}}~\] and \[\mathbf{P}({{\mathbf{E}}_{\mathbf{2}}})\text{ }=\text{ }{{\mathbf{p}}_{\mathbf{2}}}\] Now, its clearly seen that either...

Given, P(A) = \[2/5\], P(B) = \[1/3\] and P(C) = \[1/2\] \[\mathbf{P}\left( \mathbf{A}\text{ }\mathbf{C} \right)\text{ }=\text{ }\mathbf{1}/\mathbf{5}\]and \[\mathbf{P}\left( \mathbf{B}\text{...

Given, P(A) = \[1/2\], P(B) = \[1/3\] and \[\mathbf{P}\left( \mathbf{A}\text{ }\mathbf{}\text{ }\mathbf{B} \right)\text{ }=\text{ }\mathbf{1}/\mathbf{4}\] \[P\left( A \right)\text{ }=\text{ }1\text{...

As the random variable X takes \[0,1,2,3,......n\] is said to be binomial distribution having parameters n and p, if the probability is given by P(X = r) = \[^{n}{{C}_{r~}}{{p}^{r~}}{{q}^{n-r}}\]...

If two dice are thrown together, we have n(S) = \[36\] Now, let’s consider: E = A total of \[4\text{ }=\text{ }\left\{ \left( 2,\text{ }2 \right),\text{ }\left( 1,\text{ }3 \right),\text{ }\left(...

Let red marbles be presented with R and black marble with B. Also, let E be the event that at least one of the three marbles drawn be black when the first marble is red. Now, the following three...

W.k.t, \[A\cup B\] denotes that atleast one of the events occurs and \[A\cap B\] denotes that two events occur simultaneously. Therefore, the required answer is \[1.1\].

According to the solution of exercise 1, we have \[A\text{ }=\text{ }\left\{ \left( 1,\text{ }1 \right),\text{ }\left( 2,\text{ }2 \right),\text{ }\left( 3,\text{ }3 \right),\text{ }\left( 4,\text{...

Given that a loaded die is thrown such that \[\mathbf{P}\left( \mathbf{1} \right)\text{ }=\text{ }\mathbf{P}\left( \mathbf{2} \right)\text{ }=\text{ }\mathbf{0}.\mathbf{2},\text{ }\mathbf{P}\left(...