Suppose you flip a fair coin 10 times. Let X be the total number of times that you see the sequence heads-tails (HT). a. What is P(X = 0)?b. What is E(X)? (Hint: think of X as a sum.) | https://leowiki.com

Given:

• Suppose you flip a fair coin {eq}10

  {/eq} times. Let {eq}X

  {/eq} be the total of times that you see the sequence heads-tails {eq}\text{HT}

  {/eq}.

now according to the question, a fair coin is tossed { eq } 10 { /eq } times. so, in a flip of a coin, { equivalent } 10 { /eq } times, we will notice a pair sequence like { equivalent } \text { HT, TH, TT, HH } { /eq } { equivalent } 9 { /eq } times .

now we see that { eq } X { /eq } is the random variable star, which indicates the appearance of the succession ‘HT ‘ in the confuse of a fair coin { equivalent } 10 { /eq } times that is { equivalent } 9 { /eq } sequences .

therefore, we can say that { eq } X { /eq } follows a Binomial Distribution with the probability of success that is probability that { eq } X = HT { /eq } being denoted as { equivalent } p { /eq } and the total act of observations that is the total number of sequences denoted by { equivalent } n { /eq } .

{ equivalent } X \sim Bin\left ( { nitrogen, phosphorus } \right ) { /eq }

nowadays, { equivalent } P\left ( { X = HT } \right ) = p \Rightarrow \dfrac { 1 } { 4 } { /eq } since there is { equivalent } 1 { /eq } opportunity of happening of HT among HT, TH, HH, TT and { eq } nitrogen = 9 { /eq } since, { equivalent } q { /eq } sequences for { equivalent } 10 { /eq } flips. sol ,

{ equivalent } X \sim { \text { Bin } } \left ( { north = 9, p = \dfrac { 1 } { 4 } } \right ) { /eq }

(a)

now, the binomial pdf will be :

{ equivalent } P\left ( { X = x } \right ) = \left ( { \begin { matrix } n\\ ten \end { matrix } } \right ) { p^x } { q^ { nitrogen – adam } } { /eq }

Substitute all the values in { equivalent } P\left ( { X = x } \right ) = \left ( { \begin { matrix } n\\ x \end { matrix } } \right ) { p^x } { q^ { nitrogen – ten } } { /eq } as :

{ equivalent } \begin { align* } P\left ( { X = x } \right ) & = \left ( { \begin { matrix } n\\ ten \end { matrix } } \right ) { p^x } { q^ { newton – ten } } \\ P\left ( { X = 0 } \right ) & = \left ( { \begin { matrix } n\\ 0 \end { matrix } } \right ) { p^0 } { q^ { normality – 0 } } \\ P\left ( { X = 0 } \right ) & = \dfrac { { n ! } } { { 0 ! \left ( { n – 0 } \right ) ! } } 1 \times { q^n } \\ P\left ( { X = 0 } \right ) & = \dfrac { { nitrogen ! } } { { north ! } } \times { q^n } \end { align* } { /eq }

Solving farther as :

{ equivalent } \begin { align* } P\left ( { X = 0 } \right ) & = \dfrac { { normality ! } } { { nitrogen ! } } \times { q^n } \\ P\left ( { X = 0 } \right ) & = { q^n } \\ P\left ( { X = 0 } \right ) & = { \left ( { \dfrac { 3 } { 4 } } \right ) ^9 } \\ P\left ( { X = 0 } \right ) & = 0.0751 \end { align* } { /eq }

therefore, the answer is { equivalent } P\left ( { X = 0 } \right ) = 0.0751 { /eq }.

(b)

For a random variable following Binomial Distribution, we know that :

{ equivalent } E\left ( x \right ) = neptunium { /eq }

immediately substitute the respect of { equivalent } normality = 9 { /eq } and { eq } p = \dfrac { 1 } { 4 } { /eq } in { equivalent } E\left ( x \right ) = neptunium { /eq } as :

{ equivalent } \begin { align* } E\left ( x \right ) & = np\\ E\left ( x \right ) & = 9 \times \dfrac { 1 } { 4 } \\ E\left ( x \right ) & = \dfrac { 9 } { 4 } \end { align* } { /eq }

so, taking the trace into consideration, let ‘s try solving { equivalent } E\left ( x \right ) { /eq } through the previous process is more handy .

Let us denote { eq } { X_i } { /eq } as an index random variable star which indicates HT appears at the ith military position for { equivalent } iodine = 1,2,3, \ldots ,9 { /eq } ( since the coin is flipped { eq } 10 { /eq } times ) that is :

{ equivalent } { X_i } = \left\ { \begin { align* } 1 ; { \text { if } } X = HT { \text { with probability } } = \dfrac { 1 } { 4 } \\ 0 ; { \text { if } } X { \text { is not HT with probability } } = \dfrac { 3 } { 4 } \end { align* } \right. { /eq }

immediately according to hint, { equivalent } X { /eq } is expressed as a sum that is { equivalent } X { /eq } is the sum issue of occurrences of HT that is :

{ equivalent } X = \sum\limits_ { iodine = 1 } ^9 { { X_i } } { /eq }

Since { eq } { X_i } { /eq } indicates happening of HT at { equivalent } { i^ { thorium } } { /eq } positions. therefore :

{ equivalent } \begin { align* } E\left ( X \right ) & = E\left ( { \sum\limits_ { one = 1 } ^9 { { X_i } } } \right ) \\ E\left ( X \right ) & = \sum\limits_ { i = 1 } ^9 { E { X_i } } \\ E\left ( X \right ) & = \sum\limits_ { one = 1 } ^9 { \left [ { E\left ( { { X_i } } \right ) } \right ] } \\ E\left ( X \right ) & = \sum\limits_ { iodine = 1 } ^9 { \left [ { \sum\limits_ { { x_i } } { { x_i } { P_ { { x_i } } } \left ( { { x_i } } \right ) } } \right ] } \end { align* } { /eq }

Solving far as :

{ equivalent } \begin { align* } E\left ( X \right ) & = \sum\limits_ { i = 1 } ^9 { \left [ { \left ( { 1 \times \dfrac { 1 } { 4 } } \right ) + \left ( { 0 \times \dfrac { 3 } { 4 } } \right ) } \right ] } \\ E\left ( X \right ) & = \sum\limits_ { one = 1 } ^9 { \dfrac { 1 } { 4 } } \\ E\left ( X \right ) & = \dfrac { 9 } { 4 } \end { align* } { /eq }

therefore, the answer is { equivalent } P\left ( { X = 0 } \right ) = 0.0751 { /eq } and { eq } E\left ( X \right ) = \dfrac { 9 } { 4 } { /eq } .

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