Respuesta :
This is a discrete distribution with three possible cases of interest. In any such two-card hand, you can have (1) no cards, (2) one card, or (3) both cards with the heart suit.
[tex]\mathbb P(X=x)=\begin{cases}\dfrac{\dbinom{39}2}{\dbinom{52}2}=\dfrac{19}{34}&\text{for }x=0\\\\\dfrac{\dbinom{39}1\dbinom{13}1}{\dbinom{52}2}=\dfrac{13}{34}&\text{for }x=1\\\\\dfrac{\dbinom{13}2}{\dbinom{52}2}=\dfrac1{17}&\text{for }x=2\end{cases}[/tex]
As a PMF, you would say
[tex]f_X(x)=\begin{cases}\dfrac{19}{34}&\text{for }x=0\\\\\dfrac{13}{34}&\text{for }x=1\\\\\dfrac1{17}&\text{for }x=2\\\\0&\text{otherwise}\end{cases}[/tex]
The CDF is given by
[tex]\displaystyle F_X(x)=\begin{cases}0&\text{for }x<0\\\\\dfrac{19}{34}&\text{for }0\le x<1\\\\\dfrac{32}{34}&\text{for }1\le x<2\\\\1&\text{for }x\ge2\end{cases}[/tex]
The most likely value of [tex]X[/tex] is the expected value for the distribution, which is given by
[tex]\displaystyle\mathbb E(X)=\sum_xxf_X(x)=0\times f_X(0)+1\times f_X(1)+2\times f_X(2)=\dfrac{17}{34}=\dfrac12[/tex]
[tex]\mathbb P(X=x)=\begin{cases}\dfrac{\dbinom{39}2}{\dbinom{52}2}=\dfrac{19}{34}&\text{for }x=0\\\\\dfrac{\dbinom{39}1\dbinom{13}1}{\dbinom{52}2}=\dfrac{13}{34}&\text{for }x=1\\\\\dfrac{\dbinom{13}2}{\dbinom{52}2}=\dfrac1{17}&\text{for }x=2\end{cases}[/tex]
As a PMF, you would say
[tex]f_X(x)=\begin{cases}\dfrac{19}{34}&\text{for }x=0\\\\\dfrac{13}{34}&\text{for }x=1\\\\\dfrac1{17}&\text{for }x=2\\\\0&\text{otherwise}\end{cases}[/tex]
The CDF is given by
[tex]\displaystyle F_X(x)=\begin{cases}0&\text{for }x<0\\\\\dfrac{19}{34}&\text{for }0\le x<1\\\\\dfrac{32}{34}&\text{for }1\le x<2\\\\1&\text{for }x\ge2\end{cases}[/tex]
The most likely value of [tex]X[/tex] is the expected value for the distribution, which is given by
[tex]\displaystyle\mathbb E(X)=\sum_xxf_X(x)=0\times f_X(0)+1\times f_X(1)+2\times f_X(2)=\dfrac{17}{34}=\dfrac12[/tex]
Probability mass function is probability function evaluated for discrete distributions. The pmf, cdf, and expectation of random variable X are:
- PMF(probability mass function):
[tex]P(X = 0) = P(B)P(B) =\dfrac{1}{16}\\\\P(X = 1) = 2P(A)P(B) =\dfrac{6}{16}\\\\P(X = 2) = P(A)P(A) =\dfrac{9}{16}[/tex]
- (and
- probability
- is 0 for any other value of X)
- CDF(Cumulative distributive function):For X < 0, F(X) = 0
- [tex]F(X = 0) = P(X \leq 0) = P(X = 0) = \dfrac{1}{16}\\\\F(X = 1) = P(X \leq 1) = P(X = 0) + P(X = 1) = \dfrac{7}{16}\\\\F(X = 2) = P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = \dfrac{16}{16} = 1[/tex]
- For X > 2, F(X) = 1
- Most likely value of random variable X:
What is the most likely value of a random variable?
It is its expected value, denoted by E(X).
How to calculate the expectation of a discrete random variable?
Expectation can be taken as a weighted mean, weights being the probability of occurrence of that specific observation.
Thus, if the random variable is X, and its probability mass function is given to be f(x) = P(X = x), then we have:
[tex]E(X) = \sum_{i=1}^n( f(x_i) \times x_i)[/tex]
(n is number of values X takes)
What is probability mass function of a random variable X?
PMF (probability mass function) is a function describing probabilities for each value of discrete random variable X.
What is cumulative distribution function?
Suppose that for a random variable X, its probability function be f(x).
Then we have:
[tex]CDF = F(x) = f(X\leq x)[/tex]
For the given case, X is tracking the number of cards drawn belonging to heart suit.
Since there are only 2 cards drawn, thus, there can be at max 2 heart suit cards and at minimum, can be 0 heart suit cards.
Let two events be:
A = drawing a card belonging to heart suit
B = drawing a card not belonging to heart suit
Thus,
[tex]P(A) = \dfrac{^{13}C_1}{^{52}C_1} = \dfrac{13}{52} = \dfrac{1}{4}[/tex]
[tex]P(B) = \dfrac{^{39}C_1}{^{52}{C_1}} = \dfrac{39}{52} = \dfrac{3}{4}[/tex]
Then, as we're doing replacement, thus, using chain rule, we get:
[tex]P(X = 0) = P(B)P(B) =\dfrac{1}{16}\\\\P(X = 1) = 2P(A)P(B) =\dfrac{6}{16}\\\\P(X = 2) = P(A)P(A) =\dfrac{9}{16}[/tex]
(and probability is 0 for any other value of X)
P(X = 1) = 2P(A)P(B) since either first card can be heart or second can be heart. Thus, 2 orders.
For cumulative distribution function, we get:
For X < 0, F(X) = 0
[tex]F(X = 0) = P(X \leq 0) = P(X = 0) = \dfrac{1}{16}\\\\F(X = 1) = P(X \leq 1) = P(X = 0) + P(X = 1) = \dfrac{7}{16}\\\\F(X = 2) = P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = \dfrac{16}{16} = 1[/tex]
For X > 2, F(X) = 1
For getting most likely value of X, we get:
[tex]E(X) = 0.P(X=0) + 1.P(X=1) + 2.P(X=2) = \dfrac{6}{16} + \dfrac{18}{16} = 1.5[/tex]
Learn more about cumulative distribution function here:
https://brainly.com/question/19884447
