A gin hand consists of 10 cards from a deck of 52 cards. Find the probability that a gin hand has a) all 10 cards of the same suit. b) exactly 4 cards in one suit and 3 in two other suits. c) a 4, 3, 2, 1, distribution of suits.
\[ \text{Ways to choose 10 cards: } {52\choose 10}\\ \text{Ways to choose ten of the same suit: } 4 \cdot {13\choose 10}\\ \frac{4 \cdot {13\choose10}}{52\choose10} \approx 0.000000072 \]
(4*choose(13,10))/choose(52,10)## [1] 7.231342e-08\[ \text{Ways to choose 10 cards: } {52\choose 10}\\ \text{Ways to select exactly 4 cards in one suit and 3 in two other suits: } {13\choose 4}{4\choose 1}{3\choose 2}{13\choose 3}{13\choose 3}\\ \frac{{13\choose 4}{4\choose 1}{3\choose 2}{13\choose 3}{13\choose 3}}{52\choose 10} \approx 0.044362112 \]
(choose(13,4)*choose(4,1)*choose(3,2)*choose(13,3)*choose(13,3))/choose(52,10)## [1] 0.04436211\[ \text{Ways to choose 10 cards: } {52\choose 10}\\ \text{Ways to choose a 4,3,2,1 disribution of suits: } 4 \cdot {13\choose4} \cdot 3 \cdot {13\choose 3} \cdot 2 \cdot {13\choose 2} \cdot 1 \cdot {13\choose 1}\\ \frac{4 \cdot {13\choose4} \cdot 3 \cdot {13\choose 3} \cdot 2 \cdot {13\choose 2} \cdot 1 \cdot {13\choose 1}}{52\choose 10} \approx 0.3145677 \]
(4*choose(13,4)*3*choose(13,3)*2*choose(13,2)*1*choose(13,1))/choose(52,10)## [1] 0.3145677Assume that the random variables X and Y have the joint distribution given in Table 4.6.
WhatisP(X≥1andY ≤0)?
What is the conditional probability that Y ≤ 0 given that X = 2?
Are X and Y independent?
What is the distribution of Z = XY ?
I summarized the joint distribution below. To see the original table, check page 156 for Table 4.6.
library(dplyr)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionxy_df <- data.frame(
X = as.integer(c(-1, -1, -1, -1, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2)),
Y = as.integer(c(-1, 0, 1, 2, -1, 0, 1, 2, -1, 0, 1, 2, -1, 0, 1, 2)),
Probability = as.numeric(c(0, 1/18, 0, 1/12, 1/36, 0, 1/36, 0, 1/6, 1/18, 1/6, 1/12, 1/12, 0, 1/12, 1/6)))
print(xy_df)## X Y Probability
## 1 -1 -1 0.00000000
## 2 -1 0 0.05555556
## 3 -1 1 0.00000000
## 4 -1 2 0.08333333
## 5 0 -1 0.02777778
## 6 0 0 0.00000000
## 7 0 1 0.02777778
## 8 0 2 0.00000000
## 9 1 -1 0.16666667
## 10 1 0 0.05555556
## 11 1 1 0.16666667
## 12 1 2 0.08333333
## 13 2 -1 0.08333333
## 14 2 0 0.00000000
## 15 2 1 0.08333333
## 16 2 2 0.16666667\[ P(X\geq 1 \text{&} Y \leq 0) = 0 + 1/6 + 1/12 + 1/18 = 11/36 \approx0.30555 \]
prob_a <- xy_df %>%
filter(X>=1 & Y<=0)
print(prob_a)## X Y Probability
## 1 1 -1 0.16666667
## 2 1 0 0.05555556
## 3 2 -1 0.08333333
## 4 2 0 0.00000000sum(prob_a$Probability)## [1] 0.3055556\[ P(Y\leq 0 | X = 2) = \frac{P(Y\leq 2 \cap X=2)}{P(X=2)}\\ =\frac{1/12 + 0}{1/12 + 0 + 1/12 + 1/6} = \frac{1/12}{1/3} = \frac{1}{4} \]
# P(Y\leq2 \cap X =2) = b
b <- xy_df %>%
filter(Y<=0 & X==2)
prob_b <- sum(b$Probability)
# P(X=2) =b2
b2 <- xy_df %>%
filter(X==2)
prob_b2 <- sum(b2$Probability)
#solution
prob_b/prob_b2## [1] 0.25In order to show that X and Y are independent, we need to prove that for each X and Y, \(P(X=x \cap Y=y) = P(X=x) \cdot P(Y=y)\).
First we will look at \(P(X=-1 \cap Y=-1)\):
\[ (X=-1 \cap Y=-1) = (1/36 + 1/6 + 1/12) \cdot (1/18 + 1/12) = 5/18 \cdot 1/216 \neq 0 \]
c_x <- xy_df %>%
filter(X==-1)
prob_c_x <- sum(c_x$Probability)
print(prob_c_x)## [1] 0.1388889c_y <- xy_df %>%
filter(Y==-1)
prob_c_y <- sum(c_y$Probability)
print(prob_c_y)## [1] 0.2777778# test to see if equal
prob_c_y == prob_c_x## [1] FALSEWe can already confirm that X and Y are not independent because \(P(X=-1 \cap Y=-1)\) is not equal to 0.
# Create a dataframe with the given values
xy_df_z <- xy_df %>%
mutate(
Probability = as.numeric(c(0, 1/18, 0, 1/12, 1/36, 0, 1/36, 0, 1/6, 1/18, 1/6, 1/12, 1/12, 0, 1/12, 1/6)),
Z = X * Y)
# Print the dataframe
print(xy_df_z)## X Y Probability Z
## 1 -1 -1 0.00000000 1
## 2 -1 0 0.05555556 0
## 3 -1 1 0.00000000 -1
## 4 -1 2 0.08333333 -2
## 5 0 -1 0.02777778 0
## 6 0 0 0.00000000 0
## 7 0 1 0.02777778 0
## 8 0 2 0.00000000 0
## 9 1 -1 0.16666667 -1
## 10 1 0 0.05555556 0
## 11 1 1 0.16666667 1
## 12 1 2 0.08333333 2
## 13 2 -1 0.08333333 -2
## 14 2 0 0.00000000 0
## 15 2 1 0.08333333 2
## 16 2 2 0.16666667 4# find the Z=XY distribution;
Z_df <- xy_df_z %>%
select(Z, Probability) %>%
group_by(Z) %>%
summarise(Total_Probability = sum(as.numeric(Probability))) %>%
as.data.frame()
# Z=XY distribution
print(Z_df)## Z Total_Probability
## 1 -2 0.1666667
## 2 -1 0.1666667
## 3 0 0.1666667
## 4 1 0.1666667
## 5 2 0.1666667
## 6 4 0.1666667