Added 25 solutions. Changed folder hierarchy

ThreefLi · Feb 11, 2017 · bcf11f2 · bcf11f2
1 parent 0670409
commit bcf11f2
Show file tree

Hide file tree

Showing 59 changed files with 736 additions and 46 deletions.
diff --git a/... 0 - Mean, Median, and Mode/Solution.java → ... 0 - Mean, Median, and Mode/Solution.java b/... 0 - Mean, Median, and Mode/Solution.java → ... 0 - Mean, Median, and Mode/Solution.java
diff --git a/...stics/Day 0 - Weighted Mean/Solution.java → ...stics/Day 0 - Weighted Mean/Solution.java b/...stics/Day 0 - Weighted Mean/Solution.java → ...stics/Day 0 - Weighted Mean/Solution.java
diff --git a/...Day 1 - Interquartile Range/Solution.java → ...Day 1 - Interquartile Range/Solution.java b/...Day 1 - Interquartile Range/Solution.java → ...Day 1 - Interquartile Range/Solution.java
diff --git a/...tatistics/Day 1 - Quartiles/Solution.java → ...tatistics/Day 1 - Quartiles/Solution.java b/...tatistics/Day 1 - Quartiles/Solution.java → ...tatistics/Day 1 - Quartiles/Solution.java
diff --git a/.../Day 1 - Standard Deviation/Solution.java → .../Day 1 - Standard Deviation/Solution.java b/.../Day 1 - Standard Deviation/Solution.java → .../Day 1 - Standard Deviation/Solution.java
diff --git a/10 Days of Statistics/Day 2 - Basic Probability/Solution.txt b/10 Days of Statistics/Day 2 - Basic Probability/Solution.txt
@@ -0,0 +1,9 @@
+Answer: 5/6
+
+There are 6 possibilities on each die. On 2 dice, there are 6 * 6 = 36 possibilities
+
+There are 6 cases where sum >= 10: (4,6), (5,6), (6,6), (5,5), (5,6), (6,6)
+
+This gives us probability(sum >= 10) = 6/36 = 1/6
+
+That means probability(sum <= 9) = 1 - 1/6 = 5/6
diff --git a/10 Days of Statistics/Day 2 - Compound Event Probability/Solution.txt b/10 Days of Statistics/Day 2 - Compound Event Probability/Solution.txt
@@ -0,0 +1,15 @@
+Answer: 17/42
+
+Urn X has a 4/7 probability of giving a red ball
+Urn Y has a 5/9 probability of giving a red ball
+Urn Z has a 1/2 probability of giving a red ball
+
+Urn X has a 3/7 probability of giving a black ball
+Urn Y has a 4/9 probability of giving a black ball
+Urn Z has a 1/2 probability of giving a black ball
+
+P(2 red, 1 black) = P(Red Red Black) + P(Red Black Red) + P(Black Red Red)
+                  = (4/7)(5/9)(1/2) + (4/7)(4/9)(1/2) + (3/7)(5/9)(1/2)
+                  = 20/126 + 16/126 + 15/126
+                  = 51/126
+                  = 17/42
diff --git a/10 Days of Statistics/Day 2 - More Dice/Solution.txt b/10 Days of Statistics/Day 2 - More Dice/Solution.txt
@@ -0,0 +1,7 @@
+Answer: 1/9
+
+There are 6 possibilities on each die. On 2 dice, there are 6 * 6 = 36 possibilities
+
+There are 4 cases that match the desired criteria: (1,5) (5,1) (2,4) (4,2)
+
+This gives us a probability of 4/36 = 1/9
diff --git a/10 Days of Statistics/Day 3 - Cards of the Same Suit/Solution.txt b/10 Days of Statistics/Day 3 - Cards of the Same Suit/Solution.txt
@@ -0,0 +1,5 @@
+Answer: 12/51
+
+There are 13 of each suit in a deck.
+
+We can view this as 2 separate events. First we draw 1 card and see what suit it is. Whatever suit it is, there are 12 of the same matching suit remaining in the 51-card deck. Therefore, when we draw the 2nd card, there is a 12/51 chance that it is the same suit.
diff --git a/10 Days of Statistics/Day 3 - Conditional Probability/Solution.txt b/10 Days of Statistics/Day 3 - Conditional Probability/Solution.txt
@@ -0,0 +1,29 @@
+Answer: 1/3
+
+****** 2 valid answers: 1/3 or 1/2
+
+Reference: https://en.wikipedia.org/wiki/Boy_or_Girl_paradox
+
+This question is known as the "Boy or Girl Paradox". The actual answer is either 1/3 or 1/2 depending on how the question is interpreted. Here are 2 ways to interpret the question:
+
+1) From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3.
+2) From all families with two children, one child is selected at random, and the sex of that child is specified to be a boy. This would yield an answer of 1/2.
+
+
+******* Interpretation (1) gives an answer of 1/3
+
+There are 4 possible scenarios: (B, B), (B, G), (G, B), (G, G)
+
+We know that 1 child is a boy, so now we have 3 scenarios: (B, B), (B, G), (G, B)
+
+When asked, "what is the probability that both children are boys", the only scenario that matches this is: (B, B). That is, only 1 of the 3 scenarios satisfies the critera, so there is a 1/3 chance that both children are boys.
+
+
+******* Interpretation (2) gives an answer of 1/2
+
+In this scenario, a child was selected at random from a set of all children that have exactly 1 sibling. Knowing the fact that the child is a boy does not give us any information on whether his sibling is a boy or a girl. 
+
+
+******* HackerRank's correct answer
+
+The accepted answer on HackerRank is 1/3. This aligns with the fact that the problem wording more closely resembles Interpretation (1).
diff --git a/10 Days of Statistics/Day 3 - Drawing Marbles/Solution.txt b/10 Days of Statistics/Day 3 - Drawing Marbles/Solution.txt
@@ -0,0 +1,7 @@
+Answer: 2/3
+
+After drawing the first marble, we are left with 2 red marbles and 4 blue marbles. Now we calculate the probability of drawing a blue marble as :
+
+= (# of blue marbles) / (total # of marbles)
+= 4 / (2 + 4)
+= 2 / 3
diff --git a/10 Days of Statistics/Day 4 - Binomial Distribution I/Solution.java b/10 Days of Statistics/Day 4 - Binomial Distribution I/Solution.java
@@ -0,0 +1,41 @@
+public class Solution {
+
+    public static void main(String[] args) {
+        double ratio = 1.09;
+
+        double p = ratio / (1 + ratio);
+        int    n = 6;
+
+        /* Calculate result */
+        double result = 0;
+        for (int x = 3; x <= n; x++) {
+            result += binomial(n, x, p);
+        }
+        System.out.format("%.3f", result);
+    }
+
+    private static Double binomial(int n, int x, double p) {
+        if (p < 0 || p > 1 || n < 0 || x < 0 || x > n) {
+            return null;
+        }
+        return combinations(n, x) * Math.pow(p, x) * Math.pow(1 - p, n - x);
+    }
+
+    private static Long combinations(int n, int x) {
+        if (n < 0 || x < 0 || x > n) {
+            return null;
+        }
+        return factorial(n) / (factorial(x) * factorial(n - x));
+    }
+
+    private static Long factorial (int n) {
+        if (n < 0) {
+            return null;
+        }
+        long result = 1;
+        while (n > 0) {
+            result *= n--;
+        }
+        return result;
+    }
+}
diff --git a/10 Days of Statistics/Day 4 - Binomial Distribution II/Solution.java b/10 Days of Statistics/Day 4 - Binomial Distribution II/Solution.java
@@ -0,0 +1,44 @@
+public class Solution {
+
+    public static void main(String[] args) {
+        /* Values given in problem statement */
+        double p = 0.12;
+        int    n = 10;
+
+        /* "No more than 2 rejects" */
+        double result = 0;
+        for (int x = 0; x <= 2; x++) {
+            result += binomial(n, x, p);
+        }
+        System.out.format("%.3f%n", result);
+
+        /* "At least 2 rejects" */
+        result = 1 - binomial(n, 0, p) - binomial(n, 1, p);
+        System.out.format("%.3f%n", result);
+    }
+
+    private static Double binomial(int n, int x, double p) {
+        if (p < 0 || p > 1 || n < 0 || x < 0 || x > n) {
+            return null;
+        }
+        return combinations(n, x) * Math.pow(p, x) * Math.pow(1 - p, n - x);
+    }
+
+    private static Long combinations(int n, int x) {
+        if (n < 0 || x < 0 || x > n) {
+            return null;
+        }
+        return factorial(n) / (factorial(x) * factorial(n - x));
+    }
+
+    private static Long factorial (int n) {
+        if (n < 0) {
+            return null;
+        }
+        long result = 1;
+        while (n > 0) {
+            result *= n--;
+        }
+        return result;
+    }
+}
diff --git a/10 Days of Statistics/Day 4 - Geometric Distribution I/Solution.java b/10 Days of Statistics/Day 4 - Geometric Distribution I/Solution.java
@@ -0,0 +1,39 @@
+import java.util.Scanner;
+
+public class Solution {
+
+    public static void main(String[] args) {
+        /* Read and save input */
+        Scanner scan = new Scanner(System.in);
+        int numerator   = scan.nextInt();
+        int denominator = scan.nextInt();
+        int n           = scan.nextInt();
+        scan.close();
+
+        /* Geometric Series */
+        double p = (double) numerator / denominator;
+        System.out.format("%.3f", geometric(n, p));
+    }
+
+    private static double geometric (int n, double p) {
+        return Math.pow(1 - p, n - 1) * p;
+    }
+
+    private static Long combinations(int n, int x) {
+        if (n < 0 || x < 0 || x > n) {
+            return null;
+        }
+        return factorial(n) / (factorial(x) * factorial(n - x));
+    }
+
+    private static Long factorial (int n) {
+        if (n < 0) {
+            return null;
+        }
+        long result = 1;
+        while (n > 0) {
+            result *= n--;
+        }
+        return result;
+    }
+}
diff --git a/10 Days of Statistics/Day 4 - Geometric Distribution II/Solution.java b/10 Days of Statistics/Day 4 - Geometric Distribution II/Solution.java
@@ -0,0 +1,43 @@
+import java.util.Scanner;
+
+public class Solution {
+
+    public static void main(String[] args) {
+        /* Read and save input */
+        Scanner scan = new Scanner(System.in);
+        int numerator   = scan.nextInt();
+        int denominator = scan.nextInt();
+        int n           = scan.nextInt();
+        scan.close();
+
+        /* Geometric Series */
+        double p = (double) numerator / denominator;
+        double result = 0;
+        for (int i = 1; i <= 5; i++) {
+            result += geometric(i, p);
+        }
+        System.out.format("%.3f", result);
+    }
+
+    private static double geometric (int n, double p) {
+        return Math.pow(1 - p, n - 1) * p;
+    }
+
+    private static Long combinations(int n, int x) {
+        if (n < 0 || x < 0 || x > n) {
+            return null;
+        }
+        return factorial(n) / (factorial(x) * factorial(n - x));
+    }
+
+    private static Long factorial (int n) {
+        if (n < 0) {
+            return null;
+        }
+        long result = 1;
+        while (n > 0) {
+            result *= n--;
+        }
+        return result;
+    }
+}
diff --git a/10 Days of Statistics/Day 5 - Normal Distribution I/Solution.java b/10 Days of Statistics/Day 5 - Normal Distribution I/Solution.java
@@ -0,0 +1,37 @@
+public class Solution {
+
+    public static void main(String[] args) {
+        double mean = 20;
+        double std  = 2;
+        System.out.format("%.3f%n", cumulative(mean, std, 19.5));
+        System.out.format("%.3f%n", cumulative(mean, std, 22) - cumulative(mean, std, 20));
+    }
+
+    /* Calculates cumulative probability */
+    public static double cumulative(double mean, double std, double x) {
+        double parameter = (x - mean) / (std * Math.sqrt(2));
+        return (0.5) * (1 + erf(parameter));
+    }
+
+    /* Source: http://introcs.cs.princeton.edu/java/21function/ErrorFunction.java.html */
+    // fractional error in math formula less than 1.2 * 10 ^ -7.
+    // although subject to catastrophic cancellation when z in very close to 0
+    // from Chebyshev fitting formula for erf(z) from Numerical Recipes, 6.2
+    public static double erf(double z) {
+        double t = 1.0 / (1.0 + 0.5 * Math.abs(z));
+
+        // use Horner's method
+        double ans = 1 - t * Math.exp( -z*z   -   1.26551223 +
+                                            t * ( 1.00002368 +
+                                            t * ( 0.37409196 + 
+                                            t * ( 0.09678418 + 
+                                            t * (-0.18628806 + 
+                                            t * ( 0.27886807 + 
+                                            t * (-1.13520398 + 
+                                            t * ( 1.48851587 + 
+                                            t * (-0.82215223 + 
+                                            t * ( 0.17087277))))))))));
+        if (z >= 0) return  ans;
+        else        return -ans;
+    }
+}
diff --git a/10 Days of Statistics/Day 5 - Normal Distribution II/Solution.java b/10 Days of Statistics/Day 5 - Normal Distribution II/Solution.java
@@ -0,0 +1,43 @@
+public class Solution {
+
+    public static void main(String[] args) {
+        double mean = 70;
+        double std  = 10;
+
+        double result_1 = 100 * (1 - cumulative(mean, std, 80));
+        double result_2 = 100 * (1 - cumulative(mean, std, 60));
+        double result_3 = 100 * cumulative(mean, std, 60);
+
+        System.out.format("%.2f%n", result_1);
+        System.out.format("%.2f%n", result_2);
+        System.out.format("%.2f%n", result_3);
+    }
+
+    /* Calculates cumulative probability */
+    public static double cumulative(double mean, double std, double x) {
+        double parameter = (x - mean) / (std * Math.sqrt(2));
+        return (0.5) * (1 + erf(parameter));
+    }
+
+    /* Source: http://introcs.cs.princeton.edu/java/21function/ErrorFunction.java.html */
+    // fractional error in math formula less than 1.2 * 10 ^ -7.
+    // although subject to catastrophic cancellation when z in very close to 0
+    // from Chebyshev fitting formula for erf(z) from Numerical Recipes, 6.2
+    public static double erf(double z) {
+        double t = 1.0 / (1.0 + 0.5 * Math.abs(z));
+
+        // use Horner's method
+        double ans = 1 - t * Math.exp( -z*z   -   1.26551223 +
+                                            t * ( 1.00002368 +
+                                            t * ( 0.37409196 + 
+                                            t * ( 0.09678418 + 
+                                            t * (-0.18628806 + 
+                                            t * ( 0.27886807 + 
+                                            t * (-1.13520398 + 
+                                            t * ( 1.48851587 + 
+                                            t * (-0.82215223 + 
+                                            t * ( 0.17087277))))))))));
+        if (z >= 0) return  ans;
+        else        return -ans;
+    }
+}
diff --git a/10 Days of Statistics/Day 5 - Poisson Distribution I/Solution.java b/10 Days of Statistics/Day 5 - Poisson Distribution I/Solution.java
@@ -0,0 +1,29 @@
+import java.util.Scanner;
+
+public class Solution {
+
+    public static void main(String[] args) {
+        /* Read and save input */
+        Scanner scan = new Scanner(System.in);
+        double lambda = scan.nextDouble();
+        int k = scan.nextInt();
+        scan.close();
+
+        System.out.println(poisson(k, lambda));
+    }
+
+    private static double poisson(int k, double lambda) {
+        return (Math.pow(lambda, k) * Math.pow(Math.E, -1 * lambda)) / factorial(k);
+    }
+
+    private static Long factorial (int n) {
+        if (n < 0) {
+            return null;
+        }
+        long result = 1;
+        while (n > 0) {
+            result *= n--;
+        }
+        return result;
+    }
+}
diff --git a/10 Days of Statistics/Day 5 - Poisson Distribution II/Solution.java b/10 Days of Statistics/Day 5 - Poisson Distribution II/Solution.java
@@ -0,0 +1,24 @@
+import java.util.Scanner;
+
+/* Useful Formulas:
+    https://www.hackerrank.com/challenges/s10-poisson-distribution-2/forum/comments/175398
+    https://www.hackerrank.com/challenges/s10-poisson-distribution-2/forum/comments/176962
+*/
+public class Solution {
+
+    public static void main(String[] args) {
+        /* Read and save input */
+        Scanner scan = new Scanner(System.in);
+        double A = scan.nextDouble();
+        double B = scan.nextDouble();
+        scan.close();
+
+        /* E[x^2] = lambda + lambda^2. Plug this into each formula */
+        double dailyCostA = 160 + 40 * (A + (A * A));
+        double dailyCostB = 128 + 40 * (B + (B * B));
+
+        /* Print output */
+        System.out.format("%.3f%n", dailyCostA);
+        System.out.format("%.3f%n", dailyCostB);
+    }
+}