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Andrew [12]
3 years ago
13

A bag contains 4 red and 6 green gumballs. if you select 5 of them without looking, how many ways can you get exactly 3 red or e

xactly 2 green gumballs?
Mathematics
1 answer:
marusya05 [52]3 years ago
8 0
This is a combination problem. The approach to this is by using the nCr method, where n is the number of total objects and r is the number of success. For example, if you want to get 3 red out of the total 4, then that would be 4C3. That means,

4C3 = 4!/3!(4 - 3)! = 4 ways

If you want to get 2 green out of the total 6, then that would be 6C2. That means,

6C2 = 6!/2!(6-2)! = 15 ways

Therefore, there are a total of 4+15 = <em>19 ways</em>.
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Step-by-step explanation: In a 52-card deck, you have 4 Queens altogether belonging to each of the four suits. If you pull a card at random out of a complete pack, the probability of picking a Queen would be as follows;

P(Q) = Number of required outcomes/Number of all possible outcomes

P(Q) = 4/52

P(Q) = 1/13

P(Q) = 0.07692 or 7.69%

Also in a 12-pair pack of socks, you have 3 pairs with the color grey. The probability of picking a pair of grey socks at random is given as follows;

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Answer:

Step-by-step explanation:

package net.qiguang.algorithms.C1_Fundamentals.S5_CaseStudyUnionFind;

import java.util.Random;

/**

* 1.5.20 Dynamic growth.

* Using linked lists or a resizing array, develop a weighted quick-union implementation that

* removes the restriction on needing the number of objects ahead of time. Add a method newSite()

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*/

public class Exercise_1_5_20 {

public static class WeightedQuickUnionUF {

private int[] parent; // parent[i] = parent of i

private int[] size; // size[i] = number of sites in subtree rooted at i

private int count; // number of components

int N; // number of items

public WeightedQuickUnionUF() {

N = 0;

count = 0;

parent = new int[4];

size = new int[4];

}

private void resize(int n) {

int[] parentCopy = new int[n];

int[] sizeCopy = new int[n];

for (int i = 0; i < count; i++) {

parentCopy[i] = parent[i];

sizeCopy[i] = size[i];

}

parent = parentCopy;

size = sizeCopy;

}

public int newSite() {

N++;

if (N == parent.length) resize(N * 2);

parent[N - 1] = N - 1;

size[N - 1] = 1;

return N - 1;

}

public int count() {

return count;

}

public int find(int p) {

// Now with path compression

validate(p);

int root = p;

while (root != parent[root]) {

root = parent[root];

}

while (p != root) {

int next = parent[p];

parent[p] = root;

p = next;

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return p;

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// validate that p is a valid index

private void validate(int p) {

if (p < 0 || p >= N) {

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}

public boolean connected(int p, int q) {

return find(p) == find(q);

}

public void union(int p, int q) {

int rootP = find(p);

int rootQ = find(q);

if (rootP == rootQ) {

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}

// make smaller root point to larger one

if (size[rootP] < size[rootQ]) {

parent[rootP] = rootQ;

size[rootQ] += size[rootP];

} else {

parent[rootQ] = rootP;

size[rootP] += size[rootQ];

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count--;

}

}

public static void main(String[] args) {

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Random r = new Random();

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int q = r.nextInt(i+1);

if (uf.connected(p, q)) continue;

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uf.union(r.nextInt(i+1), r.nextInt(i+1));

}

}

}

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