Free Challenging Questions : M0-16-10

Challenging Questions Mathematics – Compulsory Part Learning Unit: More about probability Question M0-16-10 A bag contains 2 white balls (W) and 2 red balls (R). # More challenging questions will be updated # Answer

Free Challenging Questions : M0-16-09

Challenging Questions Mathematics – Compulsory Part Learning Unit: More about probability Question M0-16-09 In a school, \(\frac{1}{8}\) of the students travel by bus, \(\frac{3}{8}\) by car, and the remaining students walk to school. The probabilities of being late are as follows: # More challenging questions will be updated # Answer

Free Challenging Questions : M0-16-08

Challenging Questions Mathematics – Compulsory Part Learning Unit: More about probability Question M0-16-08 We have n bags with the following compositions: A ball is drawn randomly from bag 1 and placed in bag 2. Then, a ball is drawn randomly from bag 2 and placed in bag 3. This process continues until a ball is... » read more

Free Challenging Questions : M0-16-07

Challenging Questions Mathematics – Compulsory Part Learning Unit: More about probability Question M0-16-07 Ben and Ron play a badminton match consisting of multiple rounds. In each round, the probability of Ben winning is \(0.6\). # More challenging questions will be updated # Answer

Free Challenging Questions : M0-16-06

Challenging Questions Mathematics – Compulsory Part Learning Unit: More about probability Question M0-16-06 Consider \(r\) distinct balls distributed uniformly at random into \(n\) distinct boxes, where \(n>r\). Calculate the probability that at least one box contains two or more balls. # More challenging questions will be updated # Answer

Free Challenging Questions : M0-16-05

Challenging Questions Mathematics – Compulsory Part Learning Unit: More about probability Question M0-16-05 Given five line segments with lengths 1, 2, 3, 4, and 5 units, three segments are selected at random. What is the probability that these three segments can form a triangle? # More challenging questions will be updated # Answer

Free Challenging Questions : M0-16-04

Challenging Questions Mathematics – Compulsory Part Learning Unit: More about probability Question M0-16-04 Two people, Ben and Ron, take turns rolling a fair six-sided die. Ben rolls first, and Ron rolls second. The first player to roll a one (1 dot) wins the game. # More challenging questions will be updated # Answer

Free Challenging Questions : M0-16-03

Challenging Questions Mathematics – Compulsory Part Learning Unit: More about probability Question M0-16-03 In a lucky draw, there are 24 tickets. The tickets are drawn at random, one by one, without replacement. The person who draws a ticket labeled X, Y, or Z wins a prize. Please express your answers as fractions. # More challenging... » read more

Free Challenging Questions : M0-16-02

Challenging Questions Mathematics – Compulsory Part Learning Unit: More about probability Question M0-16-02 There are \(50\) teachers in a school. Among them, \(30\) teachers are male, while \(20\) teachers are female. In addition, \(8\) English teachers are male and \(5\) English teachers are female. A school committee of \(4\) teachers is formed by selecting them... » read more

Free Challenging Questions : M0-16-01

Challenging Questions Mathematics – Compulsory Part Learning Unit: More about probability Question M0-16-01 There are \(200\) students in a school. Among them, \(84\) students are members of the Biology Club, while \(108\) students are members of the Chemistry Club. In addition, \(56\) students are members of both clubs. \(U\) is the set of all students... » read more

Expected Number of Distinct Balls Drawn with Replacement

Problem Statement:A box contains 5 balls labeled from 1 to 5. If three balls are drawn with replacement, let X be the number of distinct balls that are drawn at least once in the three draws. Find E(X). Answer: \(P(X=3)=\frac{5}{5}\times\frac{4}{5}\times\frac{3}{5}=\frac{12}{25}\) \(P(X=2)=\frac{5}{5}\times\frac{1}{5}\times\frac{4}{5}\times\text{C}^3_2=\frac{12}{25}\) \(P(X=1)=\frac{5}{5}\times\frac{1}{5}\times\frac{1}{5}=\frac{1}{25}\) \(E(X)=3 \times P(X=3)+2 \times P(X=2)+1 \times P(X=1)\) \(=3\times\frac{12}{25}+2\times\frac{12}{25}+1\times\frac{1}{25}\) \(=\frac{61}{25}\) There are approximately 2.44... » read more

Understanding Independence for Three Events: Going Beyond Two

Introduction You’ve already learned about probability and how two events can be independent. Recall that two events, A and B, are independent if knowing that one occurs doesn’t change the probability of the other occurring. Mathematically, this is written as: P(A and B) = P(A) × P(B) But what happens when we have three events? Is independence... » read more