Probability - Problem 5- Question

</section>
<section style="font-size: 50px";>
<h2 id="probability">Probability</h2>
</section>
<section style="font-size: 50px";>
<h2 id="introduction">Introduction</h2>
<ul>
<li>Probability is a branch of mathematics that deals with the likelihood of events occurring.</li>
<li>It plays a significant role in various fields such as statistics, gambling, and finance.</li>
<li>In this lecture, we will learn about the basic concepts of probability.</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="terminologies">Terminologies</h2>
<ul>
<li><strong>Experiment:</strong> Any activity or process that leads to an outcome is called an experiment.</li>
<li><strong>Outcome:</strong> The result of an experiment is called an outcome.</li>
<li><strong>Sample Space:</strong> The set of all possible outcomes of an experiment is called the sample space.</li>
<li><strong>Event:</strong> Any subset of the sample space is called an event.</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="types-of-probability">Types of Probability</h2>
<ul>
<li><strong>Theoretical Probability:</strong> is the likelihood of an event occurring based on mathematical reasoning.</li>
<li><strong>Experimental Probability:</strong> is the likelihood of an event occurring based on conducting experiments or trials.</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="probability-calculation">Probability Calculation</h2>
<ul>
<li>Probability is expressed as a number between 0 and 1, where 0 represents impossibility and 1 represents certainty.</li>
<li>The probability of an event A is denoted as P(A).</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="probability-formulas">Probability Formulas</h2>
<ul>
<li><strong>Classical Probability Formula:</strong> P(A) = (Number of favorable outcomes) / (Total number of possible outcomes)</li>
<li><strong>Relative Frequency Probability Formula:</strong> P(A) = (Number of times event A occurred) / (Total number of trials or experiments)</li>
<li><strong>Complementary Probability Formula:</strong> P(A’) = 1 - P(A)

</section>
<section style="font-size: 50px";>
<h2 id="independent-and-dependent-events">Independent and Dependent Events</h2>
<ul>
<li><strong>Independent Events:</strong> When the outcome of one event does not affect the outcome of another event, they are called independent events.</li>
<li><strong>Dependent Events:</strong> When the outcome of one event affects the outcome of another event, they are called dependent events.</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="addition-rule-of-probability">Addition Rule of Probability</h2>
<ul>
<li><strong>General Addition Rule:</strong> P(A or B) = P(A) + P(B) - P(A and B)</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="multiplication-rule-of-probability">Multiplication Rule of Probability</h2>
<ul>
<li><strong>General Multiplication Rule:</strong> P(A and B) = P(A) * P(B|A)</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="permutations">Permutations</h2>
<ul>
<li>Permutations are arrangements of objects in a specific order.</li>
<li>The nPermuteR formula is used to calculate permutations: nPr = n! / (n - r)!</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="combinations">Combinations</h2>
<ul>
<li>Combinations are arrangements of objects where the order does not matter.</li>
<li>The nChooseR formula is used to calculate combinations: nCr = n! / [(n - r)! * r!]

</section>
<section style="font-size: 50px";>
<h2 id="conditional-probability">Conditional Probability</h2>
<ul>
<li>Conditional probability is the probability of an event occurring given that another event has already occurred.</li>
<li>The formula for conditional probability is: P(A|B) = P(A and B) / P(B)</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="bayes-theorem">Bayes’ Theorem</h2>
<ul>
<li>Bayes’ Theorem is used to calculate the probability of an event happening based on prior knowledge.</li>
<li>The formula for Bayes’ Theorem is: P(A|B) = [P(B|A) * P(A)] / P(B)</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="probability-distributions">Probability Distributions</h2>
<ul>
<li>A probability distribution is a mathematical function that describes the likelihood of various outcomes in an experiment.</li>
<li>Examples of probability distributions include the binomial distribution, normal distribution, and Poisson distribution.</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="random-variables">Random Variables</h2>
<ul>
<li>A random variable is a variable that takes on different values depending on the outcome of an experiment.</li>
<li>Random variables can be classified as discrete or continuous.

</section>
<section style="font-size: 50px";>
<h2 id="expected-value">Expected Value</h2>
<ul>
<li>The expected value, also known as the mean or average, represents the long-term average outcome of a random variable.</li>
<li>It is calculated by multiplying each possible outcome by its probability and summing them up.</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="variance-and-standard-deviation">Variance and Standard Deviation</h2>
<ul>
<li>Variance measures the spread or dispersion of a random variable’s values.</li>
<li>Standard deviation is the square root of the variance and provides a measure of the average deviation from the mean.</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="normal-distribution">Normal Distribution</h2>
<ul>
<li>The normal distribution is a continuous probability distribution that is symmetric and bell-shaped.</li>
<li>It is widely used in statistical analysis and is often referred to as the Gaussian distribution.</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="central-limit-theorem">Central Limit Theorem</h2>
<ul>
<li>The Central Limit Theorem states that the sampling distribution of the sample mean approaches a normal distribution as the sample size increases.</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="hypothesis-testing">Hypothesis Testing</h2>
<ul>
<li>Hypothesis testing is a statistical method used to make inferences or draw conclusions about a population based on sample data.</li>
<li>It involves formulating null and alternative hypotheses and testing them using statistical tests.

Probability - Problem 5- Question

A bag contains 5 red balls and 3 green balls.
Two balls are drawn from the bag without replacement.
What is the probability that both balls are green?

Probability - Problem 5- Solution

Step 1: Determine the total number of balls in the bag: 8 (5 red + 3 green)
Step 2: Determine the number of ways to choose 2 green balls from 3:
- Combination formula: nCr = n! / [(n - r)! * r!]
- n = 3 (number of green balls), r = 2 (number of balls drawn)
- nCr = 3! / [(3 - 2)! * 2!] = 3
Step 3: Determine the number of ways to choose any 2 balls from 8:
- Combination formula: nCr = n! / [(n - r)! * r!]
- n = 8 (total number of balls), r = 2 (number of balls drawn)
- nCr = 8! / [(8 - 2)! * 2!] = 28
Step 4: Calculate the probability: P(2 green balls) = (number of ways to choose 2 green balls) / (number of ways to choose any 2 balls)
- P(2 green balls) = 3 / 28 = 0.107

Probability - Problem 6- Question

A fair die is rolled 3 times.
What is the probability that the sum of the numbers obtained is 15?

Probability - Problem 6- Solution

Step 1: Determine the total number of outcomes when a die is rolled 3 times: 6^3 = 216
Step 2: Find the number of favorable outcomes where the sum of the numbers is 15:
- Possible combinations: (4,5,6), (5,4,6), (5,5,5), (5,6,4), (6,4,5), (6,5,4), (6,6,3)
- Total number of favorable outcomes: 7
Step 3: Calculate the probability: P(sum of numbers = 15) = (number of favorable outcomes) / (total number of outcomes)
- P(sum of numbers = 15) = 7 / 216 ≈ 0.0324

Probability - Problem 7- Question

A bag contains 5 red balls, 3 green balls, and 2 blue balls.
Three balls are drawn from the bag without replacement.
What is the probability that exactly 2 balls are red?

Probability - Problem 7- Solution

Step 1: Determine the total number of balls in the bag: 10 (5 red + 3 green + 2 blue)
Step 2: Determine the number of ways to choose exactly 2 red balls:
- Combination formula: nCr = n! / [(n - r)! * r!]
- n = 5 (number of red balls), r = 2 (number of red balls drawn)
- nCr = 5! / [(5 - 2)! * 2!] = 10
Step 3: Determine the number of ways to choose any 3 balls from 10:
- Combination formula: nCr = n! / [(n - r)! * r!]
- n = 10 (total number of balls), r = 3 (number of balls drawn)
- nCr = 10! / [(10 - 3)! * 3!] = 120
Step 4: Calculate the probability: P(2 red balls) = (number of ways to choose 2 red balls) / (number of ways to choose any 3 balls)
- P(2 red balls) = 10 / 120 = 0.0833

Probability - Problem 8- Question

In a box, there are 5 red balls, 3 green balls, and 2 blue balls.
Two balls are chosen at random from the box with replacement.
What is the probability that both balls are red?

Probability - Problem 8- Solution

Step 1: Determine the total number of balls in the box: 10 (5 red + 3 green + 2 blue)
Step 2: Calculate the probability of drawing a red ball: P(red ball) = (number of red balls) / (total number of balls)
- P(red ball) = 5 / 10 = 0.5
Step 3: Calculate the probability of drawing two red balls with replacement = P(red ball) * P(red ball)
- P(both balls are red) = 0.5 * 0.5 = 0.25

</section>
<section style="font-size: 50px";>
<h2 id="probability---problem-9--question">Probability - Problem 9- Question</h2>
<ul>
<li>A box contains 2 red balls and 3 green balls.</li>
<li>Four balls are drawn from the box without replacement.</li>
<li>What is the probability that all four balls are green?</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="probability---problem-9--solution">Probability - Problem 9- Solution</h2>
<ul>
<li>Step 1: Determine the total number of balls in the box: 5 (2 red + 3 green)</li>
<li>Step 2: Determine the number of ways to choose all four green balls:
<ul>
<li>Combination formula: nCr = n! / [(n - r)! * r!]</li>
<li>n = 3 (number of green balls), r = 4 (number of balls drawn)</li>
<li>nCr = 3! / [(3 - 4)! * 4!] = 0 (as it is not possible to choose 4 green balls from 3)</li>
</ul>
</li>
<li>Step 3: Calculate the probability: P(all four balls are green) = (number of ways to choose all four green balls) / (total number of ways to choose any four balls)
<ul>
<li>P(all four balls are green) = 0 / (5! / [(5 - 4)! * 4!]) = 0</li>
</ul>
</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="probability---problem-10--question">Probability - Problem 10- Question</h2>
<ul>
<li>A deck of cards contains 52 cards, including 4 aces.</li>
<li>Two cards are drawn from the deck without replacement.</li>
<li>What is the probability that both cards are aces?</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="probability---problem-10--solution">Probability - Problem 10- Solution</h2>
<ul>
<li>Step 1: Determine the total number of cards in the deck: 52</li>
<li>Step 2: Determine the number of ways to choose 2 aces from 4:
<ul>
<li>Combination formula: nCr = n! / [(n - r)! * r!]</li>
<li>n = 4 (number of aces), r = 2 (number of cards drawn)</li>
<li>nCr = 4! / [(4 - 2)! * 2!] = 6</li>
</ul>
</li>
<li>Step 3: Determine the number of ways to choose any 2 cards from 52:
<ul>
<li>Combination formula: nCr = n! / [(n - r)! * r!]</li>
<li>n = 52 (total number of cards), r = 2 (number of cards drawn)</li>
<li>nCr = 52! / [(52 - 2)! * 2!] = 1326</li>
</ul>
</li>
<li>Step 4: Calculate the probability: P(both cards are aces) = (number of ways to choose 2 aces) / (number of ways to choose any 2 cards)
<ul>
<li>P(both cards are aces) = 6 / 1326 ≈ 0.0045</li>
</ul>
</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="probability---problem-11--question">Probability - Problem 11- Question</h2>
<ul>
<li>A bag contains 8 red balls and 4 green balls.</li>
<li>Two balls are drawn from the bag, one by one, without replacement.</li>
<li>What is the probability that the first ball is red and the second ball is green?</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="probability---problem-11--solution">Probability - Problem 11- Solution</h2>
<ul>
<li>Step 1: Determine the total number of balls in the bag: 12 (8 red + 4 green)</li>
<li>Step 2: Calculate the probability of drawing a red ball on the first draw: P(red ball) = (number of red balls) / (total number of balls)
<ul>
<li>P(red ball) = 8 / 12 = 0.6667</li>
</ul>
</li>
<li>Step 3: Calculate the probability of drawing a green ball on the second draw, given that the first ball was red: P(green ball | first ball is red) = (number of green balls) / (total number of balls - 1)
<ul>
<li>P(green ball | first ball is red) = 4 / 11 ≈ 0.3636</li>
</ul>
</li>
<li>Step 4: Calculate the probability of the first ball being red and the second ball being green: P(first ball is red and second ball is green) = P(red ball) * P(green ball | first ball is red)
<ul>
<li>P(first ball is red and second ball is green) = 0.6667 * 0.3636 ≈ 0.2424</li>
</ul>
</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="probability---problem-12--question">Probability - Problem 12- Question</h2>
<ul>
<li>A jar contains 6 red balls and 4 blue balls.</li>
<li>Three balls are drawn from the jar without replacement.</li>
<li>What is the probability that at least two balls are blue?</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="probability---problem-12--solution">Probability - Problem 12- Solution</h2>
<ul>
<li>Step 1: Determine the total number of balls in the jar: 10 (6 red + 4 blue)</li>
<li>Step 2: Determine the number of ways to choose all three blue balls:
<ul>
<li>Combination formula: nCr = n! / [(n - r)! * r!]</li>
<li>n = 4 (number of blue balls), r = 3 (number of balls drawn)</li>
<li>nCr = 4! / [(4 - 3)! * 3!] = 4</li>
</ul>
</li>
<li>Step 3: Determine the number of ways to choose exactly two blue balls:
<ul>
<li>Combination formula: nCr = n! / [(n - r)! * r!]</li>
<li>n = 4 (number of blue balls), r = 2 (number of balls drawn)</li>
<li>nCr = 4! / [(4 - 2)! * 2!] = 6</li>
</ul>
</li>
<li>Step 4: Determine the number of ways to choose any three balls:
<ul>
<li>Combination formula: nCr = n! / [(n - r)! * r!]</li>
<li>n = 10 (total number of balls), r = 3 (number of balls drawn)</li>
<li>nCr = 10! / [(10 - 3)! * 3!] = 120</li>
</ul>
</li>
<li>Step 5: Calculate the probability: P(at least two balls are blue) = (number of ways to choose all three blue balls + number of ways to choose exactly two blue balls) / (number of ways to choose any three balls)
<ul>
<li>P(at least two balls are blue) = (4 + 6) / 120 = 0.0833</li>
</ul>
</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="probability---problem-13--question">Probability - Problem 13- Question</h2>
<ul>
<li>A box contains 5 red balls and 7 green balls.</li>
<li>Three balls are drawn from the box without replacement.</li>
<li>What is the probability that all three balls are the same color?</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="probability---problem-13--solution">Probability - Problem 13- Solution</h2>
<ul>
<li>Step 1: Determine the total number of balls in the box: 12 (5 red + 7 green)</li>
<li>Step 2: Determine the number of ways to choose all three red balls:
<ul>
<li>Combination formula: nCr = n! / [(n - r)! * r!]</li>
<li>n = 5 (number of red balls), r = 3 (number of balls drawn)</li>
<li>nCr = 5! / [(5 - 3)! * 3!] = 10</li>
</ul>
</li>
<li>Step 3: Determine the number of ways to choose all three green balls:
<ul>
<li>Combination formula: nCr = n! / [(n - r)! * r!]</li>
<li>n = 7 (number of green balls), r = 3 (number of balls drawn)</li>
<li>nCr = 7! / [(7 - 3)! * 3!] = 35</li>
</ul>
</li>
<li>Step 4: Determine the number of ways to choose any three balls:
<ul>
<li>Combination formula: nCr = n! / [(n - r)! * r!]</li>
<li>n = 12 (total number of balls), r = 3 (number of balls drawn)</li>
<li>nCr = 12! / [(12 - 3)! * 3!] = 220</li>
</ul>
</li>
<li>Step 5: Calculate the probability: P(all three balls are the same color) = (number of ways to choose all three red balls + number of ways to choose all three green balls) / (number of ways to choose any three balls)
<ul>
<li>P(all three balls are the same color) = (10 + 35) / 220 = 0.25</li>
</ul>
</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="probability---problem-14--question">Probability - Problem 14- Question</h2>
<ul>
<li>A bag contains 4 green balls, 3 red balls, and 5 blue balls.</li>
<li>Three balls are drawn from the bag without replacement.</li>
<li>What is the probability that the first ball is blue, the second ball is red, and the third ball is green?</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="probability---problem-14--solution">Probability - Problem 14- Solution</h2>
<ul>
<li>Step 1: Determine the total number of balls in the bag: 12 (4 green + 3 red + 5 blue)</li>
<li>Step 2: Calculate the probability of drawing a blue ball on the first draw: P(blue ball) = (number of blue balls) / (total number of balls)
<ul>
<li>P(blue ball) = 5 / 12 ≈ 0.4167</li>
</ul>
</li>
<li>Step 3: Calculate the probability of drawing a red ball on the second draw, given that the first ball was blue: P(red ball | first ball is blue) = (number of red balls) / (total number of balls - 1)
<ul>
<li>P(red ball | first ball is blue) = 3 / 11 ≈ 0.</li>
</ul>
</li>
</ul>
</section>

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