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372 changes: 372 additions & 0 deletions .ipynb_checkpoints/lab-intro-probability-checkpoint.ipynb
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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Lab | Intro to Probability"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**Objective**\n",
"\n",
"Welcome to this Intro to Probability lab, where we explore decision-making scenarios through the lens of probability and strategic analysis. In the business world, making informed decisions is crucial, especially when faced with uncertainties. This lab focuses on scenarios where probabilistic outcomes play a significant role in shaping strategies and outcomes. Students will engage in exercises that require assessing and choosing optimal paths based on data-driven insights. The goal is to enhance your skills by applying probability concepts to solve real-world problems."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**Challenge 1**"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### Ironhack Airlines \n",
"\n",
"Often Airlines sell more tickets than they have seats available, this is called overbooking. Consider the following:\n",
"- A plane has 450 seats. \n",
"- Based on historical data we conclude that each individual passenger has a 3% chance of missing it's flight. \n",
"\n",
"If the Ironhack Airlines routinely sells 460 tickets, what is the chance that they have a seat for all passenger?"
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The probability to have a seat with 460 passengers is: 0.88\n"
]
}
],
"source": [
"#code here\n",
"from scipy.stats import binom\n",
"\n",
"k = 10 # number of extra passengers (460 - 450)\n",
"n = 460 # total number of passengers that bought a ticket\n",
"p = 0.03 # percentage per passenger missing \n",
"\n",
"# probability to get a seat \n",
"print(f\"The probability to have a seat with 460 passengers is: {1 - binom.cdf(9, n, p): .2f}\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**Challenge 2**"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### Ironhack Call Center "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Suppose a customer service representative at a call center is handling customer complaints. Consider the following:\n",
"- The probability of successfully resolving a customer complaint on the first attempt is 0.3. \n",
"\n",
"\n",
"What is the probability that the representative needs to make at least three attempts before successfully resolving a customer complaint?"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The probability that the representative resolves the a customer complaint after three attempts is 0.15\n"
]
}
],
"source": [
"#code here\n",
"from scipy.stats import geom\n",
"\n",
"p = 0.3\n",
"geom_dist = geom(p)\n",
"\n",
"print(f\"The probability that the representative resolves the a customer complaint after three attempts is {geom_dist.pmf(3): .2f}\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**Challenge 3**"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### Ironhack Website"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Consider a scenario related to Ironhack website traffic. Where:\n",
"- our website takes on average 500 visits per hour.\n",
"- the website's server is designed to handle up to 550 vists per hour.\n",
"\n",
"\n",
"What is the probability of the website server being overwhelmed?"
]
},
{
"cell_type": "code",
"execution_count": 24,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The probability of the website server being overwhelmed is 0.01\n"
]
}
],
"source": [
"#code here\n",
"from scipy.stats import poisson\n",
"\n",
"# in the next cells we inverse the probability as we are looking when it is overwhelmed and not when it is handling it\n",
"\n",
"mu = 500\n",
"poisson_dist = poisson(mu)\n",
"p = 1 - poisson_dist.cdf(549) # n ou plus on utilise cette formule ci\n",
"\n",
"print(f\"The probability of the website server being overwhelmed is {p: .2f}\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"What is the probability of being overwhelmed at some point during a day? (consider 24hours)"
]
},
{
"cell_type": "code",
"execution_count": 29,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The probability of the website server being overwhelmed is 0.29414775667981297\n"
]
}
],
"source": [
"#code here\n",
"print(f\"The probability of the website server being overwhelmed during the day is {1-((1-p)**24)}\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**Challenge 4**"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### Ironhack Helpdesk"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Consider a scenario related to the time between arrivals of customers at a service desk.\n",
"\n",
"On average, a customers arrives every 10minutes.\n",
"\n",
"What is the probability that the next customer will arrive within the next 5 minutes?"
]
},
{
"cell_type": "code",
"execution_count": 30,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"np.float64(0.3934693402873666)"
]
},
"execution_count": 30,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"from scipy.stats import expon\n",
"\n",
"#the parameter represents the rate of events occuring per unit of time\n",
"lambda_value = 0.1 # 1 customer / 10mins\n",
"\n",
"lambda_inv = expon(scale = 1/lambda_value)\n",
"\n",
"lambda_inv.cdf(5)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"If there is no customer for 15minutes, employees can that a 5minutes break.\n",
"\n",
"What is the probability an employee taking a break?"
]
},
{
"cell_type": "code",
"execution_count": 31,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The probability that the there is no customer for 15 minutes is 0.22\n"
]
}
],
"source": [
"print(f\"The probability that the there is no customer for 15 minutes is {1 - lambda_inv.cdf(15): .2f}\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**Challenge 5**"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"The weights of a certain species of birds follow a normal distribution with a mean weight of 150 grams and a standard deviation of 10 grams. \n",
"\n",
"- If we randomly select a bird, what is the probability that its weight is between 140 and 160 grams?"
]
},
{
"cell_type": "code",
"execution_count": 32,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The probability of the bird's weight being between 140 cm and 160 cm is 0.68\n"
]
}
],
"source": [
"#code here\n",
"from scipy.stats import norm\n",
"\n",
"mean = 150\n",
"std = 10\n",
"\n",
"norm_dist = norm(loc = mean, scale = std)\n",
"\n",
"print(f\"The probability of the bird's weight being between 140 cm and 160 cm is {norm_dist.cdf(160) - norm_dist.cdf(140): .2f}\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**Challenge 6**"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"If the lifetime (in hours) of a certain electronic component follows an exponential distribution with a mean lifetime of 50 hours, what is the probability that the component fails within the first 30 hours?"
]
},
{
"cell_type": "code",
"execution_count": 33,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"np.float64(0.4511883639059735)"
]
},
"execution_count": 33,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#code here\n",
"from scipy.stats import expon\n",
"\n",
"#the parameter represents the rate of events occuring per unit of time\n",
"lambda_value = 0.02 # 1 component / 50 hours\n",
"\n",
"lambda_inv = expon(scale = 1/lambda_value)\n",
"\n",
"lambda_inv.cdf(30)"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3 (ipykernel)",
"language": "python",
"name": "python3"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.13.9"
}
},
"nbformat": 4,
"nbformat_minor": 4
}
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