Probability Mass Function (PMF) 2/2
Probability mass function over a range:
In some cases, we want to know the probability that discrete random variables are equal to numbers over a range.
Example:
Calculate the probability that we will have success with the fair coin at 4 or less times (0, 1, 2, 3 or 4) (the chance of getting head or tail is 50%) if we have 15 attempts. We assume that success means getting a head.
Python realization:
12345678910111213141516171819202122232425262728293031# Import required library import scipy.stats as stats # The probability of getting 0 successes prob_0 = stats.binom.pmf(0, n = 15, p = 0.5) # The probability of getting 1 success prob_1 = stats.binom.pmf(1, n = 15, p = 0.5) # The probability of getting 2 successes prob_2 = stats.binom.pmf(2, n = 15, p = 0.5) # The probability of getting 3 successes prob_3 = stats.binom.pmf(3, n = 15, p = 0.5) # The probability of getting 4 successes prob_4 = stats.binom.pmf(4, n = 15, p = 0.5) # The resulting probability probability = prob_0 + prob_1 + prob_2 + prob_3 + prob_4 print("The probability is", probability * 100, "%")
Explanation
We've found the probability that a discrete random variable will equal exactly 0, 1, 2, 3, or 4 using the probability mass function. Then, we summed up all probabilities using the addition rule because each of the found outcomes was permissible for us.
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Completion rate improved to 3.7Probability Mass Function (PMF) 2/2
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Probability mass function over a range:
In some cases, we want to know the probability that discrete random variables are equal to numbers over a range.
Example:
Calculate the probability that we will have success with the fair coin at 4 or less times (0, 1, 2, 3 or 4) (the chance of getting head or tail is 50%) if we have 15 attempts. We assume that success means getting a head.
Python realization:
12345678910111213141516171819202122232425262728293031# Import required library import scipy.stats as stats # The probability of getting 0 successes prob_0 = stats.binom.pmf(0, n = 15, p = 0.5) # The probability of getting 1 success prob_1 = stats.binom.pmf(1, n = 15, p = 0.5) # The probability of getting 2 successes prob_2 = stats.binom.pmf(2, n = 15, p = 0.5) # The probability of getting 3 successes prob_3 = stats.binom.pmf(3, n = 15, p = 0.5) # The probability of getting 4 successes prob_4 = stats.binom.pmf(4, n = 15, p = 0.5) # The resulting probability probability = prob_0 + prob_1 + prob_2 + prob_3 + prob_4 print("The probability is", probability * 100, "%")
Explanation
We've found the probability that a discrete random variable will equal exactly 0, 1, 2, 3, or 4 using the probability mass function. Then, we summed up all probabilities using the addition rule because each of the found outcomes was permissible for us.
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Completion rate improved to 3.7single
