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Home » This James Doakes Meme Is Constantly Reacted to—Watch the Chaos! 🤯 - AMAZONAWS

This James Doakes Meme Is Constantly Reacted to—Watch the Chaos! 🤯 - AMAZONAWS

📅 March 5, 2026 👤 scraface

Mar 05, 2026

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📌 Solution: Treat the two letters as a single entity. This gives $4!$ arrangements for the entities (the letter pair and 3 manuscripts). The letters within the pair can be ordered in $2!$ ways. Total favorable arrangements: $4! imes 2!$. Total possible arrangements: $5!$. The probability is $rac{4! imes 2!}{5!} = rac{2}{5}$. $oxed{\dfrac{2}{5}}$

📌 Question: If a researcher studies 4 random integers between 0 and 10 inclusive, what is the probability that all are even?

📌 Solution: There are 11 integers (0â10), of which 6 are even (0, 2, 4, 6, 8, 10). The number of ways to choose 4 distinct even integers is $inom{6}{4}$. The total number of ways to choose 4 distinct integers is $inom{11}{4}$. The probability is $rac{inom{6}{4}}{inom{11}{4}} = rac{15}{330} = rac{1}{22}$. $oxed{\dfrac{1}{22}}$

📌 Question: If a linguist generates a 3-letter word using distinct letters from the English alphabet, what is the probability that it contains at least one vowel (A, E, I, O, U)?

📌 Solution: Total letters: 26. Vowels: 5, consonants: 21. Total 3-letter arrangements with distinct letters: $26 imes 25 imes 24$. Unfavorable cases (no vowels): $21 imes 20 imes 19$. Favorable probability: $1 - rac{21 imes 20 imes 19}{26 imes 25 imes 24} = 1 - rac{7980}{15600} = 1 - rac{133}{260} = rac{127}{260}$. $oxed{\dfrac{127}{260}}$

📌 Question: If a historian selects 3 manuscripts at random from a collection of 15, where 5 are from the 18th century, what is the probability that all 3 selected are 18th-century manuscripts?

📌 Solution: The number of ways to choose 3 18th-century manuscripts is $inom{5}{3}$. Total ways to choose 3 manuscripts from 15 is $inom{15}{3}$. The probability is $rac{inom{5}{3}}{inom{15}{3}} = rac{10}{455} = rac{2}{91}$. $oxed{\dfrac{2}{91}}$Question: If the combined population of two marine mammal species in a region is 10 and the sum of their squared populations is 50, find the sum of their cubes.

📌 Solution: Let the populations be $ a $ and $ b $. Given $ a + b = 10 $ and $ a^2 + b^2 = 50 $. First, compute $ ab $ using $ (a + b)^2 = a^2 + 2ab + b^2 $. Substituting, $ 100 = 50 + 2ab $, so $ ab = 25 $. The sum of cubes is $ a^3 + b^3 = (a + b)^3 - 3ab(a + b) = 1000 - 3 \cdot 25 \cdot 10 = 1000 - 750 = oxed{250} $.

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