# Solved Problem

What is the probability that the problem is solved? E1 = First bag is chosen E2 = Second bag is chosen E3 = Third bag is chosen A = Ball drawn is red Since there are three bags and one of the bags is chosen at random, so P (E1) = P(E2) = P(E3) = 1 / 3 If E1 has already occurred, then first bag has been chosen which contains 3 red and 7 black balls.

Every few months, the Internet eats itself over some kind of viral riddle or illusion, each more infuriating than the last.

And so, like clockwork, this maddening math problem has gone viral, following in the grand tradition of such traumatic events as The Dress and Yanny/Laurel.

No matter how high-quality the camera, math has dictated that the curve of optical lenses would always be slightly softer than the center. González-Acuña, a doctoral student at Mexico’s Tecnológico de Monterrey, up and solved it.

The problem goes back thousands of years to the Greek mathematician Diocles.

Number of ways of getting a sum 22 are 6,6,6,4 = 4!

It's a problem that has plagued photography since its creation: soft edges.I have my own flow, sound and bars and my lyrics has a meaning, peace!Example 1: A coin is thrown 3 times is the probability that atleast one head is obtained?Sol: Probability math - Total number of ways = 6 × 6 = 36 ways. Now, these two days can be (Sat, Sun) (Sun, Mon) (Mon, Tue) (Tue, Wed) (Wed, Thur) (Thur, Friday) (Friday, Sat). = 1/7 Odds against the event = 6 : 1Example 15: Three bags contain 3 red, 7 black; 8 red, 2 black, and 4 red & 6 black balls respectively.Favorable cases = (1, 6) (6, 1) (2, 5) (5, 2) (3, 4) (4, 3) --- 6 ways. Sol: (i) honor cards = (A, J, Q, K) 4 cards from each suits = 4 × 4 = 16 P (honor card) = 16/52 = 4/13 (ii) face cards = (J, Q, K) 3 cards from each suit = 3 × 4 = 12 Cards. So there are total 7 cases out of which (Sat, Sun) (Sun, Mon) are two favorable cases. of ways in which two particular people sit together is 13! The probability of two particular persons sitting together 13! 1 of the bags is selected at random and a ball is drawn from it.Bobby: i'll work on preparing my argument now Taylor: You've got, what, 11 years to perfect it Bobby: time is on my side which is code for: I can put this off for a reeeaaaalllly long time Pat: which is code for "ask your mother"Bobby: she likes to claim she's good at math. Mathematicians try to make rules as precise as possible.She may come to rue the day she bragged about that Pat: "This won't help me win millions of dollars playing Fortnite tho"According to order of operations, you solve whatever is in the parentheses first. Then, in PEMDAS, multiplication and division take equal precedence, so you’d do the first that occurs from left to right. Thus, it’s 16 according to classic order of operations. According to strict order of operations, you’d get 16, but I wouldn’t hit someone on the wrist with a ruler if they said 1.Find the probability that first is green and second is red. Sol: A leap year can have 52 Sundays or 53 Sundays.Sol: P (G) × P (R) = (5/12) x (7/11) = 35/132Example 4: What is the probability of getting a sum of 7 when two dice are thrown? In a leap year, there are 366 days out of which there are 52 complete weeks & remaining 2 days.And educators and parents love the powerful reporting that allows them to monitor progress and celebrate success.⚠️PLEASE listen im doing you a favor reading this comment⚠️ I just released a song called '' EA\$T\$IDE \$UICIDE🔥'' and i know that you will like it and vibe with it, otherwise prove me wrong.

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List of all problems posted. Link to the post, image of the circuit, link to video if available and link to the PDFof the Solution Sheet are provided…

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• ###### List of unsolved problems in mathematics - Wikipedia

Millennium Prize Problems. Of the original seven Millennium Prize Problems set by the Clay Mathematics Institute in 2000, six have yet to be solved as of 2019 P.…