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How do you solve a probability math problem?
To solve a probability problem remember the basics: In a set of equally likely outcomes (e1, e2, e3, e4, ... , en) where e = outcome, n = last number, the probability that one will occur is 1/n. The probability that it won't occur is 1 - 1/n or (n-1)/n. If the likeliness of the outcomes vary, then another factor is accounted for. Think (l1 x e1, l2 x e2, l3 x e3, ... , ln x en) where l = how often the element shows up. The probability that an event will occur is l/(sum of set). For independent outcomes (l1 x d1, l2 x d2, ... , ln x dn) and (L1 x e1, L2 x e2, ... , Ln x en), the probability that you will reach a certain outcome in set one and a certain outcome in set two is (l/(sum of set d)) x (L/(sum of set e)). Finally you should be familiar with counting principles such as factorial (!) which means: n! = n x (n - 1) x (n - 2) x ... x 1
What is the difference between the terms 'probability' and 'relative frequency'?
If an action is repeated n times and a certain event occurred b times then the ratio b/n is called the relative frequency.Where as theoretical probability is used to determine the number of ways that the event can occur if an experiment is repeated a large number of times.
There are 38 numbers on a roulette wheel and I bet 12 of them. What is the probability of not hitting one of my numbers on 11 consecutive spins?
If the probability of an event will occur is p, then the probability that it will occur in n trials is pn.(That's p raised to the n power). So if you bet on 12 numbers, then (38-12=26) numbers are empty. The probability of the ball landing on one of these empty numbers is 26/38. So (26/38)^11 = 0.01538, which is about 1.538 % or a 1 in 65 chance.
What is the hardest math question ever?
n+n-n-n-n+n-n-n squared to the 934892547857284579275348975297384579th power times 567896578239657824623786587346378 minus 36757544.545278789789375894789572356757583775389=n solve for n! the answer is 42
Adding n to the second power to n?
n2 + n = n(n + 1)
