Probability Flashcards
What is the equation for normalization?
sum ove i of Pi = 1
What is the equation for the mean or average or expected value?
= sum over i of xi*Pi
What is the equation for the variance?
σ(x)^2 = )^2> = -^2
What is the equation for the standard or RMS deviation?
σ(x) = sqrt(variance)
What are the variables in discrete probability distribution?
x is a discrete random variable with values xi and probabilities Pi
What are the variables in continuous probability distributions?
x is a continuous random variable with probability P(x)dx of having a value between x and x+dx
What is the equation for normalization in continuous probability distribution?
integral of P(x) dx = 1
What is the equation for the mean in continuous probability distribution?
= integral of x*P(x) dx
What is a Bernoulli trial?
An experiment with two outcomes.
What are the two probabilities for a Bernoulli trial?
Pi(x) = P for xi = 1 and 1-P for xi = 0
What is a good example of a Bernoulli trial?
Tossing a coin.
What is the equation for the mean value of x for tossing a coin?
= sum of xiPi = 1P + 0*(1-P) = P
What is the equation for the mean value of the squared x values for the coin tossing example?
= sum of xi^2Pi = 1P+0*(1-P) = P
What is the equation for the standard deviation for the coin toss example?
σ(x) = sqrt(-^2) = sqrt(P-P^2) = sqrt(P*(1-P))
If we perform the same experiment n times, what is the probability of getting k successes and n-k failures?
Given by the binomial distribution: P(n,k) = nCkP^k(1-P)^(n-k), where nCk = n!/(k!*(n-k)!)
In Bayes theorem, what is the condition probability?
P(A|B) = prob of event A given that event B has occurred.
In Bayes theorem, what is the joint probability?
P(A n B) = prob that events A and B occur = P(A)P(B) if A and B are independent -> if they are not independent: = P(B|A)P(A)
What is the equated version of the joint probability?
P(A|B) = P(B|A)*P(A)/P(B) = Bayes theorem
An athlete fails a drugs test which is 95% accurate. What is the likelihood they are guilty if on average only 1% of athletes take drugs? (first step)
P(D) = 0.01, P(\D) = 0.99, so P(+|D) = 0.95, P(+|\D) = 0.05, P(-|\D) = 0.95, P(-|D) = 0.05
An athlete fails a drugs test which is 95% accurate. What is the likelihood they are guilty if on average only 1% of athletes take drugs? (second step)
Find P(+) = P(+|D)P(D)+P(+|\D)P(\D) = 0.06
An athlete fails a drugs test which is 95% accurate. What is the likelihood they are guilty if on average only 1% of athletes take drugs? (third step)
Bayes theorem: P(D|+) =P(+|D)P(D)/P(+) = 0.950.01/0.06 = 0.16, so ~ 16%