Find the probability that the student got a 'B' GIVEN they are male.?
A test was given to a group of students. The grades and gender are summarized below A B C Total Male 16 14 6 36 Female 5 17 11 33 Total 21 31 17 69 If one student is chosen at random from those who took the test,
Probability question?
This is an exercise in conditional probability. You can use Bayes' Rule to solve this. However, I find that it is much easier to just use the definition of conditional probability. In short I derived Bayes' Rule every time I do this but it prevents me from making errors in more complex problems. This is an exercise in conditional probability. For any two events A and B, where P(B) ≠ 0, you have the conditional probability: P( A | B ) = P( A ∩ B ) / P( B ) = P( B | A) * P(A) / P(B) the above is read as: the probability of A given B is equal to the probability of A and B divided by the probability of B. Use The Law of Total Probability to find P(B) For a set of events A1, A2, A3, ... , An where the Ai's are mutually exclusive and exhaustive events and for any other event B P(B) = P(B and A1) + P(B and A2) + ... + P(B and An) = P(B | A1) * P(A1) + P(B | A2) * P(A2) + ... + P(B | An) * P(An) What we know is: I will use B to be the event of a poor driver P(G) = 0.30 P(M) = 0.50 P(B) = 0.20 Let C be the event the drive as at least one citation P(C | G) = 0.1 P(C | M) = 0.3 P(C | B) = 0.5 Find P( G | C ) P( G | C) = P( G ∩ C) / P(C) = P( C | G ) * P(G) / P(C) P(C) = P( C | G ) * P(G) + P( C | M) * P(M) + P( C | B) * P(B) P(C) = 0.1 * 0.3 + 0.3 * 0.5 + 0.5 * 0.2 P(C) = 0.28 P( G | C) = 0.1 * 0.3 / 0.28 = 0.1071429 The probability that the drive is a good risk driver given they had a citation is 0.1071429
Find p for the given probabilities?
P( A or B or C ) = P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) + P(A and B and C) = 0.39 + 0.24 + 0.12 - 0.11 - 0.03 - 0.06 + (0.39 * 0.24 * 0.12) = 0.56 The principle here is that we add up the probabilities of each... but then we subtract the outcomes that were counted twice... but then add back the outcome that was removed twice.
What is the probability of (B) given (A)? (Stats question I need help!)?
P(A) = 0.3 P(B) = 0.8 Probability that A and B occur simultaneously = 0.2 If it is known that event A occured, what is the probability that event B also occured?
Statistics / probability question!!?
A) p(a or b) = p(a) + p(b) - p(a and b). Since p(a) = 0.3, p(b) = 0.4, but p(a and b) is not given, you do not have enough information to find p(a or b). B) If you know that a and b are mutually exclusive, then p(a or b) = p(a) + p(b). Since p(a) = 0.3 and p(b) = 0.4, you have enough information to find p(a or b) if you know that a and b are mutually exclusive. 2) Since (a and b) is a subset of a, p(a and b) cannot exceed p(a). So p(a and b) = .40 and p(a) =.20 is an impossible situation. Note that in this situation, p(b given a) = p(a and b)/p(a) = .40/.20 = 2, which is nonsense since 2 > 1, and probabilities cannot exceed 1. Have a blessed, wonderful day!
Find the Probability?
(a) Since it is know that a male was selected, the size of the sample space is |S| = 46+39+1 = 86 The probability of getting a Republican is P(R) = 46/86 = 0.53488 (b) If the problem is "find the probability of getting a male, given that a republican was selected," then since it is known that a republican was selected, the size of the sample space is |S| = 46 + 5 = 51 The probability of getting a male is P(M) = 46/51 = 0.90196 (c) Since it is know that am independant was selected, the size of the sample space is |S| = 1+0 = 1 The probabitity of getting a female is P(F) = 0/1 = 0 (d) As in part (a) it is known that a male was selected, so the cardinality of the sample space is again |S| = 86 so you have P(D) = 39/86 P(I) = 1/86 Since the two events are mutually exclusive, the probability of getting a Democrat or an Independant is P(D+I) = P(D) + P(I) = 39/86 + 1/86 = 40/86 = 0.46512 Note that you could also use the results of part (a): P(D+I) = 1 - P(R) = 1 - 0.53488 = 0.46512
If I know the probability of A given B, do I know anything at all about the probability of B given A?
The other answers here are basically correct, in that without a prior there's no tight linkage in general between likelihoods and posteriors. But you DO know one very important thing - if [math] P(A \mid B) = 0[/math], then observing A rules out B no matter your prior. This point seems trivial, but many inferences we make in real life are of this categorical sort. "If my coworker is reliable, I can recruit him to help out on a small task for my project as a favor. He shirked, so he must not be reliable." Many of the models we carry around in our heads about how the world works are that stark, or close to it. So when your model predicts X is impossible, and X happens, you can reject that model without assessing your prior.
What is the conditional probability of event B given event A, knowing that B is a subset of A?
Let's calculate [math] P(B) [/math][math]P(B)=P(B=1)=P(B=1|A=1)*P(A=1)+P(B=1|A=0)*P(A=0)[/math]Since [math]P(B=1|A=0) = 0[/math] then we can write [math]P(B)=P(B=1|A=1)*P(A=1)[/math] so all we can say is that [math]P(B=1|A=1) = \frac{P(B)}{P(A=1)}[/math] So no! [math]P(B)[/math] doesn't necessarily equals [math]P(B|A)[/math]. (unless of course [math]P(A=1)=1[/math])
How do u find the binomial probability?
n = 5 ----- number of trials ( 5 questions ) k = 3------ exactly 3 right n – k = 2 p = 0.25 = probability of guessing the correct answer on a question q = 0.75 = probability of guessing the wrong answer on a question (a)P = 5C3 (0.25)^3 (0.75)^2 = 0.087890625 (b) u get the probability of 3 and 4 and 5 right. or another simpler way is to get probability of 2 right + 1 right then subtract them from 1 so that u do less calculations 1 - ( 5C1 (0.25)^1 (0.75)^4 + 5C2 (0.25)^2 (0.75)^3 ) (c) here its probabilty of getting one right + that of getting 2 right. or its 1 - probability of getting ( 3 + 4 + 5 right ) and we got it while calculating (b)
Is it possible to find the probability of event A when given conditional probabilities P(A|B), P (A|B'), and P(B)?
Yes!!Why? Well P(A)=P(A|B)P(B)/P(B|A) from Baye’s law.Hence[math]P(A)=\sum_{i =1,..,n}P(A|C_i)P(C_i)[/math]This formula is valid for any set of events [math]C_i[/math] such that [math]P(C_i\cap C_j)=0,i\neq j, \sum P(C_i)=1[/math]I assume in the OP B’ denotes the complement of B? Then find your desired value using the above: just fiddle around a tad with properties of complement probabilities,Hint: P(B’)=1-P(B). So P(A|B’)(1-P(B))= ?