Event A is drawing a King first, and Event B is drawing a King second. I'm sure they're equivalent, but do you want the joint probability of B, C & D given A? the probabilities of A and B solving the problems are i.e., respectively. Probability of solving specific independently by A and B are 1/2 and 1/3 respectively. what is the probability of (AnB) if P(A) =.50; P(B) =.20; P(A|B) =.75 P(B|A) =.30 P(AuB) =.55 This gives more information that you need. No. P(B|A) means “the probability of B happening given A has occurred” If you draw two cards, without replacement, what is the probability that both cards are red? probability; cbse; class-12; Share It On Facebook Twitter Email. No, it is not. I just begging my course in probability and I have a problem with this. Thylacoleo Thylacoleo. Sorry for my bad English, any help? (i) The probabilities of A and B solving the problem are respectively. I tried to find max value of $(1-\frac{1}{2^{a}})(1-\frac{9^{20-a}}{10^{20-a}})$ (complement of event that some of players doesn't make a point), and get solution a=1, b=19, but solution in book is a=5,b=15. Formula for the probability of A and B (independent events): p(A and B) = p(A) * p(B) Remember that if the probability of one event doesn’t affect the other, then it means you have an independent event. So you can do the problem by any of three methods using different ones of those pieces of information. 1 Answer +1 vote . probability that problem is not solved by A and B = probability that the problem is solved = (ii) P(exactly one of them solves the problem) i.e. So, as mentioned earlier all you need to multiply the probability of one by the probability of another. If both try to solve the problem, independently, then find the probability that (i) The problem is solved (ii) Exactly one of them solves the problem. \underbrace{\mathsf P(A\cap B^\complement)}_{\text{what you want}} = \underbrace{\mathsf P(A\mid A\oplus B)}_{\small\text{what you are talking about}}\!\!\!\!\cdot\mathsf P(A\oplus B… 4,479 5 5 gold badges 21 21 silver badges 31 31 bronze badges $\endgroup$ $\begingroup$ Can someone provide some references to this answer? "Probability of event A and event B equals the probability of event A times the probability of event B given event A" Let's do the next example using only notation: Example: Drawing 2 Kings from a Deck . share | cite | improve this answer | follow | edited Aug 13 '10 at 6:44. answered Aug 12 '10 at 10:19. Then, the probability of only A occurring is the probability of A occurring given that only one of the events will occur.