I need to finish up a problem. if I have a 1/3 probability and and 1/4 probability, what is the chance they will both occur in the same equation, what is the chance of both of them happening with 1 chance each?​

Given a probability A, denoted by P(A), it is simple to calculate the complement, or the probability that the event described by P(A) does not occur, P(A'). If for example P(A) = 0.65 represents the probability that Bob does not do his homework, his teacher Sally can predict the probability that Bob does his homework as follows:

P(A') = 1 - P(A) = 1 - 0.65 = 0.35

Given this scenario, there is therefore a 35% chance that Bob does his homework. Any P(B') would be calculated in the same manner, and it is worth noting that in the calculator above, can be independent; i.e. if P(A) = 0.65, P(B) does not necessarily have to equal 0.35, and can equal 0.30 or some other number.

The probability of both happening at the same time, multiply the probabilities together:

1/3 x 1/4 = 1/12

The probability would be 1/12

answer: the 4 kilogram bag

step-by-step explanation: this is because if you divide the number of kilograms into the total it will tell you how much each individual kilogram would be worth. 2 kilos is 7 and change 3 would be around 6 ish and the 4 kilo would be 4.6 and change and the 5 kilos would be like 4.88

his first public project he is asked to create a small scale drawing. the garden is a right triangle with base 25m and height 30m. are you trying to solve for the last side/hypotenuse? the irrational form of the answer would be 5\sqrt{61}