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

hello :

by applying the distributive property :

6 × 4.3 = 6 × ( 4 + 0.3 )

            =   (6 ×   4) + (6 ×   0.3)

            = 24 + 1.8

6 × 4.3 = 25.8

$119.20

step-by-step explanation:

$372.50 - 60% off is $149. $149-20% off is $119.20