John and annie have two distinguishable ponds. initially, each of the ponds contains four ducks and five geese. john first picks a bird uniformly at random from the left pond and moves it to the right pond. then, annie picks a bird uniformly at random from the right pond. what is the probability that annie picks a duck?

4/9

Step-by-step explanation:

The probability that annie picks a duck Pt= the probability of picking a duck without accounting for the added one (Po)+ probability of picking the added bird and it's a duck(Pi)

Pt = Po+Pi

Since the total number of birds in the right pond is 10 after the addition of one by john

Po= 4/10

Pi= the of john adding a duck × the probability of annie picking the added bird

Pi= 4/9 × 1/10

Pi = 4/90

Pt= 4/10 + 4/90

Pt = (36+4)/90

Pt=40/90

Pt= 4/9

(This implies that the probability of picking a duck remain the same even after the addition of one bird from the left ponds because they both have equal proportions of duck and geese i.e the initial number of duck and geese in both right and left ponds are 4 and 5 respectively)

Thanks.

the answer is b

step-by-step explanation:

y = 2x

step-by-step explanation:

if you divide each number in the sequence, you'll see that they all have a multiple factor of 2. each number in the series, and any other numbers which may continue, will be multiplied by 2. therefore, y is the next number in the sequence, and x is the current one, so, our formula will be y = 2x.