Every morning, my neighbor goes out walking. I observe that 20% of the time she walks with her beagle, 70% of the time she walks with her golden retriever, and 30% of the time she walks alone. If these events are all disjoint, is this an example of a valid probability distribution?

No, This is not a valid Probability Distribution

Step-by-step explanation:

Probability [Neighbour walks with Beagle] = 20%

Probability [Neighbour walks with Golden Retriever] = 70%

Probability [Neighbour walks alone] = 30%

Disjoint Events are the events that have zero probability of occurring together. If all the three above items are disjoint, it means that it can never happen that two of them happen together.

The total (summed) probability of a valid probability distribution, with disjoint sets = 1 . In given case, total probability = 0.2 + 0.7 + 0.3 = 1.2 ; i.e > 1.

So, this probability distribution with stated disjoint events is not a Valid Probability distribution.

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~