### When to use cumulative binomial probability?

When the event of interest is a cumulative event. For example, to find the probability of getting three Heads in 8 tosses of a fair coin you would use the regular binomial distribution. But to find the probability of up to 3 Heads you would use the cumulative distribution. This is because Prob("up to 3") = Prob(0 or 1 or 2 or 3) = Prob(0) + Prob(1) + Prob(2) + Prob(3) since these are mutually…

### What is the probability of getting a sum of 5 if two dice are tossed 120 times?

Prob(5 in a throw of two dice) = 5/36Prob(Not 5 in a throw of two dice) = 1 - 5/36 = 31/36 Prob(Not 5 in 120 throws) = (31/36)120= 0.000 000 016 and so Prob(5 in 120 throws) = 1 - Prob(Not 5 in 120 throws) = 0.999 999 984 Prob(5 in a throw of two dice) = 5/36Prob(Not 5 in a throw of two dice) = 1 - 5/36 = 31/36 Prob(Not 5 in…

### Would someone help guide you about how to do Probability Math -See Discussion Area for more detailed information?

The first step is to create the write out the probability space. For each possible product, list the values on each die that will give that product. Each such outcome has a probability of 1/36 and so you can calculate the probability of each of the numbers on the card as follows: 1: (1,1) prob = 1/36 2: (1, 2), (2, 1) prob = 2/36 3: (1, 3), (3, 1) prob = 2/36 4: (1…

### What is the probability of getting a sum of 5 if the dice is tossed 120 times?

Prob(at least one 5 in 120 throws) = 1 - prob(no 5s in 120 throws)= 1 - (5/6)120= 0.999 999 999 69 = almost 1. Prob(at least one 5 in 120 throws) = 1 - prob(no 5s in 120 throws)= 1 - (5/6)120= 0.999 999 999 69 = almost 1. Prob(at least one 5 in 120 throws) = 1 - prob(no 5s in 120 throws)= 1 - (5/6)120= 0.999 999 999 69 = almost 1…

### Is doppelganger safe?

it prob depends what the doppleganger is doing to you becases I seen a vidoe on youtube about a man playing his guitar and he hears a bang and it was his doppleganger trying to kill him but it doesent kill him it hides and I have seen a doppleganger but the vidoe on youtube prob isent real its prob his twin brother

### How would you calculate the probability you'll have experienced a rainy day at some point in the last seven days if there's been a 10 percent chance of rain each day?

Rainy days are not independent events, but for theis question you will need to assume that they are. Otherwise, there is insufficient information to answer. Prob(a day being rainy) = 0.1 so Prob(a day being not rainy) = 1 - Prob(a day being rainy) = 1 - 0.1 = 0.9 Therefore, Prob(7 days being no rainy) = (0.9)7 and so, Prob(At least one rainy day in 7) = 1 - Prob(no rainy days in 7)…