Probability

b is incorrect while c is virtually meaningless.

The confidence interval consists of a central value and a margin of error around that value. If it is an X% confidence interval then there is a X% probability that the true value of the statistic in question lies inside the interval. Another way of looking at it is that if you took repeated samples and calculated the test statistic each time, you should expect X% of the test statistics to fall within the confidence interval.

Probability

The probability that an adversary will exploit a weakness in your operation, tempered by the impact to your mission, is defined as the:

A. Risk

The probability that the sum of the numbers rolled is either even or a multiple of 5 is 11/18.

False

its all about nonimportation agreement

A systematic sample is not something that you can solve!

4/6, or 66.67%.

cards are randomly chosen 7 out of 12

336

2 possibilities for the first

for each of these two you have 4 possibilities for the second - making it 8

for each of these 8 you have 6 possibilities for the third - making it 48 possibilities

for each of those 48 you have 7 possibilities for the fourth - making it 336 possible combinations

2x4x6x7 = 336

P(Heads) = 2/3

The chances of winning would be 15/3000 or 1 in 200 (or 0.5%). That means the probability of NOT winning is 99.5 %

Event 1 makes Event 2 happen.

You need to calculate your mathematical expectation.For the winning case, your gain is: 0.624 x 1

For the losing case, your gain is: (1 - 0.624) x (-2)

(So, in each case, probability multiplied by the amount you would win/lose.)

Do the calculations, and add everything up. If the sum is positive, the game is in your favor.

I assume you mean what's the chance of at least two heads showing when three fair coins are tossed.

There are 8 possible outcomes as each coin can either be head or tails.

For 3 heads, all 3 coins must show a head → 1 success

For 2 heads, one coin will be a Tail; each coin could be a tail in turn → 3 successes

→ Pr = (1+3)/8 = 4/8 = 1/2

If you are wanting the probability that the first TWO specific coins are heads and the last, third, coin is either, then:

Pr(head) = 1/2

→ Pr(1st 2 heads, 3rd anything) = 1/2 × 1/2 × 1 = 1/4

event B has something in common with event A

The answer depends on the set of cards from which the picking is done. Unfortunately, you have provided absolutely no information on that.

Chance music (Apex 2020)

It is 2/13.

The confidence intervals will increase. How much it will increase depends on whether the underlying probability model is additive or multiplicative.

false

There are 40320 ways.

Derivative Classification

Assess hazards is the step in the composite risk management process that is focused on determining the probability and severity of a hazard occurring.