Probability

HAZARD

45%

There are 10 choices for the first number, 9 for the second and 8 for the third. 10*9*8=720 possible combinations.

118.872

4

77.45%

Suppose that each time the game is played you win an amount X, with probability p or lose an amount Y with probability (1-p). To start with it is assumed that you cannot draw a game. Then each time the game is played you expect to win X with a probability p and lose Y (= win -Y) with probability (1-p).

Therefore, the expected gain for you, each time the game is played, is

X*p - Y*(1-p).

The game is fair if this value is 0. If it is greater than 0 the game is biased in your favour. Otherwise it is biased against you.

You can easily incorporate multiple outcomes - such as with a lottery: a bigger win for getting more numbers.

If X(1), X(2), ... X(n) are the n pay-outs to you (remember that what you pay will be a negative number), and P(1), p(2), ... p(n) are their respective probabilities where the sum of all the p(r) = 1, then your expected win per game is

X(1)*p(1) + X(2)*p(2) + ... + X(n)*p(n)

and the game is fair if this is 0.

A barge is the shape it is because it need to be able to push tons of things to anywhere if it wasnt that shaoe it wouldn't be ableto do anything

Odd-

3/12 ---> 1/4

Prime-

3/12 ---> 1/4

They are the same

3 out of 7

there can be only three combo's ------ head n tail,,,,,,,, tail n tail,,,,,,,,,,,,,, head n head

Describe how the Swap() instruction can be used to provide mutual exclusion that satisfies the bounded-waiting requirement.

11/16

M=0

n=0

m*n=0

Two sixes rolled on dice are commonly referred to as "double sixes." In many games, rolling double sixes can have special significance, such as providing an advantage or allowing an additional turn. In the game of craps, for example, it can be a winning roll.

Most games both, Some games only one, but you chose. BUT THE VALUE IS 1 OR 11

The area is 0.008894

It is 4/11.

Discrete controller

Look up the formula in a book of elementary statistics or on the web. The browser that we are required to use is not suitable for providing an answer.

It depends on how well you pack it, how good a condition it is in, and on how correctly you use it.

In practice, the probability is very low. About one in every 100,000 jumps in the US results in a fatality, a probability of about 0.00001, but that's not just due to parachute failure. One unconfirmed estimate is that one in a thousand (0.001) main chute deployments result in a malfunction, but it is true that very few of those result in a fatality, primarily because there is a reserve chute that is maintained and packed under different, more stringent, requirements.

Presuming that the spinner and the number cube are both "fair", then no - spinning the spinner and tossing the six-sided number cube are called statistically independent events. They do not influence each other, and it does not matter which order the events occur in.

The 5th percentile of a standard normal distribution is -1.645 (from the normal probability table).

z = (x - μ) / σ

-1.645 = (x-98.2) / .62

(0.62)(-1.645) = x-98.2

-1.0199 = x-98.2

x = 98.2-1.0199 = 97.1801

The 5th percentile is 97.18

The 95th percentile of a standard normal distribution is 1.645 (from the normal probability table).

z = (x - μ) / σ

1.645 = (x-98.2) / .62

(0.62)(1.645) = x-98.2

1.0199 = x-98.2

x = 98.2+1.0199 = 99.2199

The 95th percentile is 99.22

18