Probability

81

Here are the odds in craps: 7 vs. 6 Odds: 6 to 5 7 vs. 5 Odds: 6 to 4 or 3 to 2 7 vs. 4 Odds: 6 to 3 or 2 to 1 7 vs. 3 Odds: 6 to 2 or 3 to 1 7 vs. 2 Odds: 6 to 1

The odds of winning prizes are shown in table from the official Mega Millions website. From the table, you can calculate the cumulative probability of winning a prize and the expectation of the prize.

- If you bought 1 ticket

You have 1/75 + 1/141 + 1/844 + 1/306 + 1/13781 + 1/15313 + 1/689065 + 1/3904701 + 1/175711536 = 1/40 chances to win a prize with the expectation of $2/75 + $3/141 + $10/844 + $7/306 + $150/13781 + $150/15313 + $10,000/689065 + $250,000/3904701 + Jackpot/175711536 = $0.83. Here we assumed Jackpot prize is $12,000,000. This can be interpreted statistically as follows: You pay $1 to buy 1 ticket, and you have 2.5% of chances to get $0.83 back out of your $1 and 97.5% of chances to lose your $1. So, the expected net worth of the 1 ticket is 0.025x$0.83=$0.02, which is about 2% of the purchased ticket price.* In case the Jackpot is rolled over, the expectation of the prize will increase, but increasing the Jackpot prize from $12,000,000 to $120,000,000 only results in increasing the expectation of the prize from $0.83 to $1.44, thus the expected net worth of a ticket increases from 0.025x$0.83=$0.02 to 0.025x$1.44=$0.036, which is a trivial difference.

* The calculations are brought from PickSmarter.com.

- If you bought 2 tickets

The chances for at least one of 2 tickets to win a prize is 1-(39/40)2 =0.049 with the expectation of $1.65. So, you pay $2 to buy 2 tickets, and you have 4.9% of chances to get $1.65 back out of your $2 and 95.1% of chances to lose your $2. The expected net worth of the 2 tickets is 0.049x$1.65=$0.08, which is about 4% of the purchased tickets price.

- If you bought 40 tickets

The chances for at least one of 40 tickets to win a prize is 1-(39/40)40 =0.637 with the expectation of $33.04. So, you pay $40 to buy 40 tickets, and you have 63.7% of chances to get $33.04 back out of your $40 and 36.3% of chances to lose your $40. The expected net worth of the 40 tickets is 0.637x$33.04=$21.05, which is about 52% of the purchased tickets price.

- If you bought 100 tickets

The chances for at least one of 100 tickets to win a prize is 1-(39/40)100 =0.92 with the expectation of $83. So, you pay $100 to buy 100 tickets, and you have 92% of chances to get $83 back out of your $100 and 8% of chances to lose your $100. The expected net worth of the 100 tickets is 0.92x$83=$76.4, which is about 76% of the purchased tickets price.

So, the more tickets you buy, the chances of at least one of your tickets win a prize increase with higher expected amount of prize, and thus the chances of your getting nothing decrease, but the amount of money you lose increases when you do lose. The percentage of the expectected net worth of the ticksts gradually converges to the purchased tickets price, at the risk of losing more money with less probability.

Probable is the same as likely or that something is likely to happen. While I hope it doesn't rain, the dark clouds in the sky make it probable that it will.

1.

PCIe version 2 doubled the fequency of the PCIe bus, theoretically doubling the throughput, It also allows for up to 32 lanes on one slot

sample size, n = 140
standard deviation, s = 11.45

standard error of the mean, SE = s / n^1/2 = 11.45 / 140^1.2 = 0.9677

95% confidence interval => mean +- 1.96SE

95% CI = 86.5 - 1.96*0.9677; 86.5 + 1.96*0.9677

= 84.6; 88.4

When rolling a pair of dice there are 6∙6 = 36 possible outcomes.

The outcomes that give a sum of 5 are 4, [(1,4), (2,3), (3,2), (4,1)].

So the probability of not rolling a sum of 5 is:

P(NOT 5) = 1 - P(5) = 1 - 4/36 = 32/36 = 8/9 = 0.8888... ≈ 88.9%

The outcome is that you have pulled a marble out of the bag.

That means "Greetings from holidays/vacation"

There is a .25 chance that no heads will come up. (1/2*1/2=.25)

Therefore, there is a .75 chance that one or more heads will come up.

Value of game = .75*$2-.25*$4=$1.5-$1=$.50

Over time, you should come out ahead.

many

Your question is a bit unclear. I hope this answers your question, but if not, ask it again in a more specific manner.

The pdf of a continuous uniform random variable (X) distributed over the range a to b is f(x) = 1/(b-a).

If we consider x and y are dimensions, and each have a uniform distribution, then we can similarly define f(y) = 1/(c - d). Now, since these are independent, the joint distribution is f(x,y) = 1/(c-d) * 1/(b-a).

So, if we want to find the probability of X < x* and Y < y*, where x* is in the range of a to b and y* is in the range of c to d, we would integrate twice over f(x,y):

F(x*,y*) = integral from a to x* , integral from c to y* f(x,y) dy dx

There are ten events in a decathlon because deca- means ten

4!

Four factorial.

4 * 3 * 2

= 24 permutations

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yay

It is when you have to take on a huge role and not disapoint yourself. Say if you had to babysit a child you will need to have full consertration and need to be responsible if the child gets hurt.

No.

The highest number on a single "die" from a dice set is six. Therefore, the highest number available that one could roll on a dice set would be 6 multiplied by the number of dice thrown.

The probability of getting an even number on both dice is (2/6) * (2/6) = (4/36) = 1/9.

The probability when rolling two dice of getting one even number but not a "2" is 10/36, which is 5/18.

1-(5/6 x 5/6 x5/6)

there are 36 possible outcomes and 9 ways to get sums divible by 4, so odds are 1 in 4

0.5

1