Probability

5/36

The odds of winning the "Mega Millions" is stated as 1 out of 176,000,000 while your odds of being hit by lightning is (generally) 1 out of 576,000 -- so yeah you are 300 times more likely to be hit by lightning than to win the lottery.

It is 1/2*1/13 = 1/26.

Scientific design is always better.

Pieces in a Pizza Hut Pizza

Probability of coin heads up: 1/2

Rolling a 4 or 5 on the cube: 2/6

1/2 times 2/6 = 2/12, or 1/6.

that is like saying is your cut of ear normal looking. there is nothing normal about the unnecessary amputation of a body part.

It is a type of electrical connectors that is meant to be joined with its mating half. The spade terminal is retangular and flat appearing similar to a spade digging tool. It relies on tension not mechanical tightening. The "spade" part is the "male" and the "female" half slides over the spade. Some people refer to it as "push-on" These are only meant for small size wires usually no more than size#10. They are common in vehicle, appliance and machinery applications.

a frequency diagram.

no

Those "balls" are their scent glands. Not their actual reproductive gland, if you know what I mean. Their thingy is hidden inside, until they are ready to mate, you will see it. Which is nasty, and I dnt think you would want to see that :D Hope this helped.

It will show up any number of times from 0 to 300. The expected value is 50 but the probabiluty that is shows up exactly 50 times is around 6%.

Take out 2 aces and 2 jokers gives you 48 cards, a number divisible by four (heart, spade, clubs, diamonds). Divide 48 by 4 and you get 12 cards. So 12 out of 52.

The question is too ambiguous.

Is it about a randomly selected person having twins in 1952 as opposed to another year? The incidence of multiple births has increased over the years as a result of IVF. Does the person (parent) have to be living now so as to be selected?

Is it about an individual having twins in 1952 as opposed to other people who gave birth in 1952? For that you need data on twin and other births in 1952 for the whole world. I very much doubt if such statistics are available - certainly not reliable ones.

... then all outcomes are equally likely. Nothing more, nothing less. You can always re-define the outcomes so that they are not all equally likely.

For example, on a single roll of a fair die, the numbers 1 to 6 are equally likely. But redefine the events so that

Event A = prime number

Event B = composite number

Event C = neither prime nor composite number

then P(A) = 1/2, P(B) = 1/3 and P(C) = 1/6 : events with unequal likelihood.

Answer: there are five distinct letters in the word 'smart'. However we may rearrange them, every rearrangement will consist of all and only those five elements. The case here is just the same as if we were wanting to determine how many different ways we could seat 5 people in a five-seat row of chairs. In the first chair we could seat any one of the five--and just for fun we'll imagine this is a group of five of my former girlfriends [S]usan, [M]olly, [A]shlee, [R]ebecca, and (yes, it's a stretch) Kasey [T.]. So any one of the five chicks can take the first seat, which is to say there are five possible outcomes for seat number one. Now regardless of who the first girl is, once she sits down there are four girls left, any any one of whom can sit down next to her in seat number 2. Suppose Ashlee takes the first seat. Then there are four possibilities for who sits next to her: Susan, Molly, Rebbecca, or Kasey T. But any one of those girls could just as easily have taken have taken

The answer depends on the card game. Aces are usually at the two extremes: the highest cards or the lowest, but that is not always the case.

You can't, really, without scientific intervention. The male's sperm determines whether or not the chromosomes turn into male, XY, or female, XX.

The following figures can only lead to the sum of 10. So the permuatation of each combination is as follows.

1 1 1 1 6 = 5

1 1 1 2 5 = 20

1 1 1 3 4 = 20

1 1 2 2 4 = 30

1 1 2 3 3 = 30

1 2 2 2 3 = 20

2 2 2 2 2 = 1

That is the number of ways to get a sum of 10 is 126 and the total of number of distinct throws that can be available are (6 to the power 5) 7776 and the required probability is 126/7776 = 0.016203703703703703703703703703704.

H H

T T

H T

T H

1 in 2 chances

love

5 out of 26

The probability of rolling a sum of 2 is 1/36

The probability of rolling the value 2 on one die or the other (or both) is 11/36

The probability is 1. It must be "a red or black card".

The answer will depend on what ABCD is (or are). And since you have not bothered to provide that information, I cannot provide a sensible answer.