Probability

720

The probability of doing so, eventually, is 1. In a single random draw it is 1/52.

4

Birthdays are not uniformly distributed over the year. Also, if you were born on 29 February, for example, the probability would be much smaller. Ignoring these two factors, the probability is 0.0082

No. A probability means, how likely it is for something to happen. An observation of SEVERAL similar events can give you a good ESTIMATE of the probability.

Assuming a standard 6-sided die with sides numbered {1, 2, 3, 4, 5, 6}, then:

Number of successes = 1

Total number of outcomes = 6

→ pr(2) = 1/6

If you randomly pick a card from a standard deck of cards, that probability will be 1/4, since 1/4 of the cards are heart.

P(A)=1-(A)

• There are 475 people total
• 300 are male so 175 are female (assuming they all identify as male or female)
• 135 males like fried food
• 270 people total do not like fried food
• 105 females do not like fried food
• 175 - 105 = 70 females do like fried food
• 270 - 105 = 165 males do not like fried food

a.) Probability that the person is female is 175/475 = ~37%

b.) Probability that the person is female and likes fried food is 70/475 = ~15%

c.) Probability that the person is female or does not like fried food is (175 + 165) / 475 = 85%

In essence there are only 3 possibilities:

RRY

RYR

YRR

The sum of the odds of these three things is the possibility of one of them happening. The order does matter in this, because which one you draw first affects the odds of which one you draw second in this scenario.

There's 20 balls to begin with, and each time we remove a ball we're decreasing the total number, so the odds break down like this:

RRY:

12/20 * 11/19 * 8/18

RYR:

12/20 * 8/19 * 11/18

YRR:

8/20 * 12/19 * 11/18

Because of the specific circumstances the odds of each are actually the same (because we're decreasing the total pool and the individual pools at the same rate in all three, essentially they're the same numbers in different orders). Each has a roughly 15.2% chance, and adding those odds together gives us the odds of one of them happening (but it doesn't matter which one).

Totaling to a roughly 46.2% chance of drawing some combination of the three.

For comparison, there's a 19.2% chance that they come up as RRR, and a mere 4.9% chance that they come up YYY. The odds of two yellows and a red, meanwhile is about 29.4. Note that, because of my rounding this only totals out to 99.7%, but you get the picture. If you need the odds to a more specific decimal point, you can plug in the numbers and get the full totals.

Probability = number_of_ways_of_success/total_number_of_ways

If two events are independent, their probabilities are multiplied together, eg tossing a coin to get heads and throwing a 6 with a single [standard cube] die.

In this case, I guess you are choosing 4 marbles at random. This is the same as choosing one marble, then choosing the next WITHOUT replacing the first marble; each random choice of marble is independent of the previous in so far as the colour only depends upon what is left in the bag.

So the probability here is the product of the probability that the first marble is red, the next marble (without the first being replaced) is red and so on until 4 have been chosen:

Pr(1st red) = 5/15 (= 1/3) as there are 5 red marbles and 5+10=15 marbles in the bag

Pr(2nd red) = 4/14 (= 2/7) as there are now only 4 red and 4+10 = 14 marbles in the bag

Pr(3rd red) = 3/13 as there are now only 3 red marbles and 3+10 = 13 marbles in the bag

Pr(4th red) = 2/12 (= 1/6) as there are now 2 red marbles and 2+10 = 12 marbles in the bag.

Thus Pr(4 red) = Pr(1st red) × Pr(2nd red) × Pr(3rd red) × Pr(4th red)

→ Pr(4 red) = 1/3 × 2/7 × 3/13 × 1/6 = 1/273 ≈ 0.00366 ≈ 0.366 %

The probability is 71/489 = 0.145, approx.

A normal deck of 52 playing cards will ALWAYS contain picture cards, so the probability is 1.

This isn't a probability question since it can be checked with certainty. In 2016 there are only 52 Mondays.

It is 0.69444... recurring.

The answer depends on which 4 numbers, out of what set of numbers as well as how often the experiment is conducted, whether or not it is conducted randomly. All in all, too many unknown factors for a sensible answer.

There is no simple answer to the question because the children's genders are not independent events. They depend on the parents' ages and their genes.
However, if you assume that children's genders are independent events then, given that the probability of a girl is approx 0.48.

The probability depends on the nature of the outcomes in the set: whether or not they are mutually exclusive, independent.

One possibility: Arkansas State University - Jonesboro

If measurements are taken for two (or more) variable for a sample , then the correlation between the variables are the sample correlation. If the sample is representative then the sample correlation will be a good estimate of the true population correlation.

The answer depends on what you are trying to predict. Suppose you have a discrete random variable X with a probability density function p(X) = prob(X = x), then the expected value of a function f(X) of X is the sum of f(x)*p(x), summed over all possible values of x. For a continuous variable, the procedure is similar, except that you need to integrate rather than sum.

The answer will depend on people's access to gaming machines and to video games, the demographics of the wherever you are studying and whether or not the people have a life!

The probability is 1/6.

Very many events in daily life are uncertain. Or, given the amount of information that we can collect and process in real time, they may as well be random. The daily weather, the gender of a child when conceived, whether or not you hit a red light two miles down the road, whether the next chip made is faulty and so on. Some things are truly random - such as physics at the quantum level. Probability theory allows you to put some measure on the degree of uncertainty and work out how likely different outcomes are so that, in some case, you can plan around those uncertainties.