Probability

In a single toss? zero. In two tosses, well that's a different question. Possible combinations are TT, HT, TH and HH, of which 2 of the 4 are one head and one tail. So probability is 0.50 in two tosses.

The answer is 6C4 = 6*5/(2*1) = 15

2 out of 52

The probability is the ratio of the area of the shaded area to the area of the whole figure.

Your question is not clear, but I will attempt to interpret it as best I can. When you first learn about probability, you are taught to list out the possible outcomes. If all outcomes are equally probable, then the probability is easy to calculate. Probability distributions are functions which provide probabilities of events or outcomes. A probability distribution may be discrete or continuous. The range of both must cover all possible outcomes. In the discrete distribution, the sum of probabilities must add to 1 and in the continuous distribtion, the area under the curve must sum to 1. In both the discrete and continuous distributions, a range (or domain) can be described without a listing of all possible outcomes. For example, the domain of the normal distribution (a continuous distribution is minus infinity to positive infinity. The domain for the Poisson distribution (a discrete distribution) is 0 to infinity. You will learn in math that certain series can have infinite number of terms, yet have finite results. Thus, a probability distribution can have an infinite number of events and sum to 1. For a continuous distribution, the probability of an event are stated as a range, for example, the probability of a phone call is between 4 to 10 minutes is 10% or probability of a phone call greater than 10 minutes is 60%, rather than as a single event.

The probability that a trainee will remain with a company is 0.6. The probability that an employee earns more than Rs.10, 000 per year is 0.5. The probability that an employee is a trainee who remained with the company or who earns more than Rs.10, 000 per year is 0.7. What is the probability that an employee earns more than Rs.10, 000 per year given that he is a trainee who stayed with the company.

The odds against student X solving a Business statistics problem are 8 to 6, and odds in favour of student Y solving the problem are 14 to 16.

What is the chance that the problem will be solved if they both try independent of each other?

What is the probability that none of them is able to solve the problem?

Question 2: Marks: 1+3+2+3+1= 10

For the frequency distribution given below:

Length of Service No of employees

7.5-7.9 6

8.0-8.4 22

8.5-8.9 36

9.0-9.4 18

9.5-9.9 14

10.0-10.4 4

Calculate

Mean

Standard deviation

Co-efficient of Variation

Mean Deviation (from median)

Range

Question 3: Marks: 2+2=4

How many distinct four-digit numbers can be performed from the following integers 1, 2, 3,4,5,6 if each integer is used only once?

What would be the shape and name of the frequency distribution if

mean=median=mode

mean>median>mode

Question 4: Marks: 2+6=8

a) Describe the situation in which two variables are perfect positively correlated?

b) The cost of output at a factory is thought to depend on the number of units produced. Data have been collected for the number of units produced each month in the last six months, and the associated costs, as follows;

Output

('000s of units) X Cost

($'000) Y

2 9

3 11

1 7

4 13

3 11

5 15

Calculate the correlation coefficient and comment on your result.

Attempting Methodology / Requirements of the Assignment

This assignment covers Lesson No 9 to Lesson No 20

In Short answer question you can express your answer in 4-10 lines.

Illustrating your point you can give short examples

In numerical question, proper notation and symbols must be well written. For this use MathType Software.

Solve numerical step-by-step, giving all necessary explanation, whenever necessary.

1/6

If you rolled 2 fair dice, the probability of having a sum of 6 is 5 over 36

625

4.5%

Probabilities are calculated or estimated in a variety of methods. A non-quantitative means, used by weather forecasters, is to look at general conditions, is provide probabilies based on all indicators. In this way, the probabilities reflect their belief of certain events occurring from 0 (will not occur) to 100% (certain to occur). A second wasy probabilities are determined is to collect data, and determine the relative frequency of a particular event. Perhaps 10% of all motorists drive above the speed limit in a particular area, based on data collection, so we can state the probability of a motorist speeding in a certain area is 10%. Finally, probabilities are developed based on other known probabilites or assuming equally likely outcomes. If I have 5 outcomes, and they are equally likely, then the probability of each one occurring in 20% if these outcomes are independent and mutually exclusive. If I know the chance of coin flip coming up heads is 0.50, I can determine the probability of two coins coming up heads is 0.5 x 0.5 = 0.25.

Probability of picking purple sock first time is 6/10 or 3/5, second time, probability is 5/9.

Thus 3/5 * 5/9 = 15/45 which cancels to 1/3

7/10 X 6/9 = 42/90 it can be reduced to 7/15

polynomial functions: x2,x3,x1/2,x-1,xn if n doesn't equal 1 or 0

any trigonometric function: sin(x), cos(x), tan(x), sec(x), csc(x), cot(x)

exponential functions: 2x,3x,.5x,ex,nx if n doesn't equal 1 or 0.

any logarithmic function: log10(x), log2(x), ln(x), logn(x), for every n.

any inverse trig function: sin-1(x),sin-1(x),cos-1(x),tan-1(x),sec-1(x),csc-1(x),cot-1(x)

hyperbolic functions and their inverses: sinh(x),cosh(x),tanh(x),sech(x),csch(x),coth(x),

sinh-1(x),cosh-1(x),tanh-1(x),sech-1(x),csch-1(x),coth-1(x)

Hypergeometric function, Logarithmic integrals, Error functions, Fresnel integrals, Elliptic Integrals, and the integrals of nearly every other function.

Also, any combination of the above functions.

Theoretical probability

1/365 = 0.00274

When you use linear regression to model the data, there will typically be some amount of error between the predicted value as calculated from your model, and each data point. These differences are called "residuals". If those residuals appear to be essentially random noise (i.e. they resemble a normal (a.k.a. "Gaussian") distribution), then that offers support that your linear model is a good one for the data. However, if your errors are not normally distributed, then they are likely correlated in some way which indicates that your model is not adequately taking into consideration some factor in your data. It could mean that your data is non-linear and that linear regression is not the appropriate modeling technique.

12, because you can only get it one way: 6+6=12

And 2, because get it one way: 1+1=2.

The number of trials: n = 9,

the number of success: r = 4.

The probability of success: P(H) = 1/2,

the probability of failure: P(T) = 1/2.

P(exactly 4 heads in 9 tosses) = 9C4 x (1/2)^5 x (1/2)^4 = 126(1/2)^9 = 0.246

P(at most 2 heads) = P(none heads) + P( 1 head) + P(2 heads)

= 9C0[(1/2)^9][(1/2)^0] + 9C1[(1/2)^8][(1/2)^1] + 9C2[(1/2)^7][(1/2)^2]

= 1(1/2)^9 + 9(1/2)^9 + 36(1/2)^9

= (1 + 9 + 36)(1/2)^9

= 46(1/2)^9

= 0.09

mode

There are 52 cards in a standard playing card deck.

In such decks, there are four of each card:

Ace, Deuce (2), Trey (3), etc, through Ten,

and then there are the face (or picture) cards:

Jack, Queen, and King

(most card games remove the Joker from the deck,

so it will not included in this discussion).

Since there are a total of 12 face cards

(i.e., three types of picture cards multiplied by four of each type, or 3 x 4),

and assuming one is playing with a full deck (pun intended!),

then there are 12 out of 52 such cards,

so the probability of selecting a face card is 12-in-52,

or, reducing the terms, 3-in-13, or, roughly, 1-in-4.3.

So, if you were to randomly shuffle the deck,

and randomly remove 5 cards

(i.e., the closest whole number of cards to 4.3 that can be selected),

you could expect at least one of them to be a face card.

That does not mean that you WILL get a face card;

it only means that it is LIKELY.

As with any other math problem,

the "proof" may be found at the link, below,

from the University of Massachusetts.