For each of the following situations, state the probability rule or rules that you would use and apply it or them. Write a sentence explaining how the situation illustrates the use of the probability rules. (a) The probability of event A is 0.417. What is the probability that event A does not occur

The question is incomplete. So, the complete question is:

For each of the following situations, state the probability rule or rules that you would use and apply it or them. Write a sentence explaining how the situation illustartes the use of the probability rules. (a) The Probability of event A is 0.417. What is the probability that even A does not occur? (b) A coin is tossed 4 times. The probability of zero heads is 1/16 and the probability of zero tails is 1/16. What's the probability that all four tosses result in the same outcomes? (c) Refer to part b, what's the probability that there is at least one head and at least one tail? (d) The probability of event A is 0.4 and the probability of event B is 0.8. Events A and B are disjoint. Can this happen? (e) Event A is rare. Its probability is -0.04. Can this happen?

Answer and Explanation:

(a) P(A)=0.417

Since it's wanted to known the probability of not occuring,

P(A') = 1 - P(A)

P(A') = 1 - 0.417

P(A') = 0.583

The probability of event A not occuring is 0.583.

(b) If the coin is fair, the probability of either head or tail is 1/2. Since, there were 4 tosses, the probability for total outcomes would be 1/16.

For all heads: P(H) = 1/16;

For all tail: P(T) = 1/16:

As it wants "either all heads OR all tails", it will be

P = P(H)+P(T)

P = 1/16+1/16

P = 1/8 = 0.125

The probability of all heads or all tails is 0.125.

(c) The probability of at least one head or one tail is the probability of NOT being all heads and all tails, so:

1 - 0.125 = 0.875

The probability of at least one of each is 0.875.

(d) Yes, events A and B can be disjoint if their intersection is zero:

P(A∩B) = 0 and it also means the events are mutually exclusive.

(e) No, probability can not be negative, because an event has to occur to calculate probability and a set of numbers fo the event is positive.

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.

i think its 100.

hope this !