Probability, have yet to find a single person who knows this:(?

Im getting so frustrated, does anybody know how to do this?



A bag contains five red marbles and three blue marbles. In how many different ways can two red marbles be drawn if the first marble is not returned to the bag before the second marble is drawn?



The lock for your suitcase has four cylinders with the digits 0 to 6 on each cylinder. How many different combinations are possible for your lock?



In how many different ways can a president, vice-president, secretary, and treasurer be chosen from the twenty members of the French Club, if all are equally likely to be chosen.



The school literary magazine has 30 possible entries but can only use 25. How many different forms of the Table of Contents would be possible?



The artist, R. Locklen Jones, selects three paintings from a collection of six to display in a row. How many different arrangements of the display are possible?



The red spinner has six equal sectors labeled with numbers 1 through 6. The blue spinner has four equal sectors labeled with letters A, B, C, and D. Find the probability of spinning the two spinners and getting a 5 with a B. Please type your answer as a fraction. Do not forget to reduce



Two cards are randomly drawn from a standard deck replacing the card after the first draw. Find the probability of drawing a Jack followed by an Ace. Please type your answer as a fraction. Do not forget to reduce if needed!Probability, have yet to find a single person who knows this:(?
It's not hard. - Most students have no difficulty with these types of problems.

Where are you getting stuck? - It appears you've just posted all your math

homework for us to do... and that's just not right.





I will explain how to do the last question:



Remember that probability is defined as:



All Favorable Outcomes

- - - - divided by - - - -

All Possible Outcomes



On the first draw, there are 4 'favorable' choices (or outcomes).

There are 52 'possible' choices since there are 52 cards in a deck.



On the second draw, there are 4 'favorable' choices, (4 Aces).

Since you replaced the first card, there are still 52 'possible' choices.



Probability of drawing a Jack, followed by an Ace:

P(Jack, Ace) = (4/52) * (4/52) = (you crunch the numbers)



Best of luck in your studies,

~ Mitch ~

.Probability, have yet to find a single person who knows this:(?
Hello Morgan T,



The first thing is to understand is that probability is the ratio of the number of things that constitute a specific event to the total number of all possible events. Second thing to understand is that sometimes events affect the probabilities that follow them.



Lets look at some of your problems.



';A bag contains five red marbles and three blue marbles. In how many different ways can two red marbles be drawn if the first marble is not returned to the bag before the second marble is drawn?';

Here we are not replacing a marble so that means previous events affect probabilities of later events.

FIRST DRAW: How many red marbles? 5. How many total? 8. So the probability of our first red is 5/8 = 0.625. SECOND DRAW: NOW how many reds? 4. How many total? 7. You tell me what is the probability of the second red. 4/7 We want the first red AND the second red is going to be the product of the two probabilities (5/8)(4/7) = 5/14 ~ 0.357.



';The lock for your suitcase has four cylinders with the digits 0 to 6 on each cylinder. How many different combinations are possible for your lock?';

The is a problem on counting. There are numbers in the combination for the suitcase. Think of it as four different slots __ __ __ __. How many different possible numbers could go for the first slot? 6.

How many for the second? The third? The fourth. We multiply to find the total number of possible combinations. (6)(6)(6)(6) = 1296 [BE CAREFUL ABOUT THE WORD COMBINATION. HERE WE ARE TALKING ABOUT A COMBINATION FOR A LOCK, NOT THE FORMAL PROBABALISTIC CONCEPT OF COMBINATION.]



';In how many different ways can a president, vice-president, secretary, and treasurer be chosen from the twenty members of the French Club, if all are equally likely to be chosen.';

Here we will assume that if a person is in one office, they cannot hold another office at the same time. Think of this like chairs marked with the titles. We got four chairs (Prez, VP, Treas, Sec)

How many different ways can we put someone in that first chair? 20 ways. How many folks are left to fill that second one? 19. How about the third chair? and th last chair? (18 and 17) Like the last one, we multiply to get the total number of possible officer assignments: (20)(19)(18)(17) = 116280 [Just to check, if there is a fourth office -- grand pooh-bah -- after we have chosen the first four, how many folks could fill that post? 16, right? How would you find the total possible officer assignments? (20)(19)(18)(17)(16) = 1860480.



The rest of the problems are managed in similar fashions. Think about total number of possible events, number of events you are interested in. And yes some of the numbers may be big for the counting problems, its okay. All probabilities should be between 0 and 1, inclusive.



I am an old man, and I need to get my beauty sleep. I hope this has been of some help to you.

Good luck with your studies!