Answering these questions won't help the fact that you know nothing (so you say) about probability. You need to learn formulas and theory, quickly. I suggest picking up your DAT study guides and reading the probability sections.
A quick crash course might help you through some of these problems.
The probability of an event occurring is defined to be A/B where A = the # of successes and B = the total # of events. For instance if I roll a fair 6 sided die and want to know the probability of rolling a 4, the # of events would be 6 (there are 6 equally possible outcomes) and the # of successes would be 1 (only 1 of these is a 4). Thus the probability is 1/6. This is the most basic formula so you need to drill it into your head. It is worth mentioning here that the sum of the individual probabilities of all possible events MUST total 1. For the die example, there are 6 possible outcomes (1 through 6), each with a probability of 1/6. And 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1. In other words, if you bet all the possibilities, you will always win.
The next step up is to learn what happens when there are multiple successes wanted. Start with 'OR' - the union of two events. Let's say I want the probability of rolling a 1 or a 2 on a six sided fair die. The probability of rolling any given number is 1/6. When you have the option of more than one independent outcome, you ADD them. The key word to look for is usually 'or'. So in this example, you want either a 1 OR a 2. Each has a 1/6 chance. So together they have a 1/6 + 1/6 = 2/6 = 1/3 chance. When I say 'independent' I mean that one cannot possibly happen at the same time as the other. It is IMPOSSIBLE in this example to simultaneously roll a 1 and a 2 in the same roll. Since the events can NEVER occur together, they are independent.
An example of mutually exclusive events is if I gave you a coin and a die and told you to toss the coin and roll the die. Then I asked for the probability of the coin landing on heads and the die landing on 4. The probability of the coin landing on heads is 1/2 and the probability of the die landing on 4 is 1/6. Together they add up to 4/6 = 2/3. But that's not the right answer. Why? Because the events are not mutually exclusive. There's some overlap. There's three possible cases where you have a success:
(1) You flip heads and do NOT roll 4.
(2) You roll a 4 and do NOT flip heads.
(3) You flip heads and roll a 4.
Saying you have a 1/2 chance of flipping heads includes both (1) and (3). Saying you have a 1/6 chance of rolling a 4 includes both (2) and (3). When you add 1/2 + 1/6, you are adding (1) + (2) + (3) + (3). You've counted (3) TWICE! So you have to SUBTRACT (3) ONCE to get the right answer. That means you need to know the probability of flipping heads AND rolling a 4. That's the next step up. For now, here's the full formula for the union of two events:
P(A or B) = P(A) + P(B) - P(A and B)
Read this as "The probability of events A or B occurring equals the probability of A occurring plus the probability of B occurring minus the probability of both A and B occurring."
The intersection of events is easier since you probably won't get questions where the events are not independent. This is most often represented by the word AND. If you had a fair coin and wanted to know the probability of flipping heads twice in a row, this is the same as saying 1/2 * 1/2 = 1/4.
The formula for the intersection of two events:
P(A and B) = P(A)P(B)
Now going back to the die and coin question, you have 1/2 + 1/6 - (1/2)(1/6) = 4/6 - 1/12 = 7/12 which is your answer.
You just have to think things through. If you took a problem like #2, it wants 2 purple marbles. You take them one at a time. The first purple marble has a 3/9 chance of being chosen (there are 3 purple marbles and 9 total marbles). The second purple marble has a 2/8 chance of being chosen (with the first marble gone, there are only 2 purple marbles left and only 8 total marbles left). Since you want both of these events to occur at once, you multiply the probabilities. You get 3/9 * 2/8 which is just 1/3 * 1/4 = 1/12. I think you meant to write 5 purple marbles and 6 yellow because that would make it 5/11 for the first draw and 4/10 for the second draw which multiplies to 2/11.
You asked what (n-1)! means. The ! is short for 'factorial'. It means start with the number before it and multiply all natural numbers less than or equal to that number (a natural number is 1, 2, 3, etc - not 0, and not negative). For example 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720.
For problem 1 it wants 5 students in a row. Any of the 5 could sit in spot #1, any of the remaining 4 in spot #2, and so on. So there are 5 possibilities, then 4, then 3, then 2, then 1. Multiply them together and that's 5! = 120.
Same thing for the round table question: 6! = 720. Except now you have a round table, and for any given circular arrangement, there are 6 possible seats to place everyone in. Picture one order and then move everyone clockwise one seat. Same exact order. With 6 seats, you can do this 6 times. So only 1 out of every 6 is a unique order around the table. Thus you divide 720 by 6 to get your answer of 120.
#5 is another simple probability question. You have two dice and you want to roll a 7. There's no formula for this. You just need to count possibilities. The total # of possibilities is 6 *6 = 36 (6 per die). To get 7 you use 1+6, 6+1, 2+5, 5+2, 3+4, or 4+3. Yes you must count each twice because you have two different dice. Answer is 6/36 = 1/6.
#6 is simple and you should do on your own. It wants a 3 on one die and a 3 on the other. The trick is that one die has 9 sides and the other has 6. Given the answer, you should figure out how it was done.
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#4 is trickier. You need to know combinations and permutations. I am not going into a lecture on it. Look up and down this forum because every 10 threads someone asks the same question about them. Or google it. You absolutely need to know them. The reason you use it for this problem is because you don't know which tosses had the tails. The probability that you get 6 tails followed by 3 heads is (1/2)(1/2)(1/2)(1/2)(1/2)(1/2) * (1/2)(1/2)(1/2) = 1/512. It's tough to show here because both probabilities are 1/2, but I separated them for you to see. The first 6 represent the tails. The next 3 are the heads. But what if toss #1 was a head and toss #9 was a tail? You still have 6 tails/3 heads but the tosses are different. The combination (9 choose 6) takes this into account. This combination equals 84, which means there are 84 ways to choose the 6 tosses that will be tails. If each single round of 9 tosses has a probability of 1/512 of having 6 tails and 3 heads, and there are 84 ways to select which tosses are the tails, then there's an 84/512 = 21/128 chance of getting 6 tails and 3 heads in general. By the way there's a typo in your answer.
#7 uses the same principle of combinations. There are 10 total flowers and you want to pick 3 of them.
#8 is yours to do.
