I would like your most understandable explanation.?

Question

1. Taylor tosses three coins. The even E is "exactly two coins land heads." What is the probability, P(E), that Taylor tosses exactly two heads?

2. George is watching videos at home. In his videotape library, he has four adventure films and five comedies. If he chooses to watch two different videos, what is the probability that both are comedies?

3. This may not be a direct problem, if not answerable, that is ok. "What is a multiple or a multiple of twenty five." I don't know if that is enough information for anyone to answer that question, if it is that would be great.

Answer 1

1) Out of 8 possible outcomes (with marked coins), only 3 will have exactly two heads, so the probability is 3/8

2) The odds of picking a comedy with the first choice is 5/9. Picking another comedy with the 2nd choice is 4/8. Probability is then (5/9)(4/8), or 5/18

3) A multiple of a number is an integer times that number. Let x be a number. A multiple would be 2x, 3x, 4x, 5x, etc. So, 100 = 4*25 would be a multiple of 25.

Answer 2

1. I always get the coin ones wrong if I do them mathematically, but since you only have three coins, you could write out all the possibilities:
TTT TTH THT HTT THH HTH HHT HHH
Out of these eight possibilities, three of them have exactly two heads, so the probability would be 3/8.

2. OK, so for his first movie, the possibility that he will pick a comedy would be 5/9 (#of comedy / total).
For the second movie, there would only be 8 movies left, and only 4 comedies left, so the possibility of picking a second one would be 4/8, or 1/2.
Now, multiply the two probabilities together to get the probability that both are comedies:
(5/9)(1/2) = 5/18

3. Do you mean "What is a multiple OF a multiple of twenty five?"
Because there would be lots of answers to that, such as...
Random multiple of 25 = 100
Random multiple of 100 = 500


I don't know if the last answer is correct or not; I'm sorry if they're not, but hope I helped!

Answer 3

1. Since there is a set number of trials (3 coin flips), probability of success stays the same from trial to trial (probability of heads is 0.5) and the trials are independent, then the numbers of heads in three coin flips has a binomial distribution with n = 3 and p = 0.5. Here we want 2 successes, so

(3!/2!1!)(0.5^2)(0.5^1) = 3/8

2. P(Both are comedies) = P(1st is a comedy and the 2nd is a comedy)
This is an and probability, so you would use a multiplication rule. Keep in mind that these events will NOT be independent since if the first is a comedy, the probabilty that the second is a comedy would change.
= P(1st is comedy)P(2nd is comedy|1st is comedy)
= (5/9)(4/8)
= 5/18

3. I don't understand your question there. Multiples of 25 would be considered 0, 25, 50, 75, and so on assuming that we are multiplying by the nonegative integers. I guess I don't understand what you mean.

Answer 4

1: This is pretty easy...
There are 4 possible outcomes from the 3-coin toss:
H H H
H H T
H T T
T T T

If Taylor only tosses one time, the probability is 1 in 4 (or 25%) that the coins will land showing 2 heads.

2: This question is impossible to answer, because it depends on personal choice and George's preference in making that choice - this is not an outcome that is baseds on chance, and can therefore not be answered statistically.

However, if we BLINDFOLD George, and ask him to fondle the nine movie cases, and pick two at random, we now have a case that can be answered based on probability and chance:

There are the following outcomes:
A1 A2, A1 A3, A1 A4, A1 C1, A1 C2, A1 C3, A1 C4, A1 C5
A2 A3, A2 A4, A2 C1, A2 C2, A2 C3, A2 C4, A2 C5
A3 A4, A3 C1, A3 C2, A3 C3, A3 C4, A3 C5
A4 C1, A4 C2, A4 C3, A4 C4, A4 C5
C1 C2, C1 C3, C1 C4, C1 C5
C2 C3, C2 C4, C2 C5
C3 C4, C3 C5
C4 C5

That is 36 combinations, of which 10 are both comedies, therefore there is a 10 / 36 chance that both will be comedies - or a 27.77% chance.

3: A "multiple" is any number that is the result of multiplying the original number by a whole number.
A Multiple of 25 would be 50, or 75, or 100, etc...

Hope this helps.

Answer 5

1. The coin possibilites are:

HHH
HHT <-
HTH <-
HTT
THH <-
THT
TTH
TTT

There are 8 total (2^3) and of those 3 have exactly two heads, so P(E) = 3/8

The prob 1st is C is 5/9 (5 C's in 9 total). If the 1s is C, what remains is 4 C's and 8 total, so prob 2nd is 4/8. Prob of both = 5/9 * 4/8

Answer 6

I'd like to use math symbols.

1. P(E) = C(3, 2)(1/2)^3 = 3/18 (by binomial distribution)

2.P = C(5, 2)/C(9, 2) = 5/18 (by combination)

3. 25n, n can be any integer. (by definition of a multiple.)

Answer 7

1)
HHT,HTH,THH

P(E)=3/8

2) 5/9*4/9=20/81


3) a multiple of x: 1x,2x,3x,4x,...

multiple of 25
result=25*k
where k=1,2,3,...