This is a binomial probability experiment. 

The probability that a fair coin will come up heads in any single toss is \(\displaystyle \frac{1}{2}\). 

If the coin is tossed four times, then the probability that heads will appear three times is:

\(\displaystyle P(x=3) = C(4,3) \left (\frac{1}{2} \right )}^{3}\left (\frac{1}{2} \right )}^{4-3}\)

\(\displaystyle = \frac{4!}{3! (4-3)!}\left (\frac{1}{2} \right )}^{3}\left (\frac{1}{2} \right )}^{1}\)

\(\displaystyle = \frac{4!}{3! 1!}\left (\frac{1}{2} \right )}^{4}\)

\(\displaystyle = 4\cdot \left (\frac{1}{16} \right )}=\frac{1}{4}\)

The probability that heads will appear four times is:

\(\displaystyle P(x=4) = C(4,4) \left (\frac{1}{2} \right )}^{4}\left (\frac{1}{2} \right )}^{4-4}\)

\(\displaystyle = \frac{4!}{4! (4-4)!}\left (\frac{1}{2} \right )}^{4}\left (\frac{1}{2} \right )}^{0}\)

\(\displaystyle = \frac{4!}{4! 0!}\left (\frac{1}{2} \right )}^{4}\)

\(\displaystyle = 1\cdot \left (\frac{1}{16} \right )}=\frac{1}{16}\)

So the probability that heads will appear at least three times is 

\(\displaystyle P(x\geq 3) = \frac{1}{4} + \frac{1}{16}= \frac{5}{16}\)

The second statement is irrelevant; the result of a previous trial has no influence on that of the current trial. It has already been established in the problem that the coin is fair.

The answer is that Statement 1 alone is sufficient to answer the question, but not Statement 2.

We need both statements to get the answer.

Statement 1 alone is not enough. Having 24 total cookies does not tell us the proportions of each type of cookie. We could have 1 chocolate chip cookie and 23 oatmeal raisin, or vice versa. 

Statement 2 alone is not sufficient either. Knowing that there are 8 more chocolate chip cookies than oatmeal raisin without knowing the total amount in the bag does not help. It could be 1 oatmeal raisin cookie and 9 chocolate chip, or it could be 100 oatmeal raisin cookies and 108 chocolate chip cookies (although that is a pretty big bag!)

Only by having both statements together can we find the right proportion. If we let \(\displaystyle A\) be the number of chocolate chip cookies, and \(\displaystyle B\) be the number of oatmeal raisin, we can set two equations and find the right ratio. We know that

\(\displaystyle A+B=24\)

from statement 1, and

\(\displaystyle A=B+8\) 

from statement 2.

Substituting \(\displaystyle A\) into the first equation we get

\(\displaystyle (B+8)+B=24\)

\(\displaystyle 2B=16\)

\(\displaystyle B=8\)

 Therefore, we get 8 oatmeal raisin cookies, and 16 chocolate chip cookies. The answer is

\(\displaystyle \frac{16}{24}=\frac{2}{3}\) 

or 66.67% chance that the monster pulls out a chocolate chip cookie. 

Pizza received 24 votes from men. Since half of those who responded pizza were women, there must be 24 votes from women.

If 1 out of 5 men voted for pizza, this means we have \(\displaystyle \ 24\cdot 5 = 120\) men.  If 3 out of 10 women voted for pizza, we have \(\displaystyle \frac{10}{3}\cdot 24 = 80\) women for a total of 200 people polled.

If our population is instead given as 60% male and 40% female, we still have 24 men and 24 women that voted for pizza, but while we can determine the ratio of people who voted, we cannot get a precise number.  For example, while 80 women and 120 men works, we could also have 160 women and 240 men as our total poll (with 1 out of 10 men and 3 out of 20 women voting pizza). 

Thus the first condition is sufficient while the second is not.

This question deals with conditional probability.

Statement 1) consists of independent events. The past trials will not affect the future trials of the coin.  The odds of flipping a coin and land on heads will be \(\displaystyle \frac{1}{2}\).

Statement 2) also consists of independent events. However, given that the previous 10 days were sunny does not necessarily indicate that the next day will guarantee that it will be a sunny day.  The probability is unknown because the statement does not give any information about the percentage of a sunny sky on any particular day.

To find the probability of several events, we need to find the probability of each event separately. First, we need to know how many there are of each color pick.

Statement I relates the number of orange and blue picks. 

Statement II tells us the number of black picks.

\(\displaystyle P_{orange}+P_{black}+P_{blue}=36\)

From Statement I:

\(\displaystyle P_{orange}=P_{blue}+14\)

From Statement II:

\(\displaystyle P_{black}=12\)

Using Statement I, Statement II, and the given info, we can substitute \(\displaystyle P_{blue}+14\) from Statement I into Statement II for \(\displaystyle P_{orange}\) and get the following:

\(\displaystyle P_{blue}+14+P_{blue}+12=36\)

We can then solve for \(\displaystyle P_{blue}\):

\(\displaystyle 2P_{blue}+26=36\)

\(\displaystyle 2P_{blue}=10\)

\(\displaystyle P_{blue}=5\)

Since we are told in Statement I that there are 14 more blue picks than orange picks, we can then find the number of orange picks:

\(\displaystyle P_{orange}=P_{blue}+14\)

\(\displaystyle P_{orange}=5+14\)

\(\displaystyle P_{orange}=19\)

So, to find the odds of multiple events, multiply the probability of each event together:

\(\displaystyle \frac{12}{36}\cdot \frac{5}{35}\cdot \frac{19}{34}=\frac{1140}{42840}=\frac{19}{714}\)