When rolling a die, the following outcomes are possible:

\(\displaystyle \small 1,2,3,4,5,6\)

Of the \(\displaystyle \small 6\) outcomes, \(\displaystyle \small 4\) outcomes are \(\displaystyle \small 3\) or greater. Therefore,

\(\displaystyle \small P=\frac{4}{6}=\frac{2}{3}\)

After the removal of two red cards from a deck of 52, there are now fifty cards, of which twenty-four are red. The probability of a random card being red is therefore \(\displaystyle \frac{24}{50 }\), which as a percent is 

\(\displaystyle \frac{24}{50 } \times 100 \% =\left ( \frac{24}{50 } \times \frac{100}{1} \right ) \% =\left ( \frac{24}{1 } \times \frac{2}{1} \right ) \% = 48\%\)

A standard deck of play cards has \(\displaystyle 52\) cards. There are \(\displaystyle 4\) jacks, \(\displaystyle 2\) of which are red. Therefore, the probability is:

\(\displaystyle P=\frac{2}{52}=\frac{1}{26}\)

The mode of the data set is the most frequently occurring element. In the set given, 4 occurs three times, 5 occurs three times, 6 occurs two times, and 3 and 7 occur one time each. Therefore, the set has two modes, 4 and 5, and we want to preserve this condition.

If 3 is added to this set, it becomes

\(\displaystyle \left \{3, 3, 4,4, 4, 5, 5, 5, 6,6, 7\right \}\)

and 4 and 5 are still tied for the most frequently occurring element. The same happens if  7 is added to yield

\(\displaystyle \left \{ 3, 4,4, 4, 5, 5, 5, 6,6, 7, 7\right \}\).

If 5 is added to this set, it becomes

\(\displaystyle \left \{3, 4,4, 4, 5, 5,5, 5, 6,6, 7\right \}\)

and 5 appears more frequently than 4 or any other element. This changes the data set to one with only one mode, 5.

The correct response is therefore "I and III only".

Since this data set is arranged in ascending order and has an even number of elements, the median of the data set is the arithmetic mean of its middle two elements. Both elements are 6, so this is the median. 

6 is the mode, since it occurs most frequently.

The mean is the sum of the elements divided by the number of elements, which is 8:

\(\displaystyle \frac{2+3+ 5 + 6 + 6 + 6 + 7 +7}{8} =\frac{42}{8 } = 5.25\)

The median and the mode are equal to each other, but not to the mean, so the correct answer is "II and III only".

For a roll of "11" to occur with two dice, one die must show a "5" and the other must show a "6". For each die, since a "6" will come up with probability \(\displaystyle \frac{1}{4}\), for the other five rolls to be equally likely, each, including "5", must come up with probability 

\(\displaystyle \frac{1}{5} \left ( 1 - \frac{1}{4}\right ) = \frac{1}{5} \times \frac{3}{4} = \frac{3}{20}\).

The probability of rolling a red "5" and a blue "6" will be

\(\displaystyle \frac{3}{20} \times \frac{1}{4} = \frac{3}{80}\)

which is also the probability of rolling a blue "5" and a red "6". Add:

\(\displaystyle \frac{3}{80} + \frac{3}{80} =\frac{6}{80} = \frac{3}{40}\)

There are 4 possibilities that a 3 or higher can be rolled on a die.

The probability of rolling a 3 or higher on a fair die is:

\(\displaystyle P(Dice\geq 3) = \frac{4}{6} = \frac{2}{3}\)

Rolling a heads on a coin has a probability of one half.  Therefore,

\(\displaystyle (\frac{2}{3})(\frac{1}{2}) = \frac{1}{3}\)

Since order is irrelevant here, we are dealing with combinations. 

There are

\(\displaystyle C (10,2) = \frac{10!}{(10-2)!2!} = \frac{10!}{8!2!} = \frac{3,628,800}{40,320 \cdot 2}= 45\) 

ways to choose two out of ten students.

There are \(\displaystyle 7 \times 3 = 21\) ways to choose one boy and one girl.

The probability that one boy and one girl will be chosen is therefore

\(\displaystyle \frac{21}{45} = \frac{21 \div 3}{45 \div 3 } = \frac{7}{15}\).

Since order is irrelevant here, we are dealing with combinations. 

There are

\(\displaystyle C (10,2) = \frac{10!}{(10-2)!2!} = \frac{10!}{8!2!} = \frac{3,628,800}{40,320 \cdot 2}= 45\) 

ways to choose two out of ten students.

There are 

\(\displaystyle C (4,2) = \frac{4!}{(4-2)!2!} = \frac{4!}{2!2!} = \frac{24}{2 \cdot 2}= 6\)

ways to choose two out of four girls.

The probability that two girls will be chosen is therefore

\(\displaystyle \frac{6}{45} = \frac{6 \div 3}{45 \div 3 } = \frac{2}{15}\).

5 to 4 odds in favor of heads is equal to a probability of \(\displaystyle \frac{5}{5 + 4} = \frac{5}{9}\), which is the probability that the penny will come up heads. The probability that the penny will come up tails is \(\displaystyle 1 - \frac{5}{9} = \frac{4}{9}\).

Similarly, 4 to 3 odds in favor of heads is equal to a probability of \(\displaystyle \frac{4}{4+3} = \frac{4}{7}\), which is the probability that the nickel will come up heads. The probablity that the nickel will come up tails is \(\displaystyle 1 - \frac{4}{7} = \frac{3}{7}\).

The outcomes of the tosses of the penny and the nickel are independent, so the probabilities can be multiplied.

The probability of the penny and the nickel both coming up heads is \(\displaystyle \frac{5}{9} \times \frac{4}{7} = \frac{20}{63}\).

The probability of the penny and the nickel both coming up tails is \(\displaystyle \frac{4}{9} \times \frac{3}{7} = \frac{12}{63}\).

The probabilities are added:

\(\displaystyle \frac{20}{63}+ \frac{12}{63} = \frac{20 + 12}{63 }= \frac{32}{63}\)

Since \(\displaystyle 63-32 = 31\), this translates to 32 to 31 odds in favor of this event.