Set up an appropriate sample space for each of Problems 1.1 to 1.10 and use it to solve the problem. Use either a uniform or non-uniform sample space or try both.

A single card is drawn at random from a shuffled deck. What is the probability that it is red? That it is the ace of hearts? That it is either a three or a five? That it is either an ace or red or both?

TheIn a deck of card, there are 52 cards of 4 suits namely spade, club, diamond and heart, out of which spade and club are black and diamond and heart are red. The probability of each card is equal, that is$\frac{\mathbf{1}}{\mathbf{52}}$

It is known that there are 13 cards of each suit and 26 cards of each color. So, the sample space for the problem is all 52 cards

Each point of the obtained sample space has an equal probability of $\frac{1}{52}$.

In a deck, there are 26 red cardsand each has a probability of $\frac{1}{52}$.

Find the probability that the card drawn is red by adding the probabilities of each possible outcomes, that is 26 times $\frac{1}{52}$.

$\begin{array}{rcl}p& =& 26\times \frac{1}{52}\\ & =& \frac{1}{2}\\ & & \end{array}$

Thus, the probability that the card drawn is red is $\frac{1}{2}$.

In a deck of card, there are 4 aces out of which only one is ace of heart and has a probability of $\frac{1}{52}$

Thus,the probability that the card drawn is ace of heart $\frac{1}{52}$.

In a deck of card, there are 4 threes and 4 fives and each has a probability of$\frac{1}{52}$.

Find the probability that the card drawn is 3 or 5 by adding the probabilities of each possible outcomes.

$\begin{array}{rcl}p& =& \frac{4}{52}+\frac{4}{52}\\ & =& \frac{2}{13}\\ & & \end{array}$

Thus, the probability that the card drawn is 3 or 5 is$\frac{2}{13}$.

In a deck of card, there are 4 aces and 26 red cards out of which 2 aces are red, this implies that possibility that for a card to be an ace or red or both is as follows,

$4+26-2=28$

Find the probability that the card is either an ace or red or both by adding the probabilities of each possible outcomes.

$\begin{array}{rcl}p& =& \frac{4}{52}+\frac{26}{52}-\frac{2}{52}\\ & =& \frac{7}{13}\\ & & \end{array}$

Thus, the probability that the card is either an ace or red or both is$\frac{7}{13}$.