Question 12.6E3: Probability of A or B One card is selected from a standard d......
Probability of A or B
One card is selected from a standard deck of playing cards. Determine whether the following pairs of events are mutually exclusive and determine P (A or B).
a) A = an ace, B = a 9
b) A = an ace, B = a heart
c) A = a red card, B = a black card
d) A = a picture card, B = a red card
a) There are four aces and four 9’s in a standard deck of 52 cards. It is impossible to select both an ace and a 9 when only one card is selected. Therefore, these events are mutually exclusive.
\text{P (ace or 9) = P (ace) + P (9) = }\frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}
b) There are 4 aces and 13 hearts in a standard deck of 52 cards. One card, the ace of hearts, is both an ace and a heart. Therefore, these events are not mutually exclusive.
\text{P (ace) = }\frac{4}{52}\text{ P (heart) = }\frac{13}{52}\text{ P (ace and heart) = }\frac{1}{52}
\text{P (ace or heart) = P (ace) + P (heart) – P (ace and heart)}
= \frac{4}{52 } + \frac{13}{52} – \frac{1}{52}
= \frac{16}{52} = \frac{4}{13}
c) There are 26 red cards and 26 black cards in a standard deck of 52 cards. It is impossible to select one card that is both a red card and a black card. Therefore, the events are mutually exclusive.
\text{P (red or black) = P (red) + P (black)}
= \frac{26}{52} + \frac{26}{52} = \frac{52}{52} = 1
Since P (red or black) = 1, a red card or a black card must be selected.
d) There are 12 picture cards in a standard deck of 52 cards. Six of the 12 picture cards are red (the jacks, queens, and kings of hearts and diamonds). Thus, selecting a picture card and a red card are not mutually exclusive.
\begin{aligned} &P \left( \begin{matrix}\text{picture card}\\ \text{or red card} \end{matrix} \right) = P \left( \begin{matrix}\text{picture}\\ \text{card}\end{matrix} \right) + P \left( \begin{matrix}\text{red} \\ \text{card}\end{matrix} \right) – P \left( \begin{matrix}\text{picture card}\\ \text{and red card}\end{matrix}\right)\end{aligned}
= \frac{12}{52} + \frac{26}{52} – \frac{6}{52}
= \frac{32}{52} = \frac{8}{13}