Question

a card is selected at random from a standard deck of 52 playing cards. find the probability of each event.
(a) randomly selecting a diamond or a 5
(b) randomly selecting a black suit or an ace
(c) randomly selecting a 10 or a face card

(a) the probability of randomly selecting a diamond or a 5 is 0.308.
(type an integer or decimal rounded to three decimal places as needed.)
(b) the probability of randomly selecting a black suit or an ace is 0.538.
(type an integer or decimal rounded to three decimal places as needed.)
(c) the probability of randomly selecting a 10 or a face card is 0.308.
(type an integer or decimal rounded to three decimal places as needed.)

Explanation:

Step1: Recall probability formula

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Step2: Solve part (a)

• There are 13 diamonds ($n_{diamond}=13$), 4 cards with number 5 ($n_{5}=4$) and 1 card which is a diamond - 5 ($n_{diamond\cap5}=1$) in a 52 - card deck.
• $P(diamond)=\frac{13}{52}$, $P(5)=\frac{4}{52}$, $P(diamond\cap5)=\frac{1}{52}$.
• $P(diamond\cup5)=\frac{13 + 4-1}{52}=\frac{16}{52}\approx0.308$

Step3: Solve part (b)

• There are 26 black - suited cards ($n_{black}=26$), 4 aces ($n_{ace}=4$) and 2 black aces ($n_{black\cap ace}=2$) in a 52 - card deck.
• $P(black)=\frac{26}{52}$, $P(ace)=\frac{4}{52}$, $P(black\cap ace)=\frac{2}{52}$.
• $P(black\cup ace)=\frac{26 + 4-2}{52}=\frac{28}{52}\approx0.538$

Step4: Solve part (c)

• There are 4 cards with number 10 ($n_{10}=4$), 12 face - cards ($n_{face}=12$) and 0 cards which are both 10 and face - card ($n_{10\cap face}=0$) in a 52 - card deck.
• $P(10)=\frac{4}{52}$, $P(face)=\frac{12}{52}$, $P(10\cap face)=0$.
• $P(10\cup face)=\frac{4 + 12-0}{52}=\frac{16}{52}\approx0.308$

Answer:

(a) $0.308$
(b) $0.538$
(c) $0.308$